Question

Saccharin $(K_a=2\times {10}^{-12})$ is a weak acid represented by formula $HSaC$. A $4\times {10}^{-4}$ mole of saccharin is dissolved is 200 mL of water of $Ph=3$, Assuming no change in volume, calculate $\left[Sa{C}^{⊝}\right]$ in the solution at equilibrium.

• A. $4\times {10}^{-12}M$
• B. $4\times {10}^{-8}M$
• C. $4\times {10}^{-10}M$
• D. $4\times {10}^{-6}M$

Answer 1

The correct option is A $4\times {10}^{-12}M$

$\left[HSaC\right]=\frac{Mole}{Litre}=\frac{4\times {10}^{-4}}{200/1000}=2\times {10}^{-3}M$
The dissociation of HSaC takes place in presence of $\left[H^{\oplus }\right]={10}^{-3}M$.
$HSaC\rightleftharpoons H^{\oplus }+Sa{C}^{⊝}$
[Concentration before dissociation] $2\times {10}^{-3}$ ${10}^{-3}$ 0
In presence of $H^{\oplus }$, the dissociation of $HSaC$ is almost negligible because of the common ion effect. Thus, at equilibrium
$\left[HSaC\right]=2\times {10}^{-3};\left[H^{\oplus }\right]={10}^{-3}$
$\because K_a=\frac{\left[H^{\oplus }\right]\left[Sa{C}^{⊝}\right]}{\left[HSaC\right]}$ $\therefore 2\times {10}^{-12}=\frac{\left[{10}^{-3}\right]\left[Sa{C}^{-}\right]}{2\times 106-3}$
$\therefore \left[Sa{C}^{⊝}\right]=4\times {10}^{-12}M$