Question

Given that the Ka of benzoic acid is 6.3 x10-5, calculate the pH when 0.080M sodium hydroxide is added to 10.0 ml of 0.052M benzoic acid at the equivalence point

Answer 1

Here in the question, $K_a(C_6H_5COOH)=6.3 \times 10^{-5}$ , 0.08 M of NaOH is aded to 10

mL of 0.052 M $C_6H_5COOH$

The reaction between them is taking place as

$C_6H_5COOH + NaOH \longrightarrow C_6H_5COONa + H_2O$

and,

$C_6H_5COOH \longrightarrow C_6H_5COO^- + H^+$

$K_a= \frac{[C_6H_5COO^-][H^+]}{[C_6H_5COOH]}$

$6.3\times 10^{-5}=\frac{[0.08 +x][x]}{[0.05-x]}$

x= $3.9\times 10^{-5}$ = $[H^+]$

now, pH =-log $[H^+]$

pH= 4.408