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.


 4

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.


 0

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](http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=0.0126+x(0.1252(2x-1))%5E2%3D8*10%5E-6), 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.


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