Question

Silver chromate, Ag.CrO, has a Kp of 9.00x 1012 Calculate the solubility in mol/L of silver chromate a) 1.31 x 10M d) 2.08 x 10 M c) 2.25 x 10-12M b) 1.65 x 10" M e) 1.50 x 106*M lsa for evothermic reactions and a value for

Answer 1

Answer:  $1.31\times 10^4moles/L$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of the silver chromate is given as:

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

We are given:

Solubility of $Ag_2CrO_4$ = S mol/L

By stoichiometry of the reaction:

1 mole of $Ag_2CrO_4$ gives 2 moles of $Ag^{+}$ and 1 mole of $CrO_4^{2-}$.

When the solubility of $Ag_2CrO_4$ is S moles/liter, then the solubility of $Ag^{+}$ will be 2S moles\liter and solubility of $CrO_4^{2-}$ will be S moles/liter.

Expression for the equilibrium constant of $Ag_2CrO_4$ will be:

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

$9.00\times 10^{12}=[2s]^2[s]=4s^3$

$s=1.31\times 10^4M$

Hence, the solubility of  $Ag_2CrO_4$ is $1.31\times 10^4moles/L$