# Calculate the solubility of silver chromate, Ag_2CrO_4, in a solution that is 0.0050 M in...

## Question:

Calculate the solubility of silver chromate, {eq}\displaystyle \rm Ag_2CrO_4 {/eq}, in a solution that is {eq}\displaystyle \rm 0.0050 \ M {/eq} in potassium nitrate.

## Solubility Product:

The solubility product is a type of an equilibrium for the ionization of an ionic compound in a solution. It is expressed as the product of the concentration of the ions formed in the solution where the concentration of the ions are raised to the power of the stoichiometric coefficient of the ions in the balanced equation.

We are told to find the solubility of silver chromate in a solution that is 0.0050 M in potassium nitrate.

For this we require the solubility product of silver chromate.

{eq}\rm K_{sp}(Ag_2CrO_4) = 1.2\times10^{-12} {/eq}

Initially the concentration of silver and chromate ions is zero in the solution.

The concentration of the ions in the solution at equilibrium can be determined as follows:

{eq}\rm Ag_2CrO_4 \rightleftharpoons Ag^+ + CrO_4^{2-} {/eq}

Balanced equation: {eq}\rm Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-} {/eq}

{eq}\rm K_{sp} = [Ag^+]^2[CrO_4^{2-} ] \\ [CrO_4^{2-} ] = s \\ [Ag^+] = 2s \\ 1.2\times10^{-12} = (2s)^2(s) \\ 1.2\times10^{-12} = 4s^3 \\ s^3 = 0.3\times10^{-12} \\ \boxed{\rm s = 6.69\times10^{-5}\ M} {/eq}