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Solubility Product

15 mL of 0.05...

Question

15 mL of 0.05 M $A g N O_{3}$ is mixed with 45 mL of 0.03 M $K_{2} C r O_{4}$ . Then, which of the following is correct regarding $A g_{2} C r O_{4}$ . $K_{s p}$ of $A g_{2} C r O_{4} = 1.9 \times 10^{- 12} .$

Precipitation occurs

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Precipitation does not occur

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Can't be determined from given data

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None of these

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Solution

The correct option is A Precipitation occurs
Final volume $= 15 + 45 = 60 m L = 0.06 L$

Using these volume finding out the final concentrations:
$c = \frac{m o l e}{v o l u m e i n L}$
$[A g^{+}] = \frac{0.015 \times 0.05}{0.060} = 1.25 \times 10^{- 2} M$

$[C r O_{4}^{-}] = \frac{0.045 \times 0.03}{0.060} = 2.25 \times 10^{- 2} M$

Ionic product for $A g_{2} C r O_{4}$
$A g_{2} C r O_{4} (s) ⇌ 2 A g^{+} (s) + C r O_{4}^{- 2} (s)$

Ionic product $= [A g^{+}]^{2} [C r O_{4}^{- 2}]$
$= (1.25 \times 10^{- 2})^{2} (2.25 \times 10^{- 2})$
$= 3.51 \times 10^{- 6}$

Ionic product $> K_{s p}$
$∴$ precipitation occurs.

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Similar questions

A g N O_{3}

K_{2} C r O_{4}

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A g_{2} C r O_{4} = 1.9 \times 10^{- 12} .

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A g_{2} C r O_{4} = 1.9 \times 10^{- 12} .

K_{s p} o f A g_{2} C r O_{4} i s 1.1 \times 10^{- 12} a t 298 K .

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