In a saturated solution of the sparingly soluble strong electrolyte $\mathrm{AgIO}_{3}$ (Molecular mass $=283$) the equilibrium which sets in is $\mathrm{AgIO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{Ag}^{+}_{(\mathrm{aq})}+\mathrm{IO}_{3(\mathrm{aq})}^{-}$ If the solubility product constant $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{AgIO}_{3}$ at a given temperature is $1.0 \times 10^{-8}$, what is the mass of $\mathrm{AgIO}_{3}$ contained in 100 ml of its saturated solution?

(B) $2.83 \times 10^{-3} \mathrm{~g}$

(A) $28.3 \times 10^{-2} \mathrm{~g}$

(C) $1.0 \times 10^{-7} \mathrm{~g}$

(D) $1.0 \times 10^{-4} \mathrm{~g}$

JEE Main · 2007 · 29 Apr 2007 · Offline · Equilibrium

In a saturated solution of the sparingly soluble strong electrolyte

AgIO_{3}

(Molecular mass

= 283

) the equilibrium which sets in is

AgIO_{3 (s)} ⇌ Ag_{(aq)}^{+} + IO_{3 (aq)}^{-}

If the solubility product constant

K_{sp}

AgIO_{3}

at a given temperature is

1.0 \times 1 0^{- 8}

, what is the mass of

AgIO_{3}

contained in 100 ml of its saturated solution?

Answer: (B)

For this

1 : 1

salt the solubility

s

satisfies

K_{s p} = [A g^{+}] [I O_{3}^{-}] = s^{2}

, so

s = 1.0 \times 1 0^{- 8} = 1 0^{- 4}

mole per litre. In

100

mL this amounts to

1 0^{- 4} \times 0.1 = 1 0^{- 5}

mole, and multiplying by the molar mass

283

gives

2.83 \times 1 0^{- 3}

gram of silver iodate.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option