Question: \section{Passage $\mathrm{V}$}
A student performed qualitative tests on a series of unknown chemicals in order to identify them. The unknowns included $\mathrm{NaCl}, \mathrm{NaHCO}_{3}$, sugar $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right), \mathrm{MgSO}_{4}$, $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$, cornstarch, and chalk $\left(\mathrm{CaCO}_{3}\right)$. These are all household chemicals. $\mathrm{MgSO}_{4}$ is Epsom salts, $\mathrm{NaHCO}_{3}$ is baking soda, and $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ is photographic fixer. The student was asked to identify six unknowns: A, B, C, D, E, and F.
The qualitative tests were based on the following:
$>$ Solubility in water: starch is insoluble, as well as one of the other compounds. The rest are soluble in water.
$>$ Chalk produces a gas when treated with acid.
$>$ Starch turns blue when treated with iodine.
$>\mathrm{MgSO}_{4}$ produces a milky precipitate when treated with aqueous ammonia.
$>\mathrm{NaHCO}_{3}$ turns pink when treated with phenolphthalein.
$>\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ decolorizes iodine solution.
$>\mathrm{NaCl}$ conducts electricity in solution.
$>$ Sugar does not conduct electricity in solution.
The student prepared a flowchart that would aid in doing the experiments in a systematic manner and allow efficient identification of the unknowns.
\section{Experiment}
The experiments were conducted in the following order. The solubility of each chemical in water was tested first. From those results, the insoluble unknowns were tested next. The unknowns that were soluble in water were tested as aqueous solutions, rather than tested for conductivity.
\section{Results}
\begin{tabular}{ccccccl}
\hline Unknown & Solub & Conductivity & Acid & Phenolphthalein & $\mathbf{N H}_{\mathbf{3}}$ & $\mathbf{I}_{\mathbf{2}}$ \\
\hline A & $\mathrm{Y}$ & $\mathrm{Y}$ & $\mathrm{N}$ & $\mathrm{N}$ & $\mathrm{N}$ & $\mathrm{N}$ \\
B & $\mathrm{N}$ & $\mathrm{Y}$ & $\mathrm{Y}$ & $\mathrm{N}$ & $\mathrm{N}$ & $\mathrm{N}$ \\
C & $\mathrm{Y}$ & $\mathrm{Y}$ & $\mathrm{N}$ & $\mathrm{N}$ & $\mathrm{Y}$ & $\mathrm{N}$ \\
$\mathrm{D}$ & $\mathrm{N}$ & $\mathrm{N}$ & $\mathrm{N}$ & $\mathrm{N}$ & $\mathrm{N}$ & $\mathrm{Y}_{\mathrm{BLUE}}$ \\
E & $\mathrm{Y}$ & $\mathrm{Y}$ & $\mathrm{Y}$ & $\mathrm{Y}$ & $\mathrm{N}$ & $\mathrm{N}$ \\
F & $\mathrm{Y}$ & $\mathrm{Y}$ & $\mathrm{N}$ & $\mathrm{N}$ & $\mathrm{N}$ & $\mathrm{Y}_{\text {COLORLESS }}$ \\
\hline
\end{tabular}
From these results the student was able to identify all six unknowns.
The milky precipitate formed when $\mathrm{MgSO}_{4}$ reacts with aqueous ammonia solution is $\mathrm{Mg}(\mathrm{OH})_{2}$. Enough magnesium sulfate is used to saturate a solution that, when tested with a $\mathrm{pH}$ indicator strip, has a resulting $\mathrm{pH}$ of 10 . What is the approximate $K_{\mathrm{sp}}$ of magnesium hydroxide?

Options:
A. $10^{-12}$
B. $10^{-8}$
C. $10^{-4}$
D. $10^{-3}$
Answer: If the $\mathrm{pH}=10$, then the $\mathrm{pOH}$ is 4 because $\mathrm{pH}+\mathrm{pOH}=14$. Given the $\mathrm{pOH}=4$, the $\left[\mathrm{OH}^{-}\right]$is $1 \times 10^{-4} \mathrm{M}$. When the $K_{\mathrm{sp}}$ expression is written for magnesium hydroxide, we get $K_{\mathrm{sp}}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}$. This then becomes $K_{\mathrm{sp}}=[x][x]^{2}=[x]^{3} .\left[1 \times 10^{-4} \mathrm{M}\right]^{3}=1 \times 10^{-12}$. This is close to the actual value of $1.6 \times 10^{-12}$. The correct answer is A.