A volume of 25.0 mL of 0.100 M C₆H₅CO₂H(aq)is titrated with 0.100 M NaOH(aq).What is the pH after the addition of 12.5 mL of NaOH? (K_a of benzoic acid = 6.3 × 10^-5) A) 2.60 B) 4.20 C) 5.40 D) 7.00 E) 8.60

Question

Quizplus · Accepted Answer

The given solution contains both benzoic acid (C6H5COOH) and its conjugate base (C6H5COO-), which means it acts as a buffer solution.

The balanced equation for the acid-base reaction between benzoic acid and NaOH is:

C6H5COOH + NaOH -> C6H5COO- Na+ + H2O

Before the addition of NaOH, the moles of benzoic acid present in the solution can be calculated as:

moles of C6H5COOH = Molarity x Volume = 0.100 mol/L x 0.0250 L = 0.00250 mol

Since the initial volume is 25.0 mL and 12.5 mL of 0.100 M NaOH is added, the final volume of the solution is 37.5 mL.

The stoichiometry of the reaction shows that one mole of NaOH reacts with one mole of benzoic acid. The moles of NaOH added can be calculated as:

moles of NaOH = Molarity x Volume = 0.100 mol/L x 0.0125 L = 0.00125 mol

Therefore, the moles of benzoic acid left after the addition of NaOH are:

moles of C6H5COOH left = 0.00250 mol - 0.00125 mol = 0.00125 mol

The moles of benzoate ion (C6H5COO-) formed can be calculated from the stoichiometry of the reaction:

moles of C6H5COO- = moles of NaOH added = 0.00125 mol

The total moles of benzoic acid and benzoate ion after the addition of NaOH are:

moles of C6H5COOH + moles of C6H5COO- = 0.00125 mol + 0.00250 mol = 0.00375 mol

The volume of the solution after the addition of NaOH is 37.5 mL, so the new concentration can be calculated as:

New concentration = moles/volume = 0.00375 mol / 0.0375 L = 0.100 M

This means that the solution still has the same concentration of benzoic acid and benzoate ion as before the addition of NaOH, and therefore it is still a buffer solution.

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([base]/[acid])

where pKa is the dissociation constant of the weak acid, [base] is the concentration of the conjugate base, and [acid] is the concentration of the weak acid.

The pKa of benzoic acid is given as 6.3 × 10-5 in the question.

At the beginning of the titration, the concentration of benzoic acid and benzoate ion are both 0.100 M, so the pH of the buffer is:

pH = pKa + log([base]/[acid]) = 4.20

After the addition of 12.5 mL of 0.100 M NaOH, the concentration of benzoic acid becomes 0.0750 M (0.001875 mol/0.025 L) and the concentration of benzoate ion becomes 0.0250 M (0.001875 mol/0.075 L).

Using the Henderson-Hasselbalch equation again, the pH of the buffer after the addition of NaOH is:

pH = pKa + log([base]/[acid]) = 4.20 + log(0.0250/0.0750) = 2.60

Therefore, Choice B is the correct answer.

A Volume of 25