Big Idea #6: Equilibrium, Acids and Bases, Titrations, and Solubility
Any bond or intermolecular attraction that can be formed can be broken. These two processes are in dynamic competition, sensitive to initial conditions and external perturbations.
The Equilibrium Constant, Keq
Most chemical processes are reversible. That is, reactants react to form products, but those products can also react to form reactants. A reaction is at equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction. The relationship between the concentrations of reactants and products in a reaction at equilibrium is given by the equilibrium expression, also called the Law of Mass Action.
The Equilibrium Expression
For the reaction
aA + bB ⇌ c C + dD
1. [A], [B], [C], and [D] are molar concentrations or partial pressures at equilibrium.
2. Products are in the numerator, and reactants are in the denominator.
3. Coefficients in the balanced equation become exponents in the equilibrium expression.
4. Solids and pure liquids are not included in the equilibrium expression because they cannot change their concentration. Only gaseous and aqueous species are included in the expression.
5. Units are not given for Keq.
Let’s look at a few examples:
1. HC2H3O2(aq) ⇌ H+(aq) + C2H3O2–(aq)
Keq = Ka =
This reaction shows the dissociation of acetic acid in water. All of the reactants and products are aqueous particles, so they are all included in the equilibrium expression. None of the reactants or products have coefficients, so there are no exponents in the equilibrium expression. This is the standard form of Ka, the acid dissociation constant.
2. 2H2S(g) + 3O2(g) ⇌ 2H2O(g) + 2SO2(g)
Keq = Kc =
Keq = Kp =
All of the reactants and products in this reaction are gases, so Keq can be expressed in terms of concentration (Kc, moles/liter or molarity) or in terms of partial pressure (Kp, atmospheres). All of the reactants and products are included here, and the coefficients in the reaction become exponents in the equilibrium expression.
3. CaF2(s) ⇌ Ca2+(aq) + 2F–(aq)
Keq = Ksp = [Ca2+][F–]2
This reaction shows the dissociation of a slightly soluble salt. There is no denominator in this equilibrium expression because the reactant is a solid. Solids are left out of the equilibrium expression because the concentration of a solid is constant. There must be some solid present for equilibrium to exist, but you do not need to include it in your calculations. This form of Keq is called the solubility product, Ksp.
4. NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
Keq = Kb =
This is the acid-base reaction between ammonia and water. We can leave water out of the equilibrium expression because it is a pure liquid. This is the standard form for Kb, the base dissociation constant.
Here is a roundup of the equilibrium constants you need to be familiar with for the test.
• Kc is the constant for molar concentrations.
• Kp is the constant for partial pressures.
• Ksp is the solubility product, which has no denominator because the reactants are solids.
• Ka is the acid dissociation constant for weak acids.
• Kb is the base dissociation constant for weak bases.
• Kw describes the ionization of water (Kw = 1 × 10–14).
A large value for Keq means that products are favored over reactants at equilibrium, while a small value for Keq means that reactants are favored over products at equilibrium.
The equilibrium constant has a lot of aliases, but they all take the same form and tell you the same thing. The equilibrium constant tells you the relative amounts of products and reactants at equilibrium.
Much as Hess’s Law allows for the determination of the enthalpy for a reaction given the enthalpy values for similar reactions, you can determine the equilibrium constant of a reaction by manipulating similar reactions with known equilibrium constants. However, the rules for doing so are different than the rules for shifting enthalpy values:
1. If you flip a reaction, you take the reciprocal of the equilibrium constant to get the new equilibrium constant.
2. If you multiply a reaction by a coefficient, you take the equilibrium constant to that power to get the new constant.
3. If you add two reactions together, you multiply the equilibrium constants of those reactions to get the new constant.
2SO2(g) + O2(g) ↔ 2SO3(g) K = 4.0 × 1024
2NO(g) + O2(g) ↔ 2NO2(g) K = 0.064
According to the information above, what is the equilibrium constant for each of the following reactions?
a) 2SO3(g) ↔ 2SO2(g) + O2(g)
This reaction is the opposite of the first reaction, so K = 2.5 × 10–25.
b) SO2(g) + O2(g) ↔ SO3(g)
This reaction is the first reaction multiplied by one-half. Thus, the new equilibrium constant is:
K = (4.0 × 1024)1/2 = 2.0 × 1012
c) SO2(g) + NO2(g) ↔ SO3(g) + NO(g)
To get this reaction, we first have to flip the second reaction:
2NO2 ↔ 2NO + O2
K = 16
We can then add the flipped second reaction to the original first reaction. The O2 will cancel out, yielding the following:
2SO2(g) + 2NO2(g) ↔ 2SO3(g) + 2NO(g)
K = 4.0 × 1024(16) = 6.4 × 1025
Finally, we have to multiply that equation by half to get our desired reaction.
SO2(g) + NO2(g) ↔ SO3(g) + NO(g)
K = (6.4 × 1024)1/2 = 8.0 × 1012
LE CHÂTELIER’S PRINCIPLE
At equilibrium, the rates of the forward and reverse reactions are equal. A “shift” in a certain direction means the rate of the forward or reverse reaction increases. Le Châtelier’s Principle states that whenever a stress is placed on a system at equilibrium, the system will shift in response to that stress. If the forward rate increases, we say the reaction has shifted right, which will create more products. If the reverse rate increases, we say the reaction has shifted left, which will create more reactants.
Let’s use the Haber process, which is used in the industrial preparation of ammonia, as an example.
N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H° = –92.6 kJ/molrxn
When the concentration of a reactant or product is increased, the reaction will shift in the direction that allows it to use up the added substance. If N2 or H2 is added, the reaction shifts right. If NH3 is added, the reaction shifts left.
When the concentration of a species is decreased, the reaction will shift in the direction that allows it to create the substance that has been removed. If N2 or H2 is removed, the reaction shifts left. If NH3 is removed, the reaction shifts right.
When the external pressure on a system is increased, that will increase the partial pressures of all of the gases inside the container and cause a shift to the side with less gas molecules. In the Haber process, there are four gas molecules on the left and two on the right, so an increase in partial pressures would cause a shift to the right. Decreasing the partial pressure of all gases would have the opposite effect and cause a shift to the left.
The most common way to cause a change in pressure is by changing the size of the container in which a reaction is occurring. If the Haber process were occurring inside a balloon, and we were to decrease the total volume of that balloon by squeezing it, that would increase the partial pressures of all gases involved (assuming the temperature stayed constant). Increasing the volume of the container would have the opposite effect.
In addition to changing volume, the partial pressure of gases can be changed by adding another gas to the container while maintaining the total pressure, even if that gas does not participate in the reaction. If we were to add some helium gas to a container where the Haber process was in equilibrium, it would decrease the mole fraction of all of the gases participating in the reaction (since the helium now represents at least some percentage of the total gases). A decreased mole fraction causes a decreased partial pressure, which again causes a shift to the left.
Note that for any equilibrium, there has to be a different number of gas molecules on each side of the equilibrium for a pressure change to cause any kind of shift. In equilibria where the number of gas molecules are the same on both sides, any of the above changes would cause no shift at all.
There’s a trick to figure out what happens when the temperature changes. First, rewrite the equation to include the heat energy on the side that it would be present on. The Haber process is exothermic, so heat is generated:
N2(g) + 3H2(g) ↔ 2NH3(g) + energy
Then, if the temperature goes up, the reaction will proceed in the reverse direction (shifting away from the added energy). If the temperature goes down, the reaction will proceed in the forward direction (creating more energy). The reverse would be true in an endothermic reaction, as the energy would be part of the reactants.
One last type of shift which is caused by dilution. For an aqueous equilibrium, diluting that equilibrium with extra water can cause a shift. The Haber process won’t work for this one, so let’s examine another equilibrium:
Fe3+(aq) + SCN−(aq) ⇌ FeSCN2+(aq)
If you were to dilute the above equilibrium with extra water, it would shift to the side that has more aqueous species. In the above example, this causes a shift to the left. If some water were removed (say, by evaporation), the reaction would shift to the side with less aqueous species (right).
CHANGES IN THE EQUILIBRIUM CONSTANT
It is important to understand that shifts caused by concentration or pressure changes are temporary shifts, and do not change the value of the equilibrium constant itself. Eventually, the concentrations of the products and reactants will re-establish the same ratio as they originally had at equilibrium.
However, shifts caused by temperature changes are different. Because changing temperature also affects reaction kinetics by adding (or removing) energy from the equilibrium system, a change in temperature will also affect the equilibrium constant for the reaction itself, in addition to causing a shift. This means the ratio of the products to reactants at equilibrium will change as the temperature changes.
In the Haber process, an increase in temperature causes a shift to the left, and it would permanently affect the value of the equilibrium constant in a way that is consistent with the shift. A shift to the left causes an increase in the concentrations of the reactants (the denominator in the mass action expression) while simultaneously causing a decrease in the concentration of the product (the numerator in the mass action expression). Thus, increasing the temperature decreases the value of the equilibrium constant—that is, it causes there to be a larger amount of reactants present at equilibrium compared to the amount of products present. The reverse would be true for a temperature decrease; the shift to the right would cause an increase in the equilibrium constant.
THE REACTION QUOTIENT, Q
The reaction quotient is essentially the quantitative application of Le Châtelier’s Principle. It is determined using the Law of Mass Action. However, you can use the concentration or pressure values at any point in the reaction to calculate Q. (Remember, you can only use concentrations or pressures of reactions at equilibrium to calculate K). The value for the reaction quotient can be compared to the value for the equilibrium constant to predict in which direction a reaction will shift from the given set of initial conditions.
The Reaction Quotient
For the reaction
aA + bB ⇌ cC + dD
[A], [B], [C], and [D] are initial molar concentrations or partial pressures.
• If Q is less than K, the reaction shifts right.
• If Q is greater than K, the reaction shifts left.
• If Q = K, the reaction is already at equilibrium.
Let’s take a look at an example. The following reaction takes place in a sealed flask:
2CH4(g) ↔ C2H2(g) + 3H2(g)
(a) Determine the equilibrium constant if the following concentrations are found at equilibrium.
[CH4] = 0.0032 M [C2H2] = 0.025 M [H2] = 0.040 M
To solve this, we first need to create the equilibrium constant expression:
By plugging in the numbers, we get:
(b) Upon testing this reaction at another point, the following concentrations are found:
[CH4] = 0.0055 M [C2H2] = 0.026 M [H2] = 0.029 M
Use the reaction quotient to determine which way the reaction needs to shift to reach equilibrium.
Q uses the exact same ratios as the equilibrium constant expression, so:
0.021 is less than the equilibrium constant value of 0.16, so the reaction must proceed to the right, creating more products and reducing the amount of reactant in order to come to equilibrium.
When examining voltaic cells, the reduction potentials are always given at standard conditions; that is 25°C, 1.0 atm, and with all species having a concentration of 1.0 M. If any of those conditions deviate, it will also cause the cell potential to deviate.
Voltaic cells are all very favored, having equilibrium constants significantly greater than 1. If the reaction quotient for a voltaic cell were to ever become equal to the equilibrium constant (so, at Q = K), the voltage of the cell would drop to zero. Given that knowledge, the best way to determine how a cell’s potential will change if standard conditions are deviated from is to use the reaction quotient.
As all concentrations are equal to 1.0 M at standard conditions, we can infer that at standard conditions, the reaction quotient will also be equal to 1. Any change to the cell that would cause the reaction quotient to increase (say, an increase in the concentration of a product or the decrease in the concentration of a reactant) would cause the reaction quotient to become closer to the equilibrium constant and would thus decrease the cell potential (remember, if Q ever reaches K, the cell potential becomes zero!). Any change that would cause the reaction quotient to decrease would move it further from the equilibrium constant, and thus increase the potential of the cell.
If a cell has a gas at either the cathode or the anode, a change in pressure can also affect the reaction quotient, and a change in temperature may as well. However, concentration changes, which usually occur constantly as the reaction in a cell is progressing, are the most common reason that a voltaic cell will deviate from its standard potential.
A voltaic cell using silver and zinc is connected, and the below reaction occurs.
2Ag+ + Zn(s) → Zn2+ + 2Ag(s)
What would happen to the voltage of the above cell as the reaction progresses?
As the reaction progresses, [Ag+] decreases as it is reduced into Ag(s), and [Zn2+] increases as it is oxidized from Zn(s). Both of these changes would have the effect of increasing the reaction quotient, which would bring the cell closer to equilibrium and thus decrease the overall potential.
Roughly speaking, a salt can be considered “soluble” if more than 1 gram of the salt can be dissolved in 100 milliliters of water. Soluble salts are usually assumed to dissociate completely in aqueous solution. Most, but not all, solids become more soluble in a liquid as the temperature is increased.
Solubility Product (Ksp)
Salts that are “slightly soluble” and “insoluble” still dissociate in solution to some extent. The solubility product (Ksp) is a measure of the extent of a salt’s dissociation in solution. The Ksp is one of the forms of the equilibrium expression. The greater the value of the solubility product for a salt, the more soluble the salt.
For the reaction
AaBb(s) ⇌ a Ab+(aq) + b Ba–(aq)
The solubility expression is
Ksp = [Ab+]a[Ba–]b
Here are some examples:
|CaF2(s) ⇌ Ca2+(aq) + 2F–(aq)
||Ksp = [Ca2+][F–]2
|Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42–(aq)
||Ksp = [Ag+]2[CrO42–]
|CuI(s) ⇌ Cu+(aq) + I–(aq)
||Ksp = [Cu+][I–]
The solubility of salts can be described by the Ksp or by the molar solubility. The molar solubility of the salt describes the number of moles of salt that can be dissolved per liter of solution. The Ksp value for Ag2CrO4 is 8.0 × 10-12. We can determine the molar solubility using the following calculations:
Ksp = [Ag+]2[CrO42–]
8.0 × 10–12 = (2x)2(x)
8.0 × 10–12 = 4x3
x = [CrO42–] = 1.3 × 10–4 M
2x = [Ag+] = 2.6 × 10–4 M
The molar solubility of the salt will also be equal to the concentration of any ion that occurs in a 1:1 ratio with the salt. In the example above, the molar solubility of the salt is 1.3 × 10-4 M, the same as the concentration of the CrO42- ions.
Note that the concentration of the silver ions is both doubled (because twice as many of them will be in solution) AND squared in the Ksp expression. Thus, the 2 coefficient on the silver is represented twice in the solubility product calculations. Forgetting to account for that coefficient in the ion concentration (like using just x instead of 2x) is a very common mistake made by students; take care to avoid it!
Typically, the molar solubility of most salts will increase with rising temperatures. This is because a higher temperature has more energy available to force the water molecules apart and make room for the solute ions. There are some salts that see their solubility decrease with increasing temperature, but there is no easy way to predict when that will occur.
The Common Ion Effect
Let’s look at the solubility expression for AgCl.
Ksp = [Ag+][Cl–] = 1.6 × 10–10
If we throw a block of solid AgCl into a beaker of water, we can tell from the Ksp what the concentrations of Ag+ and Cl– will be at equilibrium. For every unit of AgCl that dissociates, we get one Ag+ and one Cl–, so we can solve the equation above as follows:
[Ag+][Cl–] = 1.6 × 10–10
(x)(x) = 1.6 × 10–10
x2 = 1.6 × 10–10
x = [Ag+] = [Cl–] = 1.3 × 10–5 M
So there are very small amounts of Ag+ and Cl– in the solution.
Let’s say we add 0.10 mole of NaCl to 1 liter of the AgCl solution. NaCl dissociates completely, so that’s the same thing as adding 0.1 mole of Na+ ions and 0.1 mole of Cl– ions to the solution. The Na+ ions will not affect the AgCl equilibrium, so we can ignore them; but the Cl– ions must be taken into account. That’s because of the common ion effect.
The common ion effect says that the newly added Cl– ions will affect the AgCl equilibrium, although the newly added Cl– ions did not come from AgCl.
Let’s look at the solubility expression again. Now we have 0.10 mole of Cl– ions in 1 liter of the solution, so [Cl–] = 0.10 M.
[Ag+][Cl–] = 1.6 × 10–10
[Ag+](0.10 M) = 1.6 × 10–10
[Ag+] = M
[Ag+] = 1.6 × 10–9 M
Now the number of Ag+ ions in the solution has decreased drastically because of the Cl– ions introduced to the solution by NaCl. So when solutions of AgCl and NaCl, which share a common Cl– ion, are mixed, the more soluble salt (NaCl) can cause the less soluble salt (AgCl) to precipitate. In general, when two salt solutions that share a common ion are mixed, the salt with the lower value for Ksp will precipitate first.
Standard Free Energy Change and the Equilibrium Constant
The amount of Gibbs free energy in any given reaction can also be calculated if you know the equilibrium constant expression for that reaction.
∆G ° = –RT ln K
R = the gas constant, 8.31 J/mol·K
T = absolute temperature (K)
K = the equilibrium constant
Note that the “R” value here is NOT the same as the “R” value that we use when dealing with ideal gases! If ∆G° is negative, K must be greater than 1, and products will be favored at equilibrium. Alternatively, if ∆G° is positive, K must be less than 1, and reactants will be favored at equilibrium.
ACIDS AND BASES DEFINITIONS
S. A. Arrhenius defined an acid as a substance that ionizes in water and produces hydrogen ions (H+). As hydrogen atoms consist of one single proton and one single electron, removing that electron when we form the ion leaves just the proton behind. For that reason, you will often hear H+ ions referred to simply as protons. For instance, HCl is an acid.
HCl → H+ + Cl–
He defined a base as a substance that ionizes in water and produces hydroxide ions (OH– ions). For instance, NaOH is a base.
NaOH → Na+ + OH–
J. N. Brønsted and T. M. Lowry defined an acid as a substance that is capable of donating a proton, which is the same as donating an H+ ion, and they defined a base as a substance that is capable of accepting a proton. This definition is the one that will be used most frequently on the exam.
Look at the reversible reaction below.
HC2H3O2 + H2O ↔ C2H3O2– + H3O+
According to Brønsted-Lowry
HC2H3O2 and H3O+ are acids.
C2H3O2– and H2O are bases.
Now look at this reversible reaction.
NH3 + H2O ↔ NH4+ + OH–
According to Brønsted-Lowry
NH3 and OH– are bases.
H2O and NH4+ are acids.
So in each case, the species with the H+ ion is the acid, and the same species without the H+ ion is the base; the two species are called a conjugate pair. The following are the acid-base conjugate pairs in the reactions above:
HC2H3O2 and C2H3O2–
NH4+ and NH3
H3O+ and H2O
H2O and OH–
Notice that water can act either as an acid or base. Any substance which has that ability is called amphoteric.
Many of the concentration measurements in acid-base problems are given to us in terms of pH and pOH.
p (anything) = –log (anything)
pH = –log [H+]
pOH = –log [OH–]
pKa = –log Ka
pKb = –log Kb
In a solution
• when [H+] = [OH–], the solution is neutral, and pH = 7
• when [H+] is greater than [OH–], the solution is acidic, and pH is less than 7
• when [H+] is less than [OH–], the solution is basic, and pH is greater than 7
It is important to remember that increasing pH means decreasing [H+], which means that there are fewer H+ ions floating around and the solution is less acidic. Alternatively, decreasing pH means increasing [H+], which means that there are more H+ ions floating around and the solution is more acidic.
Strong acids dissociate completely in water, so the reaction goes to completion and they never reach equilibrium with their conjugate bases. Because there is no equilibrium, there is no equilibrium constant, so there is no dissociation constant for strong acids or bases.
Important Strong Acids
HCl, HBr, HI, HNO3, HClO4, H2SO4
Important Strong Bases
LiOH, NaOH, KOH, Ba(OH)2, Sr(OH)2
Because the dissociation of a strong acid goes to completion, there is no tendency for the reverse reaction to occur, which means that the conjugate base of a strong acid must be extremely weak.
It’s much easier to find the pH of a strong acid solution than it is to find the pH of a weak acid solution. That’s because strong acids dissociate completely, so the final concentration of H+ ions will be the same as the initial concentration of the strong acid.
Let’s look at a 0.010-molar solution of HCl.
HCl dissociates completely, so [H+] = 0.010 M
pH = –log [H+] = –log (0.010) = –log (10-2) = 2
So you can always find the pH of a strong acid solution directly from its concentration.
When a weak acid (often symbolized with HA) is placed in water, a small fraction of its molecules will dissociate into hydrogen ions (H+) and conjugate base ions (A–). Most of the acid molecules will remain in solution as undissociated aqueous particles.
The dissociation constants, Ka and Kb, are measures of the strengths of weak acids and bases. Ka and Kb are just the equilibrium constants specific to acids and bases.
Acid Dissociation Constant
[H+] = concentration of hydrogen ions (M)
[A–] = concentration of conjugate base ions (M)
[HA] = concentration of undissociated acid molecules (M)
Base Dissociation Constant
[HB+] = concentration of conjugate acid ions (M)
[OH–] = concentration of hydroxide ions (M)
[B] = concentration of unprotonated base molecules (M)
The greater the value of Ka, the greater the extent of the dissociation of the acid and the stronger the acid. The same thing goes for Kb, but in the case of Kb, the base is not dissociating. Instead, it is accepting a hydrogen ion (proton) from an ion. So, a base does not dissociate—it protonates (or ionizes).
If you know the Ka for an acid and the concentration of the acid, you can find the pH. For instance, let’s look at 0.20-molar solution of HC2H3O2, with Ka = 1.8 × 10–5.
First, we set up the Ka equation, plugging in values.
HC2H3O2 → H+ + C2H3O2–
Therefore, the ICE (Initial, Change, Equilibrium) table for the above problem is as follows:
Because every acid molecule that dissociates produces one H+ and one C2H3O2–,
[H+] = [C2H3O2–] = x
and because, strictly speaking, the molecules that dissociate should be subtracted from the initial concentration of HC2H3O2, [HC2H3O2] should be (0.20 M – x). In practice, however, x is almost always insignificant compared with the initial concentration of acid, so we just use the initial concentration in the calculation.
[HC2H3O2] = 0.20 M
Now we can plug our values and variable into the Ka expression.
1.8 × 10–5 =
Solve for x.
x = [H+] = 1.9 × 10–3
Now that we know [H+], we can calculate the pH.
pH = –log [H+] = –log (1.9 × 10–3) = 2.7
This is the basic approach to solving many of the weak acid/base problems that will be on the test.
Another way to write out the dissociation of a weak acid is:
HA(aq) + H2O(l) ⇌ A– + H3O+
The above includes water molecules in the reaction, and it is technically more accurate overall, as an acid will not dissociate unless it has a base to give protons to. The H3O+ ion is called the hydronium ion, and you can substitute it for H+ in any acid/base reaction. So, –log [H3O+] = pH.
The primary factor when it comes to determining acid strength is that the more H+ ions that an acid can donate, the stronger that acid will be. How easily an acid dissociates is often determined in part by its structure.
Consider the binary acids composed of hydrogen and a halogen. Of those, HI, HBr, and HCl are all strong acids, meaning they dissociate completely. HF, however, is not a strong acid. The reason for this is that fluorine is extremely electronegative, and thus fluorine will “hold on” to the hydrogen more effectively.
If you consider oxoacids, though, the reverse trend is true. Consider the Lewis diagrams for HOF and HOBr:
In this case, fluorine’s very high electronegativity affects the O–F bond, drawing the shared electrons toward the fluorine. However, fluorine is so electronegative that it also attracts the shared electrons in the H–O bond as well, which weakens the overall H–O bond and makes the H more likely to dissociate. Thus, HOF is a stronger acid than HOBr.
The strength of an acid is very case-specific, but the one rule that is true no matter what the situation is that the easier it is for the H+ ion to break free, the stronger the acid will be.
The other factor that affects how easily an acid can dissociate is the concentration of the acid. The lower the concentration is for an acid, the higher the percent dissociation will be. This is because the forward reaction involves the acid donating a proton to a water molecule. In an acid, there is an overabundance of water molecules, so it is very easy for the acid to find a water molecule to donate to. However, the reverse reaction is the hydronium ion donating a proton to the conjugate base. There are a much smaller number of both hydronium and conjugate base ions in a dilute solution, so the reverse reaction is kinetically hindered from happening. This means that more H3O+ ions will stay dissociated.
The greater the concentration of the acid, the more conjugate base there will be, and the easier it will be for that reverse reaction to take place. The easier it is for the reverse reaction to take place, the more HA there will be present in solution, and the less H3O+, leading to a lower overall percent dissociation.
Some acids, such as H2SO4 and H3PO4, can give up more than one hydrogen ion in solution. These are called polyprotic acids.
Polyprotic acids are always willing to give up their first proton more easily than future protons. Looking at the three Ka values for phosphoric acid:
|H3PO4⇌ H2PO4− + H+
||Ka1 = 7.1 × 10−3
|H2PO4− ⇌ HPO42− + H+
||Ka2 = 6.3 × 10−8
|HPO42− ⇌ PO43− + H+
||Ka3 = 4.5 × 10−13
So, H3PO4 is a stronger acid than H2PO4−, which in turn is a stronger acid than HPO42−. The reason behind this is that after the first proton dissociates, there are a bunch of extra protons floating around in solution. Via Le Chatelier’s principle, in future dissociations this causes shifts to the side with no protons—in this case, to the left. That makes it more difficult for remaining protons to dissociate, weakening the acid.
This also means that the amount of each succeeding acid would decrease. In a solution of H3PO4, [H3PO4] > [H2PO4–] > [HPO42–] > [PO43–].
The Equilibrium Constant of Water (KW)
Water comes to equilibrium with its ions according to the following reaction:
H2O(l) ⇌ H+(aq) + OH–(aq) Kw = 1 × 10–14 at 25°C.
Kw = 1 × 10–14 = [H+][OH–]
pH + pOH = 14
The common ion effect tells us that the hydrogen ion and hydroxide ion concentrations for any acid or base solution must be consistent with the equilibrium for the ionization of water. That is, no matter where the H+ and OH– ions came from, when you multiply [H+] and [OH–], you must get 1 × 10–14. So for any aqueous solution, if you know the value of [H+], you can find out the value of [OH–] and vice versa.
The acid and base dissociation constants for conjugates must also be consistent with the equilibrium for the ionization of water.
Kw = 1 × 10–14 = KaKb
pKa + pKb = 14
So if you know Ka as a weak acid, you can find Kb for its conjugate base and vice versa.
It is worth noting that pH is not limited to a 0 to 14 range. While most substances will fall into that range, substances that are very acidic can have negative pHs, and substances that are very basic can have pHs that are greater than 14. While you won’t find these extreme pHs in most everyday substances, they do exist.
Most acids that you would find in an acids cabinet of a chemistry storeroom are very concentrated and may have negative pH values. For instance, nitric acid is typically manufactured at a concentration of 16 M. Doing the math:
–log(16) = –1.2
Needless to say, you should exercise extreme caution when working with concentrated acids or bases, and it should be done only while wearing the proper personal protective equipment.
Something that is important to note is that the Kw for water is 1.0 × 10-14 at a temperature of 25°C only. Like any equilibrium constant, it will change if the temperature does. In this case, the dissociation of water is an endothermic process, as bonds are being broken without any new bonds being formed. So, as temperature increases, the reaction shifts to the right, which increases the value for Kw.
This will have an effect on the pH of water. For instance, at 50°C, Kw = 5.48 × 10–14. Thus, [H+] = [OH–] = 2.34 × 10–7 M and pH = pOH = 6.63. So, the pH of pure water at 50°C is NOT 7. The Kw of water is often measured by looking at the pKw value (–log Kw). As temperature and Kw increases, pKw decreases, as illustrated in the table below.
When an acid and a base mix, the acid will donate protons to the base in what is called a neutralization reaction. There are four different mechanisms for this, depending on the strengths of the acids and bases.
1. Strong acid + strong base
When a strong acid mixes with a strong base, both substances are dissociated completely. The only important ions in this type of reaction are the hydrogen and hydroxide ions (Even though not all bases have hydroxides, all strong bases do!).
Ex: HCl + NaOH
Net ionic: H+ + OH– ↔ H2O(l)
The net ionic equation for all strong acid/strong base reactions is identical—it is always the creation of water. The other ions involved in the reaction (in the example above, Cl– and Na+) act as spectator ions and do not take part in the reaction.
2. Strong acid + weak base
In this reaction, the strong acid (which dissociates completely), will donate a proton to the weak base. The product will be the conjugate acid of the weak base.
Ex: HCl + NH3
Net Ionic: H+ + NH3(aq) ↔ NH4+
3. Weak acid + strong base
In this reaction, the strong base will accept protons from the weak acid. The products are the conjugate base of the weak acid and water.
Ex: HC2H3O2 + NaOH
Net ionic: HC2H3O2(aq) + OH– ↔ C2H3O2– + H2O(l)
This is a simple proton transfer reaction, in which the acid gives protons to the base.
Ex: HC2H3O2 + NH3
Net ionic: HC2H3O2(aq) + NH3(aq) ↔ C2H3O2– + NH4+
During neutralization reactions, the final pH of the solution is dependent on whether the excess ions at equilibrium are due to the strong acid/base or due to a weak acid/base. When a strong acid/base is in excess, the pH calculation is fairly straightforward.
Example 1: 35.0 mL of 1.5 M HCN, a weak acid (Ka = 6.2 × 10–10) is mixed with 25.0 mL of 2.5 M KOH. Calculate the pH of the final solution.
To do this, we can modify our ICE chart to determine what species are present at equilibrium. First, we must determine the number of moles of the reactants:
HCN = (1.5 M)(0.035 L) = 0.052 mol KOH = (2.5 M)(0.025 L) = 0.062 mol
Then we set up our ICE chart (leaving values for water out, as it is a pure liquid):
The reaction will continue until the HCN runs out of protons to donate to the hydroxide. While there is some weak conjugate base left in solution, it is a weak base and its contribution to the pH of the solution will be irrelevant compared to the strength of the pure hydroxide ions. We now need to determine the concentration of the hydroxide ions at equilibrium. The total volume of the solution is the sum of both the acid and the base. So:
(35.0 mL + 25.0 mL = 60.0 mL)
[OH–] = 0.010 mol/0.060 L = 0.17 M
pOH = –log (0.17) = 0.76
pH = 14 – 0.76 = 13.24
As you can see, it does not take much excess H+ or OH– to drive the pH of a solution to a fairly high or low value. Note that the Ka of the acid did not matter in this case, as the strong acid/base was in excess.
When a weak acid or base is in excess, is it easiest to use the Henderson-Hasselbalch equation:
The Henderson-Hasselbalch Equation
pH = pKa + log
[HA] = molar concentration of undissociated weak acid (M)
[A–] = molar concentration of conjugate base (M)
pOH = pKb + log
[B] = molar concentration of weak base (M)
[HB+] = molar concentration of conjugate acid (M)
Example 2: 25.0 mL of 1.0 M HCl is mixed with 60.0 mL of 0.50 M pyridine (C5H5N), a weak base (Kb = 1.5 × 10–9). Determine the pH of the solution.
Starting out, the number of moles of each reactant must be determined:
H+ = (1.0 M)(0.025 L) = 0.025 mol H+
C5H5N = (0.50 M)(0.060 L) = 0.030 mol C5H5N
Then we set up the ICE chart:
The reaction continues until the H+ runs out, leaving pyridine and its conjugate acid at equilibrium. Next, the new concentrations must be determined for both the weak base and its conjugate acid, as both will contribute to the final pH value. The total volume of the new solution is the sum of both the acid and base added, in this case, 25.0 + 60.0 = 85.0 mL. So:
[C5H5N] = 0.005 mol/0.085 L = 0.059 M
[HCH5N+]= 0.025 mol/0.085 L = 0.29 M
pOH = pKb+ log
pOH = –log (1.5 × 10–9) + log
pOH = 8.8 + log (4.9)
pOH = 8.8 + 0.69
pOH = 9.5
pH = 14 – 9.5 = 4.5
When a weak acid or base is in excess, the pH of the solution does not change as quickly as when a strong acid or base is in excess.
A buffer is a solution with a very stable pH. You can add acid or base to a buffer solution without greatly affecting the pH of the solution. The pH of a buffer will also remain unchanged if the solution is diluted with water or if water is lost through evaporation.
A buffer is created by placing a large amount of a weak acid or base into a solution along with its conjugate, in the form of salt. A weak acid and its conjugate base can remain in solution together without neutralizing each other. This is called the common ion effect.
When both the acid and the conjugate base are together in the solution, any hydrogen ions that are added will be neutralized by the base, while any hydroxide ions that are added will be neutralized by the acid, without this having much of an effect on the solution’s pH.
Let’s say we have a buffer solution with concentrations of 0.20 M HC2H3O2 and 0.50 M C2H3O2–. The acid dissociation constant for HC2H3O2 is 1.8 × 10–5. Let’s find the pH of the solution.
We can just plug the values we have into the Henderson-Hasselbalch equation for acids.
pH = pKa + log
pH = –log (1.8 × 10–5) + log
pH = –log (1.8 × 10–5) + log (2.5)
pH = (4.7) + (0.40) = 5.1
Now let’s see what happens when [HC2H3O2] and [C2H3O2–] are both equal to 0.20 M.
pH = pKa + log
pH = –log (1.8 × 10–5) + log
pH = –log (1.8 × 10–5) + log (1)
pH = (4.7) + (0) = 4.7
Notice that when the concentrations of acid and conjugate base in a solution are the same, pH = pKa (and pOH = pKb). When you choose an acid for a buffer solution, it is best to pick an acid with a pKa that is close to the desired pH. That way you can have almost equal amounts of acid and conjugate base in the solution, which will make the buffer as flexible as possible in neutralizing both added H+ and OH–.
You cannot create a buffer solution from a strong acid and its conjugate, because the conjugate base of a strong acid will be very weak. Taking HCl as an example; the Cl– ion that is left after the acid dissociates completely is a very weak base and will not readily accept protons. The same is true for strong bases; you cannot form a buffer from a strong base and its conjugate for similar reasons.
Indicators are weak acids which change colors in certain pH ranges due to Le Chatelier’s principle. All indicators are weak acids in their own right, but often have very long, very complicated chemical formulas. For simplicity’s sake, we will use the generic HIn to indicate a protonated indicator.
HIn ⇌ H+ + In−
For an indicator to be effective, the protonated (HIn) state has to be a different color than the deprotonated (In−) state. When an indicator is present in a titration and the environment in acidic, the excess H+ ions drive the equilibrium to the left, causing the solution to consist primarily of HIn and take on that color. In a basic environment, the excess OH− ions react with the H+ from the indicator and drive the reaction right, causing the solution to consist of primarily In−, changing color.
The indicator will experience a color shift right around the point that the two species (HIn and In−) are present in equal amounts. When that is the case, those two values cancel out of the equilibrium expression, leaving Ka = [H+]. A quick logarithm later and we can see that at the point of color change, pKa = pH. This means that the indicator will change colors at a pH that matches the pKa of the indicator.
When doing a titration, you only want to add a few drops of any indicator in order to avoid changing any kind of calculations involving the acid you are titrating against (or with!). It is always best to choose an indicator whose pKa matches the pH at the equivalence point of the titration you are doing in order to be able to find the most accurate endpoint.
Neutralization reactions are generally performed by titration, where a base of known concentration is slowly added to an acid (or vice versa). The progress of a neutralization reaction can be shown in a titration curve. The diagram below shows the titration of a strong acid by a strong base.
In the diagram above, the pH increases slowly but steadily from the beginning of the titration until just before the equivalence point. The equivalence point is the point in the titration when exactly enough base has been added to neutralize all the acid that was initially present. Just before the equivalence point, the pH increases sharply as the last of the acid is neutralized. The equivalence point of a titration can be recognized through the use of an indicator.
The following diagram shows the titration of a weak acid by a strong base:
In this diagram, the pH increases more quickly at first, then levels out into a buffer region. At the center of the buffer region is the half-equivalence point.
At this point, enough base has been added to convert exactly half of the acid into conjugate base; here the concentration of acid is equal to the concentration of conjugate base [HA] = [A–]). Putting that into our weak acid dissociation expression:
If [A–] and [HA] are equal and cancel out, that leaves us with Ka = [H+], which is often represented by pKa = pH. This is a good way to determine the Ka of a weak acid from a titration curve such as the one above.
The curve remains fairly flat until just before the equivalence point, when the pH increases sharply. For this titration, the pH at the equivalence point is greater than 7 because the only ion present in significant amounts at equilibrium is the conjugate base of the weak acid. Likewise, if you titrate a weak base with a strong acid, the solution will be acidic at equilibrium because the only ion present in significant amounts will be the conjugate acid of the weak base.
The following diagram shows the titration curve of a polyprotic acid:
For a polyprotic acid, the titration curve will have as many bumps as there are hydrogen ions to give up. The curve above has two bumps, so it represents the titration of a diprotic acid. There are also two half-equivalence points for a diprotic acid. At the first one, the pH will be equal to the negative log of the first dissociation constant for the polyprotic acid (pK1). At the second half-equivalence point, the pH will be equal to the negative log of the second dissociation constant (pK2).
CHAPTER 8 QUESTIONS
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Use the following information to answer questions 1-5.
A student titrates 20.0 mL of 1.0 M NaOH with 2.0 M formic acid, HCO2H (Ka = 1.8 × 10–4). Formic acid is a monoprotic acid.
1. How much formic acid is necessary to reach the equivalence point?
(A) 10.0 mL
(B) 20.0 mL
(C) 30.0 mL
(D) 40.0 mL
2. At the equivalence point, is the solution acidic, basic, or neutral? Why?
(A) Acidic; the strong acid dissociates more than the weak base
(B) Basic; the only ion present at equilibrium is the conjugate base
(C) Basic; the higher concentration of the base is the determining factor
(D) Neutral; equal moles of both acid and base are present
3. If the formic acid were replaced with a strong acid such as HCl at the same concentration (2.0 M), how would that change the volume needed to reach the equivalence point?
(A) The change would reduce the amount as the acid now fully dissociates.
(B) The change would reduce the amount because the base will be more strongly attracted to the acid.
(C) The change would increase the amount because the reaction will now go to completion instead of equilibrium.
(D) Changing the strength of the acid will not change the volume needed to reach equivalance.
4. Which of the following would create a good buffer when dissolved in formic acid?
5. CH3NH2(aq) + H2O(l) ↔ OH–(aq) + CH3NH3+(aq)
The above equation represents the reaction between the base methylamine (Kb = 4.38 × 10–4) and water. Which of the following best represents the concentrations of the various species at equilibrium?
(A) [OH–] > [CH3NH2] = [CH3NH3+]
(B) [OH–] = [CH3NH2] = [CH3NH3+]
(C) [CH3NH2] > [OH–] > [CH3NH3+]
(D) [CH3NH2] > [OH–] = [CH3NH3+]
Use the following information to answer questions 6-10.
The following reaction is found to be at equilibrium at 25°C:
2SO3(g) ↔ O2(g) + 2SO2(g) ∆H = –198 kJ/mol
6. What is the expression for the equilibrium constant, Kc?
7. Which of the following would cause the reverse reaction to speed up?
(A) Adding more SO3
(B) Raising the pressure
(C) Lowering the temperature
(D) Removing some SO2
8. The value for Kc at 25°C is 8.1. What must happen in order for the reaction to reach equilibrium if the initial concentrations of all three species was 2.0 M?
(A) The rate of the forward reactions would increase, and [SO3] would decrease.
(B) The rate of the reverse reaction would increase, and [SO2] would decrease.
(C) Both the rate of the forward and reverse reactions would increase, and the value for the equilibrium constant would also increase.
(D) No change would occur in either the rate of reaction or the concentrations of any of the species.
9. Which of the following would cause a reduction in the value for the equilibrium constant?
(A) Increasing the amount of SO3
(B) Reducing the amount of O2
(C) Raising the temperature
(D) Lowering the temperature
10. The solubility product, Ksp, of AgCl is 1.8 × 10–10. Which of the following expressions is equal to the solubility of AgCl?
(A) (1.8 × 10–10)2 molar
(C) 1.8 × 10–10 molar
11. A 0.1-molar solution of which of the following acids will be the best conductor of electricity?
12. Which of the following expressions is equal to the Ksp of Ag2CO3?
(A) Ksp = [Ag+][CO32–]
(B) Ksp = [Ag+][CO32–]2
(C) Ksp = [Ag+]2[CO32–]
(D) Ksp = [Ag+]2[CO32–]2
13. If the solubility of BaF2 is equal to x, which of the following expressions is equal to the solubility product, Ksp, for BaF2?
Use the following information to answer questions 14-16:
150 mL of saturated SrF2 solution is present in a 250 mL beaker at room temperature. The molar solubility of SrF2 at 298 K is 1.0 × 10-3 M.
14. What are the concentrations of Sr2+ and F– in the beaker?
(A) [Sr2+] = 1.0 × 10–3 M [F–] = 1.0 × 10–3 M
(B) [Sr2+] = 1.0 × 10–3 M [F–] = 2.0 × 10–3 M
(C) [Sr2+] = 2.0 × 10–3 M [F–] = 1.0 × 10–3 M
(D) [Sr2+] = 2.0 × 10–3 M [F–] = 2.0 × 10–3 M
15. If some of the solution evaporates overnight, which of the following will occur?
(A) The mass of the solid and the concentration of the ions will stay the same.
(B) The mass of the solid and the concentration of the ions will increase.
(C) The mass of the solid will decrease, and the concentration of the ions will stay the same.
(D) The mass of the solid will increase, and the concentration of the ions will stay the same.
16. How could the concentration of Sr2+ ions in solution be decreased?
(A) Adding some NaF(s) to the beaker
(B) Adding some Sr(NO3)2(s) to the beaker
(C) By heating the solution in the beaker
(D) By adding a small amount of water to the beaker, but not dissolving all the solid
17. A student added 1 liter of a 1.0 M KCl solution to 1 liter of a 1.0 M Pb(NO3)2 solution. A lead chloride precipitate formed, and nearly all of the lead ions disappeared from the solution. Which of the following lists the ions remaining in the solution in order of decreasing concentration?
(A) [NO3–] > [K+] > [Pb2+]
(B) [NO3–] > [Pb2+] > [K+]
(C) [K+] > [Pb2+] > [NO3–]
(D) [K+] > [NO3–] > [Pb2+]
18. 2HI(g) + Cl2(g) ⇌ 2HCl(g) + I2(g) + energy
A gaseous reaction occurs and comes to equilibrium, as shown above. Which of the following changes to the system will serve to increase the number of moles of I2 present at equilibrium?
(A) Increasing the volume at constant temperature
(B) Decreasing the volume at constant temperature
(C) Increasing the temperature at constant volume
(D) Decreasing the temperature at constant volume
19. 2NOBr(g) ⇌ 2NO(g) + Br2(g)
The reaction above came to equilibrium at a temperature of 100°C. At equilibrium the partial pressure due to NOBr was 4 atmospheres, the partial pressure due to NO was 4 atmospheres, and the partial pressure due to Br2 was 2 atmospheres. What is the equilibrium constant, Kp, for this reaction at 100°C?
20. Br2(g) + I2(g) ↔ 2IBr(g)
At 150°C, the equilibrium constant, Kc, for the reaction shown above has a value of 300. This reaction was allowed to reach equilibrium in a sealed container and the partial pressure due to IBr(g) was found to be 3 atm. Which of the following could be the partial pressures due to Br2(g) and I2(g) in the container?
21. A laboratory technician wishes to create a buffered solution with a pH of 5. Which of the following acids would be the best choice for the buffer?
||Ka = 5.9 × 10–2
||Ka = 5.6 × 10–3
||Ka = 1.8 × 10–5
||Ka = 3.0 × 10–8
22. Which of the following species is amphoteric?
23. How many liters of distilled water must be added to 1 liter of an aqueous solution of HCl with a pH of 1 to create a solution with a pH of 2?
(A) 0.1 L
(B) 0.9 L
(C) 2 L
(D) 9 L
24. A 1-molar solution of a very weak monoprotic acid has a pH of 5. What is the value of Ka for the acid?
(A) Ka = 1 × 10–10
(B) Ka = 1 × 10–7
(C) Ka = 1 × 10–5
(D) Ka = 1 × 10–2
25. The value of Ka for is 1 × 10–2. What is the value of Kb for ?
(A) Kb = 1 × 10–12
(B) Kb = 1 × 10–8
(C) Kb = 1 × 10–2
(D) Kb = 1 × 102
Use the following information to answer questions 26-28.
The following curve is obtained during the titration of 30.0 mL of 1.0 M NH3, a weak base, with a strong acid:
26. Why is the solution acidic at equilibrium?
(A) The strong acid dissociates fully, leaving excess [H+] in solution.
(B) The conjugate acid of NH3 is the only ion present at equilibrium.
(C) The water which is being created during the titration acts as an acid.
(D) The acid is diprotic, donating two protons for every unit dissociated.
27. What is the concentration of the acid?
(A) 0.5 M
(B) 1.0 M
(C) 1.5 M
(D) 2.0 M
28. What ions are present in significant amounts during the first buffer region?
(A) NH3 and NH4+
(B) NH3 and H+
(C) NH4+ and OH–
(D) H3O+ and NH3
Use the information below to answer questions 29-31.
Silver sulfate, Ag2SO4, has a solubility product constant of 1.0 × 10-5. The below diagram shows the products of a precipitation reaction in which some silver sulfate was formed.
29. What is the identity of the excess reactant?
30. If the beaker above were left uncovered for several hours:
(A) Some of the Ag2SO4 would dissolve.
(B) Some of the spectator ions would evaporate into the atmosphere.
(C) The solution would become electrically imbalanced.
(D) Additional Ag2SO4 would precipitate.
31. Which ion concentrations below would have led the precipitate to form?
(A) [Ag+] = 0.01 M [SO42-] = 0.01 M
(B) [Ag+] = 0.10 M [SO42-] = 0.01 M
(C) [Ag+] = 0.01 M [SO42-] = 0.10 M
(D) This is impossible to determine without knowing the total volume of the solution.
32. In a voltaic cell with a Cu (s) | Cu2+ cathode and a Pb2+ | Pb (s) anode, increasing the concentration of Pb2+ causes the voltage to decrease. What is the reason for this?
(A) The value for Q will increase, causing the cell to come closer to equilibrium.
(B) The solution at the anode becomes more positively charged, leading to a reduced electron flow.
(C) The reaction will shift to the right, causing a decrease in favorability.
(D) Cell potential will always decrease anytime the concentration of any aqueous species present increases.
33. Which of the following could be added to an aqueous solution of weak acid HF to increase the percent dissociation?
34. A bottle of water is left outside early in the morning. The bottle warms gradually over the course of the day. What will happen to the pH of the water as the bottle warms?
(A) Nothing; pure water always has a pH of 7.00.
(B) Nothing; the volume would have to change in order for any ion concentration to change.
(C) It will increase because the concentration of [H+] is increasing.
(D) It will decrease because the auto-ionization of water is an endothermic process.
1. The value of the solubility product, Ksp, for calcium hydroxide, Ca(OH)2, is 5.5 × 10–6, at 25°C.
(a) Write the Ksp expression for calcium hydroxide.
(b) What is the mass of Ca(OH)2 in 500 mL of a saturated solution at 25°C?
(c) What is the pH of the solution in (b)?
(d) If 1.0 mole of OH– is added to the solution in (b), what will be the resulting Ca2+ concentration? Assume that the volume of the solution does not change.
2. For sodium chloride, the solution process with water is endothermic.
(a) Describe the change in entropy when sodium chloride dissociates into aqueous particles.
(b) Two saturated aqueous NaCl solutions, one at 20°C and one at 50°C, are compared. Which one will have higher concentration? Justify your answer.
(c) Which way will the solubility reaction shift if the temperature is increased?
(d) If a saturated solution of NaCl is left out overnight and some of the solution evaporates, how will that affect the amount of solid NaCl present?
3. A student tests the conductivity of three different acid samples, each with a concentration of 0.10 M and a volume of 20.0 mL. The conductivity was recorded in microsiemens per centimeter in the table below:
(a) The three acids are known to be HCl, H2SO4, and H3PO4. Identify which sample is which acid. Justify your answer.
(b) The HCl solution is then titrated with a 0.150 M solution of the weak base methylamine, CH3NH2.
(Kb = 4.38 × 10–4)
(i) Write out the net ionic equation for this reaction.
(ii) Determine the pH of the solution after 20.0 mL of methylamine has been added.
4. N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H = –92.4 kJ
When the reaction above took place at a temperature of 570 K, the following equilibrium concentrations were measured:
[NH3] = 0.20 mol/L
[N2] = 0.50 mol/L
[H2] = 0.20 mol/L
(a) Write the expression for Kc and calculate its value.
(b) Calculate ∆G for this reaction.
(c) Describe how the concentration of H2 at equilibrium will be affected by each of the following changes to the system at equilibrium:
(i) The temperature is increased.
(ii) The volume of the reaction chamber is increased.
(iii) N2 gas is added to the reaction chamber.
(iv) Helium gas is added to the reaction chamber.
5. In an acidic medium, iron (III) ions will react with thiocyanate (SCN–) ions to create the following complex ion:
Fe3+(aq) + SCN–(aq) ↔ FeSCN2+(aq)
Initially, the solution is a light yellow color due to the presence of the Fe3+ ions. As the FeSCN2+ forms, the solution will gradually darken to a golden yellow. The reaction is not a fast one, and generally after mixing the ions the maximum concentration of FeSCN2+ will occur between 2–4 minutes after mixing the solution.
A student creates four solution with varying concentration of FeSCN2+ and gathers the following data at 298 K using a spectrophotometer calibrated to 460 nm:
|1.1 × 10–4 M
|1.6 × 10–4 M
|2.2 × 10–4 M
|2.5 × 10–4 M
(a) (i) On the axes below, create a Beer’s Law calibration plot for [FeSCN2+]. Draw a best-fit line through your data points.
(ii) The slope of the best-fit line for the above set of data points is 879 and the y-intercept is –0.024. Write out the equation for this line.
To determine the equilibrium constant for the reaction, a solution is made up in which 5.00 mL of 0.0025 M Fe(NO3)3 and 5.00 mL of 0.0025 M KSCN are mixed. After 3 minutes, the absorbance of the solution is found to be 0.134.
(b) (i) Using your Beer’s Law best-fit line from (a), calculate [FeSCN2+] once equilibrium has been established.
(ii) Calculate [Fe3+] and [SCN–] at equilibrium.
(iii) Calculate Keq for the reaction.
After equilibrium is established, the student heats the solution and observes that it becomes noticeably lighter.
(c) (i) Did heating the mixture increase the equilibrium constant, decrease it, or have no effect on it? Why?
(ii) Is the equilibrium reaction exothermic or endothermic? Justify your answer.
6. HA + OH ⇌ A– + H2O (l)
A student titrates a weak acid, HA, with some 1.0 M NaOH, yielding the following titration curve:
(a) Which chemical species present in solution dictates the pH of the solution in each of the volume ranges listed below?
(i) 1.0 mL–14.0 mL
(ii) 15.0 mL
(iii) 16.0 mL–30.0 mL
(b) At which volumes is:
(i) [HA] > [A–]?
(ii) [HA] = [A–]?
(iii) [HA] < [A–]?
(c) At which point in the titration (if any) would the concentration of the following species be equal to zero? Justify your answers.
(d) If the titration were performed again, but this time with 2.0 M NaOH, name two things that would change about the titration curve, and explain the reasoning behind your identified changes.
7. A student performs an experiment to determine the concentration of a solution of hypochlorous acid, HOCl (Ka = 3.5 × 10–8). The student starts with 25.00 mL of the acid in a flask and titrates it against a standardized solution of sodium hydroxide with a concentration of 1.47 M. The equivalence point is reached after the addition of 34.23 mL of NaOH.
(a) Write the net ionic equation for the reaction that occurs in the flask.
(b) What is the concentration of the HOCl?
(c) What would the pH of the solution in the flask be after the addition of 28.55 mL of NaOH?
(d) The actual concentration of the HOCl is found to be 2.25 M. Quantitatively discuss whether or not each of the following errors could have caused the error in the student’s results.
(i) The student added additiona