Question

What is the concentration of the silver ion in silver chromate, Ag₂CrO₄, if its solubility product constant (Kₛₚ) is 1.2 x 10⁻¹². Hint: write the equation first! *

2 points

1.4 x 10⁻⁵

1.1 x 10⁻⁹

1.6 x 10⁻¹²

2.4 x 10⁻¹²

Answer 1

`[Ag^+]=1.3x10^(-4)M`

Step-by-step explanation:

Hello,

In this case, given the solubility product of silver chromate:

`Ag_2CrO_4(S)\rightleftharpoons 2Ag^+(aq)+CrO_4^(-2)(aq)`

Now, the law of mass action excluding silver chromate as it is solid, turns out:

`Ksp=[Ag]^2[CrO_4^(-2)]`

Thus, given the change
`x` due to the dissolution of silver chromate, we obtain:

`1.2x10^(-12)=(2x)^2* x`

Hence, we solve for
`x`:

`x=\sqrt[3]{(1.2x10^(-12))/(2^2) } = 6.7x10^(-5)M`

Thus, the concentration of silver ions will be:

`[Ag^+]=2x=2*6.7x10^(-5)M=1.3x10^(-4)M`

Best regards.