Ionic EquilibriumHard

Question

Saccharin (Ka = 2 × 10–12) is a weak acid represented by the formula HSac. A 4 × 10–4 mole amount of saccharin is dissolved in 200 ml water of pH 3.0. Assuming no change in volume, the concentration of Sac− ions in the resulting solution at equilibrium is

Options

A.4 × 10–12 M
B.2 × 10–12 M
C.8 × 10–13 M
D.6.32 × 10–8 M

Solution

$\lbrack HSaC\rbrack_{o} = \frac{4 \times 10^{- 4}}{200/1000} = 2 \times 10^{- 3}\text{ M}$

and $P^{H} = 3.0 \Rightarrow \left\lbrack H^{+} \right\rbrack = 10^{- 3}\text{ M}$

Now, $2 \times 10^{- 12} = \frac{\left\lbrack SaC^{-} \right\rbrack \times 10^{- 3}}{2 \times 10^{- 3}} \Rightarrow \left\lbrack SaC^{-} \right\rbrack = 4 \times 10^{- 12}\text{ M}$

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