SAT Subject Test Chemistry

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 11

Acids, Bases, and Salts

TITRATION—VOLUMETRIC ANALYSIS

Knowledge of the concentrations of solutions and the reactions they take part in can be used to determine the concentrations of “unknown” solutions or solids. The use of volume measurement in solving these problems is called titration.

A common example of a titration uses acid-base reactions. If you are given a base of known concentration, that is, a standard solution, let us say 0.10 M NaOH, and you want to determine the concentration of an HCl solution, you could titrate the solutions in the following manner.

First, introduce a measured quantity, 25.0 milliliters, of the NaOH into a flask by using a pipet or burette in a setup like the one in the accompanying diagram. Next, introduce 2 drops of a suitable indicator. Because NaOH and HCl are considered a strong base and a strong acid, respectively, an indicator that changes color in the middle pH range would be appropriate. Litmus solution would be one choice. It is blue in a basic solution but changes to red when the solution becomes acidic. Slowly introduce the HCl until the color change occurs. This point is called the end point. The point at which enough acid is added to neutralize all the standard solution in the flask is called the equivalence point.

Burette Setup for Titration

Suppose 21.5 milliliters of HCl was needed to produce the color change. The reaction that occurred was

H+(aq)+OH(aq)→H2O(l)

until all the OH was neutralized; then the excess H+ caused the litmus paper to change color.

To solve the question of the concentration of NaOH, this equation is used:

Macid+Vacid=Mbase+Vbase

Substituting the known amounts in this equation gives

xMacid × 21.5 mL = 0.1 M × 25.0 mL
           x=0.116 M

TIP 

For a titration:
Macid+Vacid=Mbase+Vbase

In choosing an indicator for a titration, we need to consider whether the solution formed when the end point is reached has a pH of 7. Depending on the types of acid and base used, the resulting hydrolysis of the salt formed may cause the solution to be slightly acidic, slightly basic, or neutral. If a strong acid and a strong base are titrated, the end point will be at pH 7, and practically any indicator can be used because adding 1 drop of either reagent will change the pH at the end point by about 6 units. For titrations of strong acids and weak bases, we need an indicator, such as methyl orange, that changes color between 3.1 and 4.4 in the acid region. When titrating a weak acid and a strong base, we should use an indicator that changes in the basic range. Phenolphthalein is the suitable choice for this type of titration because it changes color in the pH 8.3 to 10.0 range.

The process of the neutralization reaction can be represented by a titration curve like the one below, which shows the titration of a strong acid with a strong base.

EXAMPLE 1: Find the concentration of acetic acid in vinegar if 21.6 milliliters of 0.20 M NaOH is needed to titrate a 25-milliliter sample of the vinegar.
Solution
Using the equation Macid × Vacid = Mbase × Vbase, we have

Another type of titration problem involves a solid and a titrated solution.

EXAMPLE 2: A solid mixture contains NaOH and NaCl. If 10.0 milliliters of 0.100 M HCl is required to titrate a 0.100-gram sample of this mixture to its end point, what is the percent of NaOH in the sample?

Solution

Since


then



Substituting the HCl information in the equation,
we have

(Note: this is 10 mL expressed in liters)

Since 1 mol of HCl neutralizes 1 mol of NaOH, 0.001 mol of NaOH must be present in the mixture. Since 1 mol NaOH = 40.0 g

then

               0.001mol×40.0 g/mol=0.04 g NaOH

Therefore 0.04 g of NaOH was in the 0.100-g sample of the solid mixture. The percent is 0.04 g/0.100 g × 100 = 40%.

In the explanations given to this point, the reactions that took place were between monoprotic acids (single hydrogen ions) and monobasic bases (one hydroxide ion per base). This means that each mole of acid had 1 mole of hydrogen ions available, and each mole of base had 1 mole of hydroxide ions available, to interact in the following reaction until the end point was reached:

H+(aq)+OH(aq)→2H2O(l)

This is not always the case, however, and it is important to know how to deal with acids and bases that have more than one hydrogen ion and more than one hydroxide ion per formula. The following is an example of such a problem.

EXAMPLE 3: If 20.0 milliliters of an aqueous solution of calcium hydroxide, Ca(OH)2, is used in a titration, and an appropriate indicator is added to show the neutralization point (end point), the few drops of indicator that are added can be ignored in the volume considerations. Therefore, if 25.0 milliliters of standard 0.050 M HCl is required to reach the end point, what was the original concentration of the Ca(OH)2 solution?
    The balanced equation for the reaction gives the relationship between the number of moles of acid reacting and the number of moles of base:

The mole relationship here is that the number of moles of acid is twice the number of moles of base:

    No. of moles of acid = 2 × No. of moles of base
                                        ↑ mole factor

Since the molar concentration of the acid times the volume of the acid gives the number of moles of acid:

               Ma × Va= moles of acid

and the molar concentration of the base times the volume of the base gives the number of moles of base:

Mb × Vb=mole of base

                           then, substituting these products into the mole relationship, we get

MaVa=2MbVb

                           Solving for Mb gives

                           Substituting values, we get