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Solubility Product

The solubilit...

Question

The solubility product of $A g_{2} C r O_{4}$ is $1.9 X 10^{- 12}$ .The volume of water in mL that can dissolve 4mg of $A g_{2} C r O_{4}$ is about:

150 mL

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1000 mL

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250 mL

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500 mL

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Solution

The correct option is A 150 mL
$A g_{2} C r O_{4} ⇌ 2 A g^{+} + C r O_{4}^{2 -}$
Let S mol/L be the solubility of $A g_{2} C r O_{4}$
$[A g_{2} C r O_{4}] = [C r O_{4}^{2 -}] = S$
$[A g^{+}] = 2 S$
$K_{s p} = [A g^{+}]^{2} [C r O_{4}^{2 -}]$
$1.9 \times 10^{- 12} = (2 S)^{2} \times S = 4 S^{3}$
$S = 7.8 \times 10^{- 5} M$
The molar mass of $A g_{2} C r O_{4} = 331.7 g / m o l$
Convert solubility of $A g_{2} C r O_{4}$ from mol/L to g/L.
$S = 7.8 \times 10^{- 5} m o l / L \times 331.7 g / m o l$
$S = 0.0259 g / L$
Convert solubility of $A g_{2} C r O_{4}$ from g/L to mg/mL.
$S = 0.0259 g / L \times \frac{1000 m g}{1 g} \times \frac{1 L}{1000 m L}$
$S = 0.0259 m g / m L$
To dissolve 4 mg of $A g_{2} C r O_{4}$ , volume of water required will be $4 m g \times \frac{1 m L}{0.0259 m g} = 154 m L ≃ 150 m L$ .

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