How to calculate pH of basic salt solution mixed with strong base? Calculate $\mathrm{ pH }$ of $\pu{ 0.05 mol }$ of $\ce{Zn(CH3COO)2}$ and $\pu{ 0.025  mol }$ of $\ce{ NaOH }$ dissolved in $\pu{ 1 L }$ of distilled water?

I tried to use the systematic method by forming the following equations to extract cubic equation which can  be solved by computer :
\begin{array}{   }
[\ce{ CH3COO- }]_I &amp;= 2\times {0.05} = \pu{ 0.1 M }\\
[\ce{Zn^{2+}}] &amp;=\pu{ 0.05 M }\\
[\ce{Na+}] &amp;=\pu{ 0.025 M }\\
0.1 = [\ce{CH3COO-}]_I &amp;= [\ce{CH3COO-}]_e + [\ce{CH3COOH}]_e\\
2\ce{[Zn^{2+}] + [Na+] +[ H+]_e &amp;= [CH3COO-]_e +[ OH-]_e} \\
K_\mathrm{a} &amp;=\frac{ [\ce{ H+}]_e\cdot{ [\ce{CH3COO-}]_e}}{[\ce{CH3COOH}]_e}\\
[\ce{OH-}]_e &amp;=\frac{ K_\mathrm{w}}{[\ce{H+}]_e}
\end{array}

First question: Are the equations correct?

Second question: Is there a simple approximate method to solve the above problem? How?

I appreciate any help? 



$\large \text{ --------- Maxw Notes ---------}$

Editing the answers with all the markup is painfully slow, so I added different analysis in a number of different answers to this problem. No doubt that will be confusing to someone looking at this problem. So let me summarize the answers.


  Note: I am going to carry "extra" significant figures in the  calculations to gauge the mathematical exactness, but this is of course is ridiculous for the chemistry since the equilibrium constants are not known to that precision. If the true solution values are within 5% of the mathematically calculated values that's really good.


Exact Solution

Setup an ICE table and solved the nasty linear equations with Wolfram Alpha to get an "exact solution." 

Iterative Solution

Setup an ICE table and then simplified the problem to create a "Simplified ICE Table". Went through some head pounding to create an iterative solution. Realized the original ICE table could be used for iterated solution so calculated that solution too. 

Reducing the Problem to a Single Variable

I had tried to solve the problem first with what I am calling the Iterative Solution. I could tell all the species other than $\ce{[Na+]}$ were independently pH controlled, but I couldn't figure out how to iterate the solution. Using the charge balance finally occurred to me so I tabulated the data to show that the approach would work. 

Solution with Cubic Equation

Used charge balance of solution, and one assumption to solve the simplified problem with a cubic equation with $\ce{[OH-]}$ as the variable. Solved the cubic equation with Wolfram Alpha.

Assumption:


$\ce{[CH3COO-]_{equil} \gg [OH-]_{equil}}$ 


Solution with Quadratic Equation

Used charge balance of solution, and two assumptions to solve the simplified problem with a quadratic equation with $\ce{[OH-]}$ as the variable. Solved the quadratic equation with Wolfram Alpha, but is reasonable to assume that solution could have been done by hand. 

Assumptions:


$\ce{[CH3COO-]_{equil} \gg [OH-]_{equil}}$ 
$\ce{[CH3COO-]_{equil} \gg K_b }$
That the normalization tweak works. (I know that it did from the exact solution, but the calculations showed that it worked too.)
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.