Question

Calculate the pH of a solution prepared by mixing 15 mL of 0.1M NaOH and 30 mL of 0.1 M benzoic acid solution. (Benzoicacid is monoprotic; its dissociation constant is 6.5 x10-5)

Answer 1

First write the chemical equation for the reaction of benzoic acid and NaOH:

NaOH < = > Na+ OH-

C6H5COOH (aq) + OH- (aq) --> C6H5COO- (aq) + H2O (l)
benzoic acid                                    conjugate base


Now calculate the moles of both reagent as follows:.

Moles of NaOH = (0.015 L)(0.10 M) = 0.0015 mol
Moles of C6H5COOH = (0.030 L)(0.10 M) = 0.030 mol

Now un-reacted C6H5COOH = 0.030 mol - 0.0015 mol = 0.0015 mol acid

According to reaction the number of moles of CH3COO- is 0.0015 mol

total volume = 15.0 mL + 30.0 mL = 45.0 mL, or 0.0450L.

Now determined the molarities of acid and its conjugates base:

[benzoic acid] = 0.0015 mol / 0.0450L = 0.0333 M
[benzoate ion] = 0.0015 mol / 0.0450L = 0.0333 M

Now to find the pH use the Henderson Hasselbach equation.

pH = pKa + log{[benzoic acid][benzoate ion]}

here; pH = pKa = -log 6.5^10-5 = 4.19.

pH = pKa + log{[benzoic acid][benzoate ion]}

pH = 4.19 + log{[0.0333]/0.0333}

pH = 4.19 + log 1.0

pH = 4.19 + 0.0

pH = 4.19