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

Create a free account to view solution

View Solution Free
Topic: Ionic Equilibrium·Practice all Ionic Equilibrium questions

More Ionic Equilibrium Questions

Fear or excitement, generally cause one to breathe rapidly and it results in the decrease of concentration of CO2 in blo...A salt of strong acid and a weak base is dissolved in water. Its hydrolysis in solution is...Ka for the acid HA is 1 × 10-6. The value of K for the reaction A- + H3O+ ⇋ HA + H2O is...The pH value of ordinary water is :...Find the number of P - O - P linkage in (HPO3)n where n = 3...