Question

If the concentration of $Cr{O}_4^{2-}$ ions in a saturated solution of silver chromate is $2\times {10}^{-4}$; solubility product of silver chromate will be:

• A. $4\times {10}^{-8}$
• B. $8\times {10}^{-12}$
• C. $16\times {10}^{-12}$
• D. $32\times {10}^{-12}$

Answer 1

Answer: (D)

$32\times {10}^{-12}$

The equilibrium is:
${Ag}_{2}Cr{O}_4\rightleftharpoons 2{Ag}^{+}+Cr{O}_4^{2-}$

On the basis of this equation, the concentration of $Cr{O}_4^{2-}$ ion will be half of the concentration of ${Ag}^{+}$ ions.

Thus, $\left[{Ag}^{+}\right]=4\times {10}^{-4}M$

and $\left[Cr{O}_4^{2-}\right]=2\times {10}^{-4}M$

$K_{sp}={\left[A{g}^{+}\right]}^2\left[Cr{O}_4^{2-}\right]=32\times {10}^{-12}$