Answer
Since $Q_{sp} \gt K_{sp}$, there will be precipitation of zinc hydroxide.
Work Step by Step
1. Calculate the molar mass $(NaOH)$:
22.99* 1 + 16* 1 + 1.008* 1 = 39.998g/mol
2. Calculate the number of moles $(NaOH)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.0040}{ 39.998}$
$n(moles) = 1.0 \times 10^{- 4}$
3. Find the concentration in mol/L $(NaOH)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 1.0\times 10^{- 4}}{ 0.01} $
$C(mol/L) = 0.010$
Since $NaOH$ is 100% ionizated in water, the concentration of $[OH^-]$ is equal to $0.010M$ too.
4. Write the $K_{sp}$ expression:
$ Zn(OH)_2(s) \lt -- \gt 1Zn^{2+}(aq) + 2OH^{-}(aq)$
$ K_{sp} = [Zn^{2+}]^ 1[OH^{-}]^ 2$
5. Find the $Q_{sp}$ value
$ Q_{sp} = (1.63 \times 10^{-4})^ 1 \times (0.010)^ 2$
$ Q_{sp} = 1.63 \times 10^{-4} \times (1 \times 10^{-4})$
$ Q_{sp} = 1.63 \times 10^{-8}$
$K_{sp} (Zn(OH)_2) = 3 \times 10^{-17}$
Since $Q_{sp} \gt K_{sp}$, there will be precipitation
