# 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 : $$\begin{array}{ } [\ce{ CH3COO- }]_I &= 2\times {0.05} = \pu{ 0.1 M }\\ [\ce{Zn^{2+}}] &=\pu{ 0.05 M }\\ [\ce{Na+}] &=\pu{ 0.025 M }\\ 0.1 = [\ce{CH3COO-}]_I &= [\ce{CH3COO-}]_e + [\ce{CH3COOH}]_e\\ 2\ce{[Zn^{2+}] + [Na+] +[ H+]_e &= [CH3COO-]_e +[ OH-]_e} \\ K_\mathrm{a} &=\frac{ [\ce{ H+}]_e\cdot{ [\ce{CH3COO-}]_e}}{[\ce{CH3COOH}]_e}\\ [\ce{OH-}]_e &=\frac{ K_\mathrm{w}}{[\ce{H+}]_e} \end{array}$$

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:

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

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:

1. $$\ce{[CH3COO-]_{equil} \gg [OH-]_{equil}}$$
2. $$\ce{[CH3COO-]_{equil} \gg K_b }$$
3. That the normalization tweak works. (I know that it did from the exact solution, but the calculations showed that it worked too.)
• You algebraic equations look fine. My main concern is that whoever wrote this question for you does not know basic chemistry. Zinc acetate will react with NaOH, it will first form a hydroxide, that hydroxide will dissolve in base forming zincate ions. I don't know what will be the final pH. These equilibrium calculations assume no side reaction. – M. Farooq Apr 22 at 0:37
• Yes if you know chemistry the problem is pretty simple. Zinc hydroxide has a $K_{sp}$ of $3.0\cdot10^{−17}$. – MaxW Apr 22 at 0:38
• There isn't enough base to form an appreciable amount of the zincate anion ($\ce{Zn(OH)4^{2-}}$. It is reasonable to assume that $\ce{Zn(OH)2}$ forms quantitatively. // An acetate solution would have a pH somewhere near 9 with no $\ce{CO2}$, somewhere between 7 and 8 with $\ce{CO2}$. // So how much chemistry are you supposed to know? – MaxW Apr 22 at 1:26
• @MaxW thanks. you neglect the $\ce{Zn(OH)2}$ contribution,and consider $[\ce{CH3COO-}]=\pu{ 0.05 M}$ in calculating $\mathrm{pH}=9$ – Adnan AL-Amleh Apr 22 at 2:06
• @MaxW Correct my solution.thanks in advance. – Adnan AL-Amleh Apr 23 at 1:57

# 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 $$\ce{CH3COOH}$$.

$$\begin{array}{c | c c c c c} &\ce{[Zn^{2+}]} & \ce{[CH3COO-]} & \ce{[CH3COOH]} & \ce{[OH-]} & \ce{Zn(OH)2(s)} \\\hline \text{I} &\pu{0.050000 M} & \pu{0.100000 M} & \pu{0.000000 M} & \pu{ 0.025000 M} & \pu{0 moles} \\ \text{C} & -x & -y & y & \pu{- 2x + y M} & \pu{x moles} \\ \text{E} &\pu{0.050000 - x M} & \pu{0.100000 - y M} & y & \pu{0.025000 - 2x + y M} & \pu{x moles} \\ \end{array}$$

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:

$$K_b = \dfrac{\ce{[CH3COOH][OH-]}}{\ce{[CH3COO-]}}$$

$$5.623\cdot10^{−10} = \dfrac{(y)\cdot(0.025000 - 2x + y)}{0.100000 - y} = \dfrac{0.025000y - 2xy + y^2}{0.100000 - y}$$

• x = 0.0134623
• y = 0.00192466

$$\ce{[Zn^{+2}]_{equil}} = 0.0500000 - 0.0134623 = 0.0365377$$

$$\ce{[OH-]_{equil}} = \sqrt{\dfrac{ K_{sp}}{\ce{[Zn^{2+}]_{equil}}}} = \sqrt{\dfrac{3.000\cdot10^{−17}}{0.0365377}} = 2.86543\cdot10^{-8}$$

$$\pu{pOH} = 7.542810$$

$$\pu{pH} = 14.000000 - 7.542810 = 6.457190 \ce{->[Round to 2 SF]} 6.46$$

$$\ce{[CH3COO-]_{equil}} = 0.100000 - 0.00192466 = 0.09807534$$

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.

Taking into my consideration to the above comment, I solved the problem. I used the following table to determine the amount of each species at the end of the reaction: $$\begin{array}{c | c c c c c c} \text{RXN} &\ce{Zn(CH3COOH)2}_\text{(aq)} & + &\ce{2NaOH}_\text{(aq)} &\ce{->}& \ce{2CH3COONa}_\text{(aq)} &+& \ce{Zn(OH)2}_\mathrm{(s)} \\\hline \text{I} &(\pu{0.05 mol}) & &(\pu{0.025 mol}) & &(\pu{0 mol}) &&(\pu{0 mol}) \\ \text{C} &(\pu{\frac{-0.025}{ 2} mol }) & &(\pu{-0.025 mol}) & &(\pu{0.025 mol}) &&(\pu{0.0125 mol})\\ \text{F} &(0.05-0.0125)=\pu{0.0375 mol} & &(\pu{0 mol}) & &(\pu{0.025 mol})&&(\pu{0.0125 mol}) \end{array}$$ Ignoring the contrbition of the insoluble $$\ce{Zn(OH)2}$$ ,and taking into account the ionization of both $$\ce{Zn(CH3COO)2}$$ and $$\ce{CH3COONa}$$ ,so: $$[\ce{CH3COO-}]_\mathrm{I}= 2\times{0.0375} + 0.025 = \pu{0.1 M }$$. The acetate ion reacts with water as the following : $$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$ so, the equilibrium expression as the following : $$K_\mathrm{b} =\frac{K_\mathrm{w}}{K_\mathrm{a}}=\frac{[\ce{CH3COOH}]\cdot{[\ce{OH-}]}}{[\ce{CH3COO-}]}$$ Ignoring the autoionization of water,so : $$[\ce{OH-}] = [\ce{CH3COOH}]$$ As $$K_\mathrm{b}$$ of acetate ion equal $$5.55\cdot{10^{-10}}$$ it is aweak base, so ,we assume : $$[\ce{ CH3COO-}]_\mathrm{e} =[\ce{ CH3COO- }]_\mathrm{I}=\pu{ 0.1 M }$$ Substitute in the above equilibrium expression to find :$$[\ce{ OH- }]=\pu{ 7.45\cdot{ 10^{ -6 } } M}\quad\to\quad \mathrm{pOH}= 5.13 \quad\to\quad \mathrm{pH}\approx{8.9}$$

• @MaxWF: final(at the end of the reaction). The amount of reacted" zinc acetate" $=\pu{0.0125 mol}$, and $\pu{ 0.0375 mol }$ ionized in solution. May I have a misconception that $\pu{0.0125 mol}$ of zinc was quantitatively ppt' ed as $\ce{Zn(OH)2}$. – Adnan AL-Amleh Apr 23 at 20:14
• @MaxW really, thanks to your interest and for your fruitful knowledge.The above problem for the group of a chemistry teacher. – Adnan AL-Amleh Apr 23 at 20:27

# Reducing the Problem to a Single Variable

$$\ce{Na+}$$ is a spectator ion and a constant, but $$\ce{[OH-]}$$, $$\ce{[CH3COO-]}$$, and $$\ce{[Zn^{2+}]}$$ are all pH dependent. We can define the charge balance, $$\pu{C.B.}$$ as:

\begin{align} \pu{C.B.} &= \pu{cations - anions} \\ &= \ce{([Na+] + 2*[Zn^{2+}]) - ([OH-] + [CH3COO-])} \\ \end{align}

If there are too much cationic charge then the C.B value is positive, and if there are too much anionic charge the C.B value is negative. This will allow a one-dimensional search of pH vs C.B looking for pH value where C.B. = 0.

We can simplify the CB equation by noting that $$\ce{[CH3COO-] \gg [OH-]}$$ so:

$$\pu{C.B.} \approx \ce{([Na+] + 2*[Zn^{2+}]) - ([CH3COO-])}$$

Just to show that this works the values were calculated in a spreadsheet. Showing the results to 6 decimal places gives:

$$\begin{array}{c | c | c | c | c| c |} \pu{pH} &\ce{[OH-]} & \ce{[CH3COO-]} & \ce{[Na+]} & \ce{[Zn^{2+}]} & C.B \\\hline 5.0000 & 0.000000 & 0.064008 & 0.025000 & 30.000000 & 59.960992 \\\hline 5.2000 & 0.000000 & 0.073812 & 0.025000 & 11.943215 & 23.837618 \\\hline 5.4000 & 0.000000 & 0.081709 & 0.025000 & 4.754680 & 9.452650 \\\hline 5.6000 & 0.000000 & 0.087624 & 0.025000 & 1.892872 & 3.723120 \\\hline 5.8000 & 0.000000 & 0.091817 & 0.025000 & 0.753566 & 1.440314 \\\hline 6.0000 & 0.000000 & 0.094676 & 0.025000 & 0.300000 & 0.530324 \\\hline 6.2000 & 0.000000 & 0.096574 & 0.025000 & 0.119432 & 0.167291 \\\hline 6.4000 & 0.000000 & 0.097810 & 0.025000 & 0.047547 & 0.022283 \\\hline 6.6000 & 0.000000 & 0.098607 & 0.025000 & 0.018929 & -0.035750 \\\hline 6.8000 & 0.000000 & 0.099117 & 0.025000 & 0.007536 & -0.059045 \\\hline 7.0000 & 0.000000 & 0.099441 & 0.025000 & 0.003000 & -0.068441 \\\hline 7.2000 & 0.000000 & 0.099646 & 0.025000 & 0.001194 & -0.072258 \\\hline 7.4000 & 0.000000 & 0.099777 & 0.025000 & 0.000475 & -0.073826 \\\hline 7.6000 & 0.000000 & 0.099859 & 0.025000 & 0.000189 & -0.074481 \\\hline 7.8000 & 0.000001 & 0.099911 & 0.025000 & 0.000075 & -0.074761 \\\hline 8.0000 & 0.000001 & 0.099944 & 0.025000 & 0.000030 & -0.074885 \\\hline 8.2000 & 0.000002 & 0.099965 & 0.025000 & 0.000012 & -0.074942 \\\hline 8.4000 & 0.000003 & 0.099978 & 0.025000 & 0.000005 & -0.074971 \\\hline 8.6000 & 0.000004 & 0.099986 & 0.025000 & 0.000002 & -0.074986 \\\hline 8.8000 & 0.000006 & 0.099991 & 0.025000 & 0.000001 & -0.074996 \\\hline 9.0000 & 0.000010 & 0.099994 & 0.025000 & 0.000000 & -0.075004 \\\hline \end{array}$$

Using very conservative starting guesses of $$\ce{[OH-]}$$ $$1\cdot10^{-9}$$ and $$1\cdot10^{-5}$$ (pH 5 and 9), for the search of the sign change of C.B., the final solution could be obtained (somewhat painfully...). Rather than a binary search, it would be worth the extra effort to interpolate the subsequent guesses. So using a calculator or four place log tables you could get an overly precise value for the final pH.

• Thanks for your powerful effort and the tangible knowledge – Adnan AL-Amleh Apr 27 at 20:50

# Solution with Cubic Equation

As noted in a previous answer, $$\ce{[OH-]}$$, $$\ce{[CH3COO-]}$$, and $$\ce{[Zn^{+2}]}$$ are all pH dependent. So we can define the charge balance, $$\pu{C.B.}$$ as:

$$\pu{C.B.} = \pu{cations - anions} = \ce{([Na+] + 2*[Zn^{2+}]) - ([OH-] + [CH3COO-])}$$

Obviously in the equilibrium solution the C.B. = 0. Thus:

$$0 = \ce{([Na+] + 2*[Zn^{2+}]) - ([OH-] + [CH3COO-])}$$

or

$$\ce{[Na+] + 2*[Zn^{2+}] = [OH-] + [CH3COO-]}$$

But the equilibrium is going to be "near" neutral so $$\ce{[CH3COO-] \gg [OH-]}$$ and thus the $$\ce{[OH-]}$$ term can be neglected.

$$\ce{[Na+] + 2*[Zn^{2+}] \approx [CH3COO-]}$$

rearranging

$$\ce{2*[Zn^{2+}] \approx [CH3COO-] - [Na+]}$$

But:

$$\ce{[Zn]_{equil} = \dfrac{K_{sp}}{\ce{[OH-]^2_{equil}}}}$$

$$\pu{Fraction(\ce{CH3COO-})} = \dfrac{\ce{[OH-]}}{K_b + \ce{[OH-]}}$$

$$\ce{[CH3COO-]_{equil}} = \ce{[CH3COO-]_{init}}\cdot\dfrac{\ce{[OH-]_{equil}}}{\ce{K_b + [OH-]_{equil}}}$$

$$\ce{[Na+] = 0.025}$$

making the substitutions:

$$2\cdot\dfrac{K_{sp}}{\ce{[OH-]^2_{equil}}} = 0.1\cdot\dfrac{\ce{[OH-]_{equil}}}{\ce{K_b + [OH-]_{equil}}} - 0.025$$

working through the factoring yields

$$\ce{0 = 0.0375[OH-]^3 - 0.0125\cdot K_b\cdot[OH-]^2 - K_{sp}\cdot[OH-] - K_b\cdot K_{sp}}$$

Solving with Wolffram Mathword's again: $$\ce{[OH-] = 2.86543\cdot10^{-8}}$$

Sometimes when solving cubic equations in chemistry you know about what the unknown is, and you can simplify by eliminating the cubic term. But in this case that is clearly not possible since the cubic term is the only positive one, and $$\ce{OH-}$$ must be real and positive. Given the Wolffram answer, the terms evaluate to $$\pu{8.823e-25, - 0.058e-25, -8.596e-25,}$$ and $$\pu{-0.169e-25}$$.

Note that the cubic equation above only assumes that $$\ce{[CH3COO-] \gg [OH-]}$$.

# Finally !!!

I've never known a text book problem to require more than a quadratic equation. I finally figured out how to do that.

So the notion for the second ICE table is that to 1L decarbonated distilled water there is added 0.037500 moles $$\ce{Zn(CH3COO)2}$$ and 0.025000 moles $$\ce{Na(CH3COO)}$$.

Simplified ICE Table

Let's let $$y = \ce{[CH3COOH]_{equil}}$$

Assume:

• $$\ce{[CH3COOH]_{equil} \gg 1\cdot10^{-7}}$$

Thus the autoionization of water will contribute a negligible quantity of $$\ce{OH-}$$ which is precipitated by the zinc.

• Obviously each molecule of $$\ce{CH3COOH}$$ must create an anion of $$\ce{OH−}$$ in the solution. The precipitation $$\ce{Zn{OH}_2}$$ removes two $$\ce{OH−}$$ anions from the solution.

Thus $$\Delta\ce{[Zn^{2+}]} = \frac{1}{2}\cdot y$$

$$\begin{array}{c | c c c c c} &\ce{[Zn^2+]} & \ce{[CH3COO-]} & \ce{[CH3COOH]} & \ce{[OH-]'} & \ce{Zn(OH)'_2(s)} \\\hline \text{I} &\pu{0.037500 M} & \pu{0.100000 M} & \pu{0.000000 M} & &\pu{0 moles} \\ \text{C} & \pu{-y/2} & \pu{-y} & \pu{y} & & y/2 \ \pu{moles}\\ \text{E} & \pu{0.037500 - y/2 M} & \pu{0.100000 - y M} & \pu{y} & & y/2 \ \pu{moles} \\ \end{array}$$

The fraction of the acetate species which is acetic acid is

$$\large f( \normalsize \ce{CH3COOH}\large ) \normalsize \ = \ \dfrac{\ce{K_b}}{\ce{K_b + [OH-]_{equil}}}$$

and the equilibrium concentration of $$\ce{CH3COOH}$$ is going to be:

\begin{align} \ce{[CH3COOH]_{equil} &= [CH3COO-]_{init}\cdot} \large f( \normalsize \ce{CH3COOH}\large ) \normalsize \\ &= 0.1000000\cdot \dfrac{\ce{K_b}}{\ce{K_b + [OH-]_{equil}}} \\ \end{align}

We know that $$\ce{[Zn^{2+}]_{init} = 0.0375}$$ and that $$\ce{[Zn^{2+}]_{equil}} = \dfrac{\ce{K_{sp}}}{\ce{[OH]^2_{equil}}}$$ thus:

\begin{align} \Delta \ce{[Zn^{2+}] &= [Zn^{2+}]_{init} - [Zn^{2+}]_{equil}} \\ &= 0.0375 - \dfrac{\ce{K_{sp}}}{\ce{[OH]^2_{equil}}}\\ \end{align} Starting with $$\Delta\ce{[Zn^{2+}]} = \frac{1}{2}\cdot y$$, then substituting and factoring:

$$0.0375 - \dfrac{\ce{K_{sp}}}{\ce{[OH]^2_{equil}}} = \dfrac{1}{2}\cdot 0.100000\cdot \dfrac{\ce{K_b}}{\ce{K_b + [OH-]_{equil}}}$$

$$0.0375 - \dfrac{\ce{K_{sp}}}{\ce{[OH]^2_{equil}}} = 0.050000\cdot \dfrac{\ce{K_b}}{\ce{K_b + [OH-]_{equil}}}$$

multiplying by $$\ce{[OH-]^2_{equil}}$$ and rearranging we get:

$$0.0375 \cdot \ce{[OH-]^2_{equil}} - 0.05\cdot \dfrac{\ce{K_b}} {\ce{K_b + [OH-]_{equil}}}\cdot \ce{[OH]^2_{equil}} - \ce{K_{sp}} = 0$$

Now if we multiply the equation by $$\ce{K_b + [OH-]_{equil}}$$, then we'll end up with a cubic equation. But let's be a bit creative and factor the equation thus:

$$0.0375 \cdot \ce{[OH-]^2_{equil}} - 0.05\cdot \dfrac{\ce{[OH]_{equil}}} {\ce{K_b + [OH-]_{equil}}}\cdot \ce{K_b} \cdot\ce{[OH]_{equil}} - \ce{K_{sp}} = 0$$

We know the pH is going to be near neutral so $$\ce{[OH-] \approx 1\cdot10^{-7}}$$ and we know that $$\ce{K_b} = 5.623\cdot10^{-10}$$ so let's assume that $$\ce{[OH-]_{equil} \gg K_b}$$ which means that:

$$\dfrac{\ce{[OH-]_{equil}}}{\ce{K_b + [OH-]_{equil}}} \approx 1$$

so the equation reduces to a simple quadratic:

$$0.0375 \cdot \ce{[OH-]^2_{equil}} - 0.050000\cdot \ce{K_b} \cdot \ce{[OH]_{equil}} - \ce{K_{sp}} = 0$$

Solving with Wolfram gives $$\ce{[OH]_{equil}} =\pu{2.86616e-8}$$.

The known value of $$\ce{[OH]_{equil}} = \pu{2.86543e-8}$$

Although the assumption was "good enough" to get nearly four figure agreement with our known value, one can't really just look at the quadratic and instinctively understand the implications. Now we can see that the assumption was not quite true. So we'll calculate a normalization factor:

$$\dfrac{\ce{[OH-]_{equil}}}{\ce{K_b + [OH-]_{equil}}} \approx \dfrac{\pu{2.86616e-8}}{\pu{0.05623e-8 + 2.86616e-8 }} = \dfrac{\pu{2.86616e-8}}{\pu{2.92239e-8}} = 0.980759$$

Since that is about a 2% error in the factor we'll calculate a new quadratic:

$$0.0375 \cdot \ce{[OH-]^2_{equil}} - 0.049038\cdot \ce{K_b} \cdot \ce{[OH]_{equil}} - \ce{K_{sp}} = 0$$

Solving with Wolfram gives $$\ce{[OH]_{equil}} =\pu{2.86543e-8}$$.

Since the value $$\pu{2.87e-8}$$ is in agreement for both calculations it gives some confidence that the value is legitimate.

# Iterative Solution

In the Exact Solution 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. So let's try to simplify the problem.

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 $$\ce{CH3COOH}$$.

$$\begin{array}{c | c c c c c} &\ce{[Zn^{2+}]} & \ce{[CH3COO-]} & \ce{[CH3COOH]} & \ce{[OH-]} & \ce{Zn(OH)2(s)} \\\hline \text{I} &\pu{0.050000 M} & \pu{0.100000 M} & \pu{0.000000 M} & \pu{ 0.025000 M} & \pu{0 moles} \\ \text{C} & -x & -y & y & \pu{- 2x + y M} & \pu{x moles} \\ \text{E} &\pu{0.050000 - x M} & \pu{0.100000 - y M} & y & \pu{0.025000 - 2x + y M} & \pu{x moles} \\ \end{array}$$

Considerations for Simplification

Comparing $$\ce{[Zn^{2+}]_{init}}$$ and $$\ce{[NaOH]_{init}}$$, $$\ce{Zn^{2+}}$$ is clearly in excess. Let's assume that the $$\ce{[NaOH]_{init}}$$ is quantitatively precipitated as $$\ce{Zn(OH)2}$$.

So the notion for the simplified ICE table is that to 1L decarbonated distilled water there is added 0.037500 moles $$\ce{Zn(CH3COO)2}$$ and 0.025000 moles $$\ce{Na(CH3COO)}$$.

• This ignores the fact that the solution would also have a precipitate of $$\ce{Zn(OH)2}$$.

• Pronation of acetate anion and the dissolution of $$\ce{Zn(OH)2}$$ both make the solution more basic. Thus it is assumed that the precipitate of $$\ce{Zn(OH)2}$$ would contribute a negligible quantity of $$\ce{Zn^{2+}}$$ or $$\ce{OH-}$$. (This could be solved exactly using a quadratic equation if it was really needed.)

• Use a "?" for the initial concentration of $$\ce{OH-}$$ but it will be near $$\pu{1e-7}$$.

The precipitation of $$\ce{Zn(OH)2}$$ is trying to make the solution acidic. Pronation of the acetate anion is trying to make the solution basic. Considering that $$\ce{Zn(OH)2}$$ is very insoluble, and that the acetate anion is a weak base, at equilibrium the solution should be slightly acidic.

Simplified ICE Table

$$\begin{array}{c | c c c c c} &\ce{[Zn^2+]'} & \ce{[CH3COO-]'} & \ce{[CH3COOH]'} & \ce{[OH-]'} & \ce{Zn(OH)'_2(s)} \\\hline \text{I} &\pu{0.037500 M} & \pu{0.100000 M} & \pu{0.000000 M} &\pu{?} \approx \pu{ 1.000e-7} &\pu{0 moles} \\ \text{C} & \pu{-x'} & \pu{-y'} & \pu{y'} & \pu{ y' -2x'} & x' \pu{moles}\\ \text{E} &\pu{0.037500 - x' M} & \pu{0.100000 - y'M} & \pu{y'} &\pu{? + y' -2x'} & x'\ \pu{moles} \\ \end{array}$$

Ignoring the problem with the initial concentration of $$\ce{OH-}$$ for the moment, at equilibrium there are four species , $$\ce{[Zn^{2+}], [CH3COOH] , [CH3COO-]}$$ and $$\ce{[OH-]}$$ to consider. A direct solution would require a quartic equation.

Let's simplify this further by assuming that each acetate anion which is converted to acetic acid creates a hydroxide anion which is precipitated by zinc. Thus $$x' = \frac{1}{2}y'$$. This reduces the problem to a cubic equation which is still painful to solve by hand. So let's use an iterative solution.

Now we assume that $$\frac{y'}{2}$$ is small and we can assume 0.0375 for $$\ce{[Zn]_{equil}}$$ in the $$\ce{K_{sp}}$$ formula to calculate the $$\ce{[OH-]}$$.

$$\ce{[OH-]^2_{equil}} = \dfrac{K_{sp}}{\ce{[Zn]_{equil}}} = \dfrac{3.00000\cdot10^{-17}}{0.0375}= 2.828427\cdot10^{-08}$$

$$\pu{Fraction(\ce{CH3COOH})} = \dfrac{\ce{K_b}}{K_b + \ce{[OH-]}}$$

$$\ce{[CH3COOH]} = 0.1\times \dfrac{\ce{K_b}}{K_b + \ce{[OH-]}} = 0.001949278$$

$$\Delta\ce{Zn^{2+}} = \dfrac{\Delta\ce{[CH3COOH]}}{2} = 0.000974639$$

But now

$$\ce{[Zn^{2+}]_{init}} = \ce{[Zn^{2+}]_{equil}} + \dfrac{y'}{2} = 0.037500 + 0.000974639 = 0.038474639$$.

So let's use a correction factor CF to normalize $$\ce{[Zn]_{init}} + \dfrac{y'}{2}$$ to 0.037500.

$$\ce{CF} = \dfrac{0.03750000}{\ce{[Zn]_{init}} + \frac{y'}{2}}$$

The table below shows the results of the iterative solution. The first value for $$\ce{Zn^{2+}}$$ of 0.0365377 is the exact solution from Wolfram, which wouldn't be known of course. The %Err column is calculated based on the true value from Wolfram.

• The second group of calculations with $$\ce{Zn^{2+}}$$ from 0.037500 to 0.0365377 is for the simplified ICE Table.

• The third group of calculations with $$\ce{Zn^{2+}}$$ from 0.0500000 to 0.0365377 is for the original ICE Table.

$$\begin{array}{c | c c c c c} \ce{[Zn^{+2}]} guess & \ce{[OH]} & \ce{[CH3COOH]} & \ce{CF} & \%Err \\ \hline 0.0365377 & 2.865431E-08 & 0.001924590 & & \\ \\ 0.0375000 & 2.828427E-08 & 0.001949278 & 0.974668 & 0.026337 \\ 0.0365501 & 2.864947E-08 & 0.001924909 & 0.999667 & 0.000338 \\ 0.0365379 & 2.865425E-08 & 0.001924594 & 0.999996 & 0.000005 \\ 0.0365377 & 2.865431E-08 & 0.001924590 & 1.000000 & 0.000000 \\ \\ 0.0500000 & 2.449490E-08 & 0.002244066 & 0.733539 & 0.368450 \\ 0.0366769 & 2.859987E-08 & 0.001928183 & 0.996253 & 0.003811 \\ 0.0365395 & 2.865360E-08 & 0.001924637 & 0.999951 & 0.000050 \\ 0.0365377 & 2.865430E-08 & 0.001924591 & 0.999999 & 0.000001 \\ \end{array}$$

$$\ce{[OH-]_{equil}} = \sqrt{\dfrac{3.0000\cdot10^{-17}}{0.0365377}}2.865\cdot10^{-8}$$

$$\pu{pOH} = 7.5428$$

$$\pu{pH} = 14.0000 - \pu{pOH} = 14.0000 - 7.5428 = 6.4572 \ce{->[Round to 2 SF]} 6.46$$

Comparison Wolfram Mathword's solution

For the original problem Wolffram Mathword's solution is:

• $$x = 0.0134623$$
• $$y = 0.00192466$$

In comparison to the starting problem,$$x' = x - 0.0125$$ and $$y' = y$$ so the simplified problem gets:

• $$x = x' + 0.0125 = 0.000962 + 0.0125 = 0.01346229$$
• $$y = y' = 0.001924590$$