Question

chromate is sparingly soluble in aqueous solutions. The Ksp of Ag2CrO4 is 1.12×10−12 . What is the solubility (in mol/L) of silver chromate in 1.00 M potassium chromate aqueous solution?

Answer 1

Answer:

$6.5X10^-^5~\frac{mol}{L}$

Explanation:

For this question, we have to start with the ionization equation for $Ag_2CrO_4_(_s_)$, so:

$Ag_2CrO_4_(_s_)~~2Ag^+~_(_a_q_)~+~CrO_4^-^2_(_a_q_)$

With this in mind we can write the Ksp expression:

$Kps~=~[Ag^+]^2[CrO_4^-^2]$

Additionally, for every mole of $CrO_4^-^2$  formed, 2 moles of $Ag^+$ are formed. We can use "X" for the unknown concentration of each ion, so:

$[CrO_4^-^2]~=~X$  and  $[Ag^+]~=~2X$

Now, we can plug the values into the Ksp expression:

$1.12x10^-^1^2~=~(2X)^2(X)$

Now we can solve for "X" :

$1.12x10^-^1^2~=~4X^3$

$X^3=\frac{1.12X10^-^1^2~}{4}$

$X=(\frac{1.12X10^-^1^2~}{4})^(^1^/^3^)$

$X=6.5X10^-^5~\frac{mol}{L}$

I hope it helps!