Question
A volume of 50 $\mathrm{mL}$ of 1.8 $\mathrm{M} \mathrm{NH}_{3}$ is mixed with an equal volume of a solution containing 0.95 $\mathrm{g}$ of $\mathrm{MgCl}_{2}$ What mass of $\mathrm{NH}_{4} \mathrm{Cl}$ must be added to the resulting solution to precipitation of $\mathrm{Mg}(\mathrm{OH})_{2} ?$
Step 1
The maximum amount of $\mathrm{OH}^{-}$ that can exist in the solution can be calculated using the solubility product constant ($K_{sp}$) expression for $\mathrm{Mg(OH)}_{2}$, which is $K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{OH}^{-}]^{2}$. Show more…
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