Question

Saccharin $(K_a=2\times {10}^{-12})$ is a weak acid represented by formula HSaC. $4\times {10}^{-4}$ moles of saccharin are dissolved in 200 cc buffer of pH 3. Assuming no change in volume, calculate the concentration of $sa{C}^{-1}$ ions in the resulting solution at equilibrium.
(Assume jpH of buffer is maintained)

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Concentration of saccharin $=\frac{4\times {10}^{-4}}{200}\times 1000=2\times {10}^{-3}M$
$\left[H^{+}\right]={10}^{-pH}={10}^{-3}M$
$HSaC\rightleftharpoons H^{+}+SaC$
$2\times {10}^{-3}-x$ ${10}^{-3}$ x
$K_a=\frac{\left[H^{+}\right]\left[Sa{C}^{-}\right]}{\left[HSaC\right]}=\frac{\left({10}^{-3}\right)\left(x\right)}{(2\times {10}^{-3}-x)}$
$\left[X\right]=\frac{2\times {10}^{-12}\times (2\times {10}^{-3}-x)}{{10}^{-3}}$
$\Rightarrow x=4\times {10}^{-12}-2\times {10}^{-9}x$
$\Rightarrow x\cong 4\times {10}^{-12}M$