Grade 11 Chemistry Effective Homework 8: Precipitation and Dissolution Equilibrium | Brief Solutions

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Let the molar solubility be $s$. From $4s^3=5.6\times10^{-12}$, $s=1.12\times10^{-4}\,\mathrm{mol\,L^{-1}}$. Taking the dilute-solution density as approximately $1\,\mathrm{g\,mL^{-1}}$, this is about $6.5\times10^{-4}\,\mathrm g$ per $100\,\mathrm g$ of water.

$c(\mathrm{OH^-})=\sqrt{5.6\times10^{-12}/0.02}=1.67\times10^{-5}\,\mathrm{mol\,L^{-1}}$.

$\mathrm{Cu(OH)_2}$ precipitates first. With both solids present, $\dfrac{c(\mathrm{Mg^{2+}})}{c(\mathrm{Cu^{2+}})}=\dfrac{5.6\times10^{-12}}{2.2\times10^{-20}}=2.55\times10^8$.

$\mathrm{AgCl}$.

$K_{sp}(\mathrm{AgCl})=10^{-9.8}$,$K_{sp}(\mathrm{Ag_2CrO_4})=10^{-11.7}$;$K=\dfrac{K_{sp}(\mathrm{Ag_2CrO_4})}{K_{sp}(\mathrm{AgCl})^2}=10^{7.9}\approx7.94\times10^7$.

$c(\mathrm{H_2CO_3})=\dfrac{8.4\times10^{-9}\times10^{-4}}{4.2\times10^{-7}}=2.0\times10^{-6}\,\mathrm{mol\,L^{-1}}$ and $c(\mathrm{CO_3^{2-}})=6.67\times10^{-7}\,\mathrm{mol\,L^{-1}}$. Therefore $c(\mathrm{Ca^{2+}})=\dfrac{2.8\times10^{-9}}{6.67\times10^{-7}}=4.2\times10^{-3}\,\mathrm{mol\,L^{-1}}$.

Neglecting hydrolysis, $c(\mathrm{F^-})=2c(\mathrm{Mg^{2+}})$, so $c(\mathrm{F^-})=(2K_{sp})^{1/3}=2.0\times10^{-4}\,\mathrm{mol\,L^{-1}}$. At low pH, fluoride forms HF, lowering free fluoride and favoring dissolution of $\mathrm{MgF_2}$.

$0.1\,\mathrm{mol\,L^{-1}}$. Here $c(\mathrm{OH^-})^3=10^{-33}$, giving an aluminum ion product of $10^{-34}<K_{sp}[\mathrm{Al(OH)_3}]$; aluminum has not started to precipitate.

$K=K_{a1}/K_{a2}=10^{-7}/10^{-13}=10^6$.

$c(\mathrm{H^+})^2=K_{a1}K_{a2}=10^{-20}$, hence $\mathrm{pH}=10$.

$c(\mathrm{S^{2-}})=2.0\times10^{-15}/0.01=2.0\times10^{-13}$ and $c(\mathrm{H^+})=K_{a2}c(\mathrm{HS^-})/c(\mathrm{S^{2-}})=10^{-4}$; thus $\mathrm{pH}=4$.

$8.0\times10^{-10}\,\mathrm{mol\,L^{-1}}$

Increase: $\mathrm{H^+}$ and $\mathrm{HS^-}$. Decrease: $\mathrm{OH^-}$.

$\mathrm{pH}\approx3$

$K=\dfrac{2.0\times10^{-10}}{6.0\times10^{-36}}=3.3\times10^{25}$

$\mathrm{SCN^-+H_2O\rightleftharpoons HSCN+OH^-}$

$K_a(\mathrm{HF})=4.0\times10^{-4}$ while $K_h(\mathrm{F^-})=K_w/K_a=2.5\times10^{-11}$. HF ionization outweighs fluoride hydrolysis, so the mixture is acidic.

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Grade 11 Chemistry Effective Homework 8: Precipitation and Dissolution Equilibrium | Brief Solutions

1Assess statements about dissolution equilibrium, solubility products and the common-ion effect.

2Which operation leaves the barium-ion concentration unchanged in a barium sulfate suspension?

3Compare conductivity during dissolution of calcium carbonate and sucrose; choose the incorrect interpretation.

4Choose the incorrect statement about dilution, silver chloride solubility and precipitate conversion.

5Use the calcium sulfate concentration curve to assess temperature changes, salt addition and precipitate amounts.

6Remove copper(II) using manganese sulfide; choose the incorrect statements.

7Compare dissolution equilibria of magnesium hydroxide and magnesium fluoride.

8.1Calculate the solubility of magnesium hydroxide at 25 degrees Celsius using $K_{sp}=5.6\times10^{-12}$.

8.2Find the hydroxide concentration at the onset of precipitation from $0.02\,\mathrm{mol\,L^{-1}}$ magnesium chloride.

8.3Add NaOH to equal-concentration magnesium and copper(II) ions; identify the first precipitate and the ion ratio when both solids coexist.

9.1Use the AgCl and silver chromate equilibrium lines to determine which precipitates first from equal anion concentrations.

9.2Use the plotted coordinates $(1.7,5)$ and $(4.8,5)$ to calculate the conversion constant from silver chromate to silver chloride.

10Use the carbonic acid constants and given hydrogen-ion and bicarbonate concentrations to find carbonic acid concentration and the calcium carbonate precipitation threshold.

11.1Precipitate magnesium using a 1:2 magnesium-to-fluoride mole ratio; find residual fluoride and explain the effect of low pH.

11.2Iron(III) and aluminum initially both have concentration 0.1 mol/L; find aluminum concentration when iron has just precipitated completely.

11.3①Use the pH 7 and 13 crossings on the sulfide distribution graph to find the constant for $\mathrm{H_2S+S^{2-}\rightleftharpoons2HS^-}$.

11.3②Find the pH when hydrogen sulfide and sulfide-ion concentrations are equal.

11.3③At the onset of precipitation of $0.01\,\mathrm{mol\,L^{-1}}$ manganese(II), $c(\mathrm{HS^-})=2.0\times10^{-4}$; find pH. The original introduction has Mg in place of Mn.

12.1Under the usual school-level treatment, identify the solution that leaves the original solute after evaporation and heating.

12.2At a temperature where pure water has hydrogen-ion concentration $4\times10^{-7}$, add acid to reach $2\times10^{-4}$; find hydrogen ions arising from water ionization.

12.3Select the correct relation among particle concentrations in the four solutions.

12.4Compress hydrogen sulfide gas above its saturated aqueous solution; identify ions whose concentrations increase and decrease.

12.5Find the pH of 1 mol/L HA with $K_a=10^{-6}$.

12.6Find the sulfide-conversion constant using $K_{sp}(\mathrm{CuS})=6.0\times10^{-36}$ and $K_{sp}(\mathrm{MnS})=2.0\times10^{-10}$.

12.7Explain the weak alkalinity of KSCN solution using an ionic equation.

12.8Mix equal volumes of equal-concentration HF and NaF; use equilibrium constants to explain the acidity.