Question

Saccarin ($K_a=2\times {10}^{-12}$) is a weak acid represented by formula $HSac$. A $4\times {10}^{-4}$ mole amount of saccharin is dissolved in $200{cm}^3$ water of pH$=3$. Assuming no change in volume, calculate the concentration of ${Sac}^{-}$ ions in the resulting solution at equilibrium.

Answer 1

$pH=3$

$\left[H^{+}\right]=0.001M$

Concentration of saccharin$=\frac{4\times {10}^{-4}}{200}\times 1000=2\times {10}^{-3}M$

$\left[H^{+}\right]={10}^{-pH}={10}^{-3}M$

$HSac\rightleftharpoons H^{+}+{Sac}^{-}$

$0.002-x$ $0.001+x$

$K_a=\frac{\left[H^{+}\right]\left[{Sac}^{-}\right]}{\left[HSac\right]}$

$\left[{Sac}^{-}\right]=\frac{2\times {10}^{-12}\times (2\times {10}^{-3}-x)}{{10}^{-3}-x}$

Since, $x$ is very small, it can be neglected.

$\left[{Sac}^{-}\right]=\frac{2\times {10}^{-12}\times (2\times {10}^{-3})}{{10}^{-3}}=4\times {10}^{-12}M$