Question

The Ksp for a sparingly soluble Ag2Cro4, is
4 * 10-12. The molar solubility of the salt is
(1) 2.0 10-6 mol L-1 (2) 1.0 * 10-4 mol L-1
(3) 2.0 * 10-12 mol L-1 (4) 1.0 * 10-15 mol L

Answer 1

Answer:

2) 1.0* 10^-4

Explanation:As mentioned in the attachment.

Solubility is cube root of ksp/4.

See the attached image solution carefully and proceed accordingly.

Answer 2

Answer:

2.$1.0\times 10^{-4} molL^{-1}$

Explanation:

We are given that $K_{sp}$ for  a sparingly soluble  $Ag_2CrO_4=4\times 10^{-12}$

We have to find the molar solubility of the salt

Let x be the molar  solubility of salt

$Ag_2CrO_4\rightleftharpoons  2Ag^+ +CrO_4^{-2}$

                                                            2x          x

$K_{sp}=[Ag^+]^2[CrO_4^{-2}]$

Substitute the values then we get

$4\times 10^{-12}=(2x)^2(x)$

$4\times 10^{-12}=4x^3$

$x^3=\frac{4\times 10^{-12}}{4}$

$x=\sqrt[3]{10^{-12}}$

$x=10^{-4}$

Hence, the molar solubility of salt=$10^{-4}molL^{-1}$

Answer:2.$1.0\times 10^{-4} molL^{-1}$