An analytical chemist is titrating 60.5mL of a 0.8700M solution of benzoic acid HC6H5CO2 with a 0.3600M solution of KOH. The pKa of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 172. mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places.

The correct answer is 12.6.

Explanation:

The formation of benzoate takes place when potassium hydroxide reacts with benzoic acid, due to the presence of a weak acid and its conjugated base the solution will act as a buffer. In the given question, the molarity of benzoic acid given is 0.8700 M and its volume is 60.5 ml. Therefore, the moles of benzoic acid will be,

Moles = molarity * volume of solution

= 0.8700 M * 60.5 ml = 52.365 m mol or 0.052365 moles

On the other hand, the molarity of KOH given is 0.3600 M and the volume given is 172 ml. Therefore, the moles of KOH added will be,

Moles = 0.3600 * 172 = 61.92 m moles or 0.06192 moles

Out of this 61.92 m mol, only 52.365 m mol of KOH will react with the benzoic acid. The moles of KOH, which remain unreactive is,

61.92 m moles - 52.365 m moles = 9.285 m moles or 0.009285 moles

The formula for calculating molarity is number of moles / volume of solution in liters

The total volume of the solution is 172 ml + 60.5 ml = 232.5 ml or 0.2325 L

The molarity of KOH will be,

Molarity = 0.009285 moles / 0.2325 L = 0.0395 M

The dissociation of KOH takes place completely to produce hydroxide ion.

pOH = - log[0.0395] = 1.4

pH + pOH = 14

pH = 14 - 1.4 = 12.6