Question

Calculate the pH of a solution prepared by mixing 15.0 mL of 0.10 M NaOH and 30.0 mL of a 0.10 M benzoic acid solution. (Benzoic acid is monprotic; its dissociation constant is 6.5 x 10^-5.).

Answer 1

moles = molarity x mass
Moles of NaOH = (0.10 mol/L) (0.015 L) = 0.0015 mol 
Moles of benzoic acid = (0.100 mol/L) (0.0300 L) = 0.003 mol

NaOH and benzoic acid react in a 1:1 molar ratio. This means they are a buffer. 

After the reaction

moles of sodium benzoate = 0.00150 mol 
moles of benzoic acid = 0.00150 mol

pH = pKa + log [base / acid]

pH = -log(6.5 x 10^-5)

pH = 4.187

pH = 4.19

Answer 2

The pH of the resulting solution is 4.2.

The equation of the reaction is;

C6H5COOH(aq) + NaOH(aq)  ---------> C6H5COONa(aq) + H2O(l)

We can see that this reaction occur between a weak acid and a strong base. The reaction is 1:1

Number of moles of Benzoic acid = 30/1000 L × 0.10 M = 0.003 moles

Number of moles of sodium hydroxide = 15/1000 L × 0.10 M = 0.0015 moles

We can see that there are more moles of benzoic acid than sodium hydroxide

Number of moles of benzoic acid left over = 0.0285 moles

Total volume of solution =  15.0 mL + 30.0 mL = 45 mL or 0.045 L

Concentration of benzoic acid left over =  0.0285 moles/ 0.045 L

= 0.633 M

Using the Henderson - Hasselbalch equation;

pH = pKa + log[conjugate base]/[Weak acid]

Ka of benzoic acid = 6.5 x 10^-5

pKa = -log Ka = - log(6.5 x 10^-5) = 4.2

Note that [C6H5COO-] = [C6H5COOH] = 0.633 M

We can see that log[conjugate base]/[Weak acid] = 0

Substituting values;

pH = pKa

pH = 4.2

Hence the pH of the solution = 4.2

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