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

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?

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.

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.

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.

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.

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!

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.