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