Question

Saccharin $\left(K_a=2\times 10^{-12}\right)$ is a weak acid respectively by formula $HSaC$. A $4\times 10^{-4}mol$ of saccharin is dissolved in $200cc$ water of $pH3$. Assuming no change in volume. Calculate the concentration of saccharin ions in the resulting solution at equilibrium.

• A. $2\times 10^{-12}M$
• B. $3\times 10^{-12}M$
• C. $4\times 10^{-12}M$
• D. $5\times 10^{-12}M$

Answer 1

Answer: (C)

$[HSaC]=\frac{4\times 10^{-4}}{200/1000}=2\times 10^{-3}M$
The dissociation of $HSaC$ takes places in the presence of $\left[H^{+}\right]$ $=10^{-3}$

Conc. before $2\times 10^{-3}10^{-3}0$ dissociation
In the presence of $H^{+}$, the dissociation of $HSaC$ is almost negligible because of common ion effect. Thus at equilibrium
$[HSaC]=2\times 10^{-3},H^{+}=10^{-3}$
$K_a=\frac{\left[H^{+}\right]\left[Sa{C}^{-}\right]}{[HSaC]}$
$\therefore 2\times 10^{-12}=\frac{\left[10^{-3}\right]\left[Sa{C}^{-}\right]}{\left[2\times 10^{-3}\right]}$
$\therefore \left[Sa{C}^{-}\right]=4\times 10^{-12}M$