The solubility product of Ag2CrO4 is 32× 10-12 . What is the concentration of [CrO4]2- ions in that solution?

Question

The solubility product of Ag2CrO4 is 32× 10-12 . What is the concentration of [CrO4]2- ions in that solution? (A)  8× 10-8mol L-1 (B)  16× 10-4mol L-1 (C)  2× 10-4mol L-1 (D)  8× 10-4mol L-1

Tardigrade Admin · Accepted Answer

First write the dissociation reaction for

A g_{2} C r O_{4}

.
Then find relationship between concentration of ions and solubility product.
Now do the calculation to find the concentration of

[C r O_{4}]^{2 -}

.
Let the solubility of

A g_{2} C r O_{4} = x m o l / L A g_{2} C r O_{4} \to 2 A g^{+} C r O_{4}^{2 -}

ions

2 xx

∴ K_{s p} = [A g^{+}]^{2} [C r O_{4}^{- 2}] = (2 x)^{2} (x)

or

K_{s p} = 4 x^{3}

or

x = 3 \frac{K _{s p}}{4}

Given,

K_{s p} = 3 \frac{32 \times 1 0 ^{- 2}}{4}

= 3 8 \times 1 0^{- 12} = 2 \times 1 0^{- 4} m o l / L

Q. The solubility product of $A g_{2} C r O_{4}$ is $32 \times 1 0^{- 12}$ . What is the concentration of $[C r O_{4}]^{2 -}$ ions in that solution?

$8 \times 1 0^{- 8} m o l L^{- 1}$

$16 \times 1 0^{- 4} m o l L^{- 1}$

$2 \times 1 0^{- 4} m o l L^{- 1}$

$8 \times 1 0^{- 4} m o l L^{- 1}$

Q. The solubility product of Ag2​CrO4​ is 32×10−12 . What is the concentration of [CrO4​]2− ions in that solution?

8×10−8molL−1

16×10−4molL−1

2×10−4molL−1

8×10−4molL−1

Q. The solubility product of $A g_{2} C r O_{4}$ is $32 \times 1 0^{- 12}$ . What is the concentration of $[C r O_{4}]^{2 -}$ ions in that solution?

$8 \times 1 0^{- 8} m o l L^{- 1}$

$16 \times 1 0^{- 4} m o l L^{- 1}$

$2 \times 1 0^{- 4} m o l L^{- 1}$

$8 \times 1 0^{- 4} m o l L^{- 1}$