How to calculate pH of basic salt solution mixed with strong base?
Exact Solution

I've fouled up this problem badly. Here is a more carefully thought out solution.

So let's start at the beginning with the ICE table. Making the normal assumptions:

Concentrations will be used instead of activities.
The $\ce{Zn(CH3COO)_2}$ dissociates completely in aqueous solution, thus the $\ce{Zn^{2+}}$ and $\ce{CH3COO-}$ ions can be considered separately.
The $\ce{NaOH}$ dissociates completely in aqueous solution, thus the $\ce{Na+}$ and $\ce{OH-}$ ions can be considered separately.
The $\ce{Na+}$ is a spectator ion and is left out of the table.

The problem only calls for the pH of the final solution, but a solution to the whole table will be calculated as a check.
Initial ICE Table
Let's let:

$x = $ the moles of $\ce{Zn(OH)2}$ formed. (The numerical value is the same as the change in molarity of $\ce{Zn^{2+}}$ since there is 1.0 liter of solution.)

$y = $ the molarity of the $\ce{CH3COOH}$.


\begin{array}{c | c c c c c}
               &amp;\ce{[Zn^{2+}]}     &amp; \ce{[CH3COO-]}        &amp; \ce{[CH3COOH]}   &amp; \ce{[OH-]}                         &amp; \ce{Zn(OH)2(s)}  \\\hline
\text{I}  &amp;\pu{0.050000 M}     &amp; \pu{0.100000 M}           &amp; \pu{0.000000 M}               &amp; \pu{  0.025000 M}               &amp; \pu{0 moles}  \\
\text{C} &amp;  -x                           &amp;  -y                               &amp;   y                              &amp; \pu{- 2x + y M}                  &amp; \pu{x moles}  \\
\text{E} &amp;\pu{0.050000 - x M} &amp; \pu{0.100000 - y M}     &amp;   y                              &amp; \pu{0.025000 - 2x + y M}   &amp; \pu{x moles}   \\
\end{array}
Applicable equilibriums
Wikipedia gives the $K_{sp}$ of zinc hydroxide is $K_{sp}=3.0\cdot10^{−17}$. So:
$K_{sp}=3.000\cdot10^{−17} = \ce{[Zn^{2+}][OH-]^2}$
Wikipedia gives the $\pu{pK_b}$ for sodium acetate as 9.25, which yields $K_b = 5.623\cdot10^{−10}$.  So:
$K_b = 5.623\cdot10^{−10} = \dfrac{\ce{[CH3COOH][OH-]}}{\ce{[CH3COO-]}}$
Exact Solution
Starting with the $K_{sp}$ equation for zinc hydroxide and making the appropriate substitutions for the final concentrations:
$K_{sp} = \ce{[Zn^{2+}][OH-]^2}$
$3.000\cdot10^{−17} = (0.050000 - x)\cdot(0.025000 - 2x + y)^2$
Starting with the $K_b$ equation for the acetate anion:
$K_b =  \dfrac{\ce{[CH3COOH][OH-]}}{\ce{[CH3COO-]}}$
$5.623\cdot10^{−10} = \dfrac{(y)\cdot(0.025000 - 2x + y)}{0.100000 - y} = \dfrac{0.025000y - 2xy + y^2}{0.100000 - y}$
Wolfram Mathworld's solution is:

x = 0.0134623
y = 0.00192466

$\ce{[Zn^{+2}]_{equil}} = 0.0500000 - 0.0134623 = 0.0365377$
$\ce{[OH-]_{equil}} = \sqrt{\dfrac{ K_{sp}}{\ce{[Zn^{2+}]_{equil}}}} = \sqrt{\dfrac{3.000\cdot10^{−17}}{0.0365377}} = 2.86543\cdot10^{-8}$
$\pu{pOH} = 7.542810$
$\pu{pH} = 14.000000 - 7.542810 =  6.457190 \ce{-&gt;[Round to 2 SF]} 6.46$
$\ce{[CH3COO-]_{equil}} = 0.100000 - 0.00192466 = 0.09807534$
There are two equations and two unknowns. Solving the equations yielded an exact solution. This is no problem with a computer but a mess even for a good calculator.