Question

For $Cu(OH{)}_2,K_{sp}=1.6\times {10}^{-19}$, the molar solubility of $Cu(OH{)}_2$ is:

• A. $3.4\times {10}^{-7}$M
• B. $6.4\times {10}^{-7}$M
• C. $2.7\times {10}^{-11}$M
• D. $5.1\times {10}^{-10}$M

Answer 1

The correct option is A $3.4\times {10}^{-7}$M

For $Cu(OH{)}_2$, the expression of solubility product is:

$K_{sp}=\left[C{u}^{2+}\right][O{H}^{-}{]}^2=(s\left)\right(2s{)}^2=4s^3$

Substitute $K_{sp}=1.6\times {10}^{-19}$ in the above expression:

$1.6\times {10}^{-19}=4s^3$

$s=3.4\times {10}^{-7}$

Hence, option A is correct.