Cracking the AP Chemistry Exam

Part IV

Content Review for the AP Chemistry Exam

Chapter 8

Big Idea #6: Equilibrium, Acids and Bases, Titrations, and Solubility

STRONG ACIDS

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.

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 for a weak acid, you can find Kb for its conjugate base and vice versa.

Neutralization Reactions

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)

    4.  Weak acid+weak base

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

Using Henderson-Hasselbalch:

 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.