Question

$40$ ml of $\frac{M}{4}$ solution of a weak monobasic acid (K${}_a$ = 2 x 10${}^{-12}$ at ${25}^{0}C$) is treated with $\frac{M}{12}$ sodium hydroxide solution. The concentration of H${}^{+}$ ions at the equivalence point is :

• A. $1.41\times {10}^{-12}$ M
• B. $\sqrt{5}\times {10}^{-13}$ M
• C. $4\sqrt{2}\times {10}^{-13}$ M
• D. $2\sqrt{5}\times {10}^{-13}$ M

Answer 1

Answer: (A)

$1.41\times {10}^{-12}$ M

• 40 ml of $\frac{M}{4}$ solution corresponds to $\frac{40\times \frac{M}{4}}{1000}=0.01$ moles.
  

Thus 0.01 moles of sodium hydroxide solution is required for the equivalence point.

This corresponds to 0.12 L or 120 mL ($V=\frac{n}{M}=0.01\times 12=0.12L$) of sodium hydroxide solution.

Thus at the equivalence point, the concentration of salt will be 0.01 M.

For a salt of weak acid with strong base, the expression for the hydrogen ion concentration is $\left[H^{+}\right]=\sqrt{\frac{k_w\times k_a}{c}}$.

Substitute values in the above expression.

$\left[H^{+}\right]=\sqrt{\frac{{10}^{-14}\times 2\times {10}^{-12}}{0.01}}=1.41\times {10}^{-12}$.