In a saturated solution of the sparingly soluble strong electrolyte ${AgIO}_{3}$ (molecular mass = 283) the equilibrium which sets is ${AgIO}_{3} (s) ⇌ {Ag}^{+} (aq) + {IO}_{3}^{-} (aq)$ If the solubility product constant $K_{sp} of {AgIO}_{3}$ at a given temperature $1 .0 \times 10^{- 8}$ what is the mass of ${AgIO}_{3}$ contained in 100 ml of its saturated solution?

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$1 .0 \times 10^{- 4} g$

$28 .3 \times 10^{- 2} g$

$2 .83 \times 10^{- 3} g$

$1 .0 \times 10^{- 7} g$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let s = solubility
${AgIO}_{3} ⇌ {Ag}^{+} + {IO}_{3}^{-} K_{sp} = [{Ag}^{+}] [{IO}_{3}^{-}] = s \times s = s^{2} Given K_{sp} = 1 \times 10^{- 8} ∴ s = \sqrt{K_{sp}} = \sqrt{1 \times 10^{- 8}} = 1 .0 \times 10^{- 4} mol / lit = 1 .0 \times 10^{- 4} \times 283 g / lit (∵ Molecular mass of Ag {IO}_{3} = 283) = \frac{1 .0 \times 10^{- 4} \times 283 \times 100}{1000} gm / 100 ml = 2 .83 \times 10^{- 3} gm / 100 ml$