Question

Calculate the equilibrium constants of each of the indicated species necessary to reduce an inital 0.2 M $Z{n}^{2+}$ solution to $1.0\times {10}^{-4}Z{n}^{2+}$.

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

• A. $1.3\times {10}^{-5}M$
• B. $1.3\times {10}^{-3}M$
• C. $1.3\times {10}^{-7}M$
• D. none of these

Answer 1

The correct option is A $1.3\times {10}^{-5}M$

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

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

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

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