How much calcium hydroxide will precipitate after addition of sodium hydroxide into saturated calcium hydroxide solution?

Question

Below was question 34 in the USNCO 2017 exam:

If $\pu{0.10 mol}$ of solid $\ce{NaOH}$ is added to $\pu{1.00 L}$ of a saturated solution of $\ce{Ca(OH)2}$ $(K_\mathrm{sp} = \pu{8.0 \times 10^-6})$, what percentage of the calcium hydroxide will precipitate at equilibrium?

(A) Roughly 50%
(B) Roughly 75%
(C) Roughly 95%
(D) Over 99%

My solution is as follows:

1. Find concentration of $\ce{Ca^2+}$ $(\pu{0.02 M})$ and $\ce{OH-}$ $(\pu{0.04 M})$ ions from dissolved calcium hydroxide using $K_\mathrm{sp}$.

2. Add hydroxide ion concentration from sodium hydroxide (assuming full dissolution) to get total hydroxide concentration of $\pu{0.14 M}$

3. Find reaction quotient $Q = 0.02 \times 0.14^2 = 3.92 \times 10^{-4}$

4. Find amount of calcium $(x)$ and hydroxide ions $(2x)$ that will precipitate at equilibrium by using algebraic equation: $$ (0.02 - x)(0.14 - 2x)^2 = 8.0 \times 10^{-6}, x = \pu{0.019 M}$$

5. Find percentage of calcium hydroxide precipitated: $$\frac{0.019}{0.02} \times 100\% = 95\%,$$ hence (C)

I am unsure about step 4, where a cubic equation appears, and would not be able to be solved in exam conditions (use of graphing calculator is not permitted).

Is there a simpler method?

Answer 1

A less analytic aproach:

Initial concentrations:

$K_{\mathrm{sp}}=\ce{[Ca^{2+}][OH^-]^2}=x\cdot (2x)^2=8\cdot10^{-6}$

$\ce{[Ca^{2+}]=0.0126}$ M; $\ce{[OH^-]=0.0252}$ M.

Now 0.1 mol NaOH is added. Let $p$ be the fraction that precipitates. The concentrations that remain in solution are:

$\ce{[Ca^{2+}}]=0.0126(1-p)$

$\ce{[OH^-]}=(0.10 + 0.0252 -2\cdot0.0126p)^2$

Since the solubility product remains the same:

$K_{\mathrm{sp}}=0.0126(1-p)(0.1252 -2\cdot0.0126p)^2$

Now, since there are only four scenarios, try them out substituting every value of $p$ in the equation and take the closest one to $8\cdot10^{-6}$:

\begin{array}{|c|c|}\hline p&K_\mathrm{sp}\\\hline 0.50&\pu{7.99E-05}\\ 0.75&\pu{3.56E-05}\\ 0.95&\pu{6.46E-06}\\ 0.99&\pu{1.27E-06}\\\hline \end{array}

The closest one is for roughly 95% precipitation, without solving any equation.

Answer 2

When you run into a cubic in evaluating concentrations, one approach you can use to solve it is the method of successive approximations. $$(0.0126-x)(0.1252-2x)^2=8\cdot10^{-6}$$ $$(0.0126-x_0)(0.1252)^2=8\cdot10^{-6}$$ $$(0.0126-x_1)(0.1252-2x_0)^2=8\cdot10^{-6}$$ For example, moving from the first line to the 2nd line, I guess that $2x=0$ and solve for $x_0$. This first guess is not great, but I can use the solution from that approximate equation to generate a better guess. So in the 3rd line, I approximate $2x=x_0$ and solve for $x_1$. You can continue this procedure until $x_{n+1}\approx x_n$, which in this case occurs very quickly. I obtain $x_0=0.01209$, $x_1=0.0119747$, and $x_2=0.011976$ after which the value doesn't change. Checking the ratio, we see that it gives just about $95\ \%$. $$\frac{0.011976}{0.0126}\times100=95.047\ \%$$

An important side note for this case is why I approximated $2x$ rather than $x$. This is because approximating $x$ will cause you to converge towards the two complex solutions to the cubic equation.

Answer 3

Note what the question actually asks you, "roughly". Approximation is an important tool in ionic equilibrium calculations.

Here, you have $0.0126M \text{ }\ce{Ca}^{+2}$ and $0.0252M\text{ } \ce{OH-}$ in the original solution. By addition of $0.1M \text{ }\ce{NaOH}$, which is a strong electrolyte, you'll not only add $0.1M\text{ } \ce{OH-}$ to the solution, but also cause $Q$ to exceed $K_{sp}$, hence, the salt will be precipitated out.

Now, the concentration of our ions before precipitation is $0.0126M \text{ }\ce{Ca}^{+2}$ and $0.1252M\text{ } \ce{OH-}$. Let $y$ be the fraction of existing ions that are not precipated out. So, we'll be left with $0.0126 \cdot y M \text{ }\ce{Ca}^{+2}$ and $0.1252 \cdot (2y-1) M\text{ } \ce{OH-}$ ions. Equating their ionic product to the salt's $K_{sp}$, we get:

$$0.0126 \cdot y\cdot(0.1252\cdot (2y-1))^2=8\times10^{-6}$$

$$y\cdot(2y-1)^2\approx0.0405$$ $$y\cdot(1+4y^2-4y)\approx0.0405$$

Magic trick$^\text{TM}$: Neglect $y^3$ at this step. If $y$ comes out be $\le5\%$, we'll assume this neglection to be correct.

$$4y^2-y+0.0405\approx0$$ $$y\approx0.050838$$

That means nearly $(100-5)\% = 95\%$ of the salt was precipitated out. Hence, your answer. Without a graphing calculator.

PS: If you actually solve the original equation, you'll get $y=0.05$. Hence, our approximate answer is very reasonably close.

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How did I get $(2y-1)$ as fraction of final concentration of $\ce{OH-}$ ions?

Fairly easy. Observe that $y$ is the fraction of concentration of ions not precipitated out, while $x$ is the fraction of concentration of ions precipitated out. Then $x+y=1$. So, $1-x=y=\text{ concentration of }\ce{Ca}^{+2}$ ions. And $1-2x=1-2\cdot(1-y)=2y-1=\text{ concentration of }\ce{OH-}$ ions.

But why did I take $y$ anyway? What was wrong with the original $x$?

Our aim was to reduce this problem into an equation which can be solved without a calculator, i.e., by approximation. Taking $y$ ensured that in the end we could neglect higher powers of $y$ (note that we cannot neglect higher powers of $x$) and solve the question easily.

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Crux of my answer: if while solving an ionic equilibrium question you're stuck in a tough calculation/unsolveable equation, consider modifying your approach to allow the use of neglection.

Hope it helps!

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^{Ok, I admit, it wasn't any magic trick :P But just a clever observation that if $x\le 0.05$, then $x^3\le0.000125$, which is too small to affect our final result significantly.}