Question

$\stackrel{⊝}{O}H$ in equilibrium with $Zn\left(OH{)}_2\right(s)$.

• A. $2.6\times {10}^{-5}M$
• B. $1.3\times {10}^{-5}M$
• C. $7.2\times {10}^{-5}M$
• D. $2.3\times {10}^{-5}M$

Answer 1

Answer: (B)

$1.3\times {10}^{-5}M$

$Zn(OH{)}_2⟶Z{n}^{2+}+2\stackrel{⊝}{O}H$

$K_{sp}=\left[Z{n}^{2+}\right][\stackrel{⊝}{O}H{]}^2$

$[\stackrel{⊝}{O}H{]}^2=\frac{1.8\times 106-14}{106-4}=1.8\times {10}^{-10}$

$\left[\stackrel{⊝}{O}H\right]=1.3\times {10}^{-5}M$