Calculating Molar Solubility from [latex]K_{sp}[/latex] Calculate the molar solubility of [latex]PbCl_2[/latex] in pure water.

Begin by writing the reaction by which solid[latex]PbCl_2[/latex]dissolves into its constituent aqueous ions and write the corresponding expression for [latex]K_{sp}[/latex]. [latex]PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 \ Cl^-(aq) [/latex] [latex]K_{sp} = [Pb^{2+}][Cl^-]^2[/latex] Refer to the stoichiometry of the reaction and prepare an ICE table, showing the equilibrium concentrations of [latex]Pb^{2+}[/latex] and [latex]Cl^-[/latex] relative to S, the amount of [latex]PbCl_2[/latex] that dissolves. [latex]PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 \ Cl^-(aq) [/latex] [latex][Pb^{2+}][/latex] [latex][Cl^−][/latex] Initial 0.00 0.00 Change +S +S Equil S 2S Substitute the equilibrium expressions for [latex][Pb^{2+}][/latex] and [latex][Cl^−][/latex] from the previous step into the expression for [latex]K_{sp}[/latex]. [latex]K_{sp} = [Pb^{2+}][Cl^-]^2[/latex] [latex]= S(2S)^2 = 4S^3[/latex] Solve for S and substitute the numerical value of [latex]K_{sp}[/latex] (from Table 17.2) to calculate S. [latex]S=\sqrt[3]{\frac{K_{sp}}{4} } [/latex] [latex]S=\sqrt[3]{\frac{1.17 \times 10^{-5}}{4} } = 1.43 \times 10^{-2 } \ M[/latex]

Question 17.8: Calculating Molar Solubility from Ksp Calculate the molar so...

Principles of Chemistry: A Molecular Approach
Chemistry: A Molecular Approach
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Calculating Molar Solubility from $K_{sp}$
Calculate the molar solubility of $PbCl_2$ in pure water.

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Begin by writing the reaction by which solid $PbCl_2$ dissolves into its constituent aqueous ions and write the corresponding expression for $K_{sp}$ .

$PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 \ Cl^-(aq)$

$K_{sp} = [Pb^{2+}][Cl^-]^2$

Refer to the stoichiometry of the reaction and prepare an ICE table, showing the equilibrium concentrations of $Pb^{2+}$ and $Cl^-$ relative to S, the amount of $PbCl_2$ that dissolves.

$PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 \ Cl^-(aq)$

	$[Pb^{2+}]$	$[Cl^-]$
Initial	0.00	0.00
Change	+S	+S
Equil	S	2S

Substitute the equilibrium expressions for $[Pb^{2+}]$ and $[Cl^-]$ from the previous step into the expression for $K_{sp}$ .

$K_{sp} = [Pb^{2+}][Cl^-]^2$
$= S(2S)^2 = 4S^3$

Solve for S and substitute the numerical value of $K_{sp}$ (from Table 17.2) to calculate S.

$S=\sqrt[3]{\frac{K_{sp}}{4} }$

$S=\sqrt[3]{\frac{1.17 \times 10^{-5}}{4} } = 1.43 \times 10^{-2 } \ M$

TABLE 17. 2 Selected Solubility-Product Constants $(K_{sp})$ at 25 °C
Compound	Formula	$K_{sp}$	Compound	Formula	$K_{sp}$
Barium fluoride	$BaF_2$	$2.45 × 10^{-5}$	Lead(II) chloride	$PbCl_2$	$1.17 × 10^{-5}$
Barium sulfate	$BaSO_4$	$1.07 × 10^{-10}$	Lead(II) bromide	$PbBr_2$	$4.67 × 10^{-6}$
Calcium carbonate	$CaCO_3$	$4.96 × 10^{-9}$	Lead(II) sulfate	$PbSO_4$	$1.82 × 10^{-8}$
Calcium fluoride	$CaF_2$	$1.46 × 10^{-10}$	Lead(II) sulfide*	PbS	$9.04 × 10^{-29}$
Calcium hydroxide	$Ca(OH)_2$	$4.68 × 10^{-6}$	Magnesium carbonate	$MgCO_3$	$6.82 × 10^{-6}$
Calcium sulfate	$CaSO_4$	$7.10 × 10^{-5}$	Magnesium hydroxide	$Mg(OH)_2$	$2.06 × 10^{-13}$
Copper(II) sulfide*	CuS	$1.27 × 10^{-36}$	Silver chloride	AgCl	$1.77 × 10^{-10}$
Iron(II) carbonate	$FeCO_3$	$3.07 × 10^{-11}$	Silver chromate	$Ag_2CrO_4$	$1.12 × 10^{-12}$
Iron(II) hydroxide	$Fe(OH)_2$	$4.87 × 10^{-17}$	Silver bromide	AgBr	$5.36 × 10^{-13}$
Iron(II) sulfide*	FeS	$3.72 × 10^{-19}$	Silver iodide	AgI	$8.51 × 10^{-17}$