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 $200 \,cc$ water of $pH \,3$. 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
$[ HSaC ]=2 \times 10^{-3}, H ^{+}=10^{-3}$
$K_{ a }=\frac{\left[ H ^{+}\right]\left[ SaC ^{-}\right]}{[ HSaC ]}$
$\therefore 2 \times 10^{-12}=\frac{\left[10^{-3}\right]\left[ SaC ^{-}\right]}{\left[2 \times 10^{-3}\right]}$
$\therefore {\left[ SaC ^{-}\right]=4 \times 10^{-12} M }$