Question

Calculate the pH at the equivalence point during the titration of 0.1 M, 25 ml $C{H}_3$ COOH with 0.05 M NaOH solution $k_a$ $\left(C{H}_{3}COOH\right)=1.8\times {10}^{-5}$

• A. 9.63
• B. 8.63
• C. 10.63
• D. 11.63

Answer 1

The correct option is B 8.63

Since at equivalence point (for acid) $N_1{V}_1=N_2{V}_2$ (for base)
$\therefore$ Volume of NaOH required to reach equivalence point
$=\frac{0.1\times 25}{0.05}=50ml$
$\therefore \text{Concentration of salt formed}=\frac{\text{millimoles of acid}}{\text{total volume in ml}}$
$=\frac{25\times 0.1}{75}=\frac{0.1}{3}$
$Since\left[H^{+}\right]=\sqrt{\frac{Kw\times Ka}{c}}=\sqrt{\frac{{10}^{-14}\times 1.8\times {10}^{-5}}{\frac{0.1}{3}}}=\therefore pH=8.68$