Calculate the molar solubility of BaSO4.

Question

Calculate the molar solubility of [latex]BaSO_4[/latex].

Accepted Answer

Begin by writing the reaction by which solid [latex]\mathrm{BaSO_{4}}[/latex] dissolves into its constituent aqueous ions.

[latex]\mathrm{BaSO_{4}}(s)\xrightleftharpoons[]{}\mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)[/latex]

Next, write the expression for [latex]K_{sp}[/latex].

[latex] K_{\mathrm{sp}}\,=\,[\mathrm{B}\mathrm{a}^{2+}][\mathrm{SO}_{4}^{\mathrm{2-}}\,][/latex]

Define the molar solubility (S) as [latex][\mathrm{B}\mathrm{a}^{2+}][/latex] or [latex][\mathrm{SO}_{4}^{\mathrm{2-}}\,] [/latex] at equilibrium.

[latex]S=\,[\mathrm{B}\mathrm{a}^{2+}]= [\mathrm{SO}_{4}^{\mathrm{2-}}\,][/latex]

Substitute S into the equilibrium expression and solve for it.

[latex]K_{\mathrm{sp}}\,=\,[\mathrm{B}\mathrm{a}^{2+}][\mathrm{SO}_{4}^{\mathrm{2-}}\,] \ = S imes S \ = S^2[/latex]

Therefore

[latex]S=\sqrt{K_{sp}}[/latex]

Finally, look up the value of [latex]K_{sp}[/latex] in Table 15.2 and calculate S. The molar solubility of [latex]\mathrm{BaSO_{4}} is 1.03 × 10^{-5} mol/L[/latex]

[latex]S=\sqrt{K_{sp}}\ = \sqrt{1.07 imes10^{-10}} \ = 1.03 imes 10^{-5} M[/latex]

TABLE 15.2 Selected Solubility-Product Constants ( $K_{sp}$ )
Compound	Formula	$K_{sp}$
barium sulfate	$BaSO_4$	$1.07 \times 10^{-10}$
calcium carbonate	$CaCO_3$	$4.96 \times 10^{-9}$
calcium fluoride	$CaF_2$	$1.46 \times 10^{-10}$
calcium hydroxide	$Ca(OH)_2$	$4.68 \times 10^{-6}$
calcium sulfate	$CaSO_4$	$7.10 \times 10^{-5}$
copper(II) sulfide	$CuS$	$1.27 \times 10^{-36}$
iron(II) carbonate	$FeCO_3$	$3.07 \times 10^{-11}$
iron(II) hydroxide	$Fe(OH)_2$	$4.87 \times 10^{-17}$
lead(II) chloride	$PbCl_2$	$1.17 \times 10^{-5}$
lead(II) sulfate	$PbSO_4$	$1.82 \times 10^{-8}$
lead(II) sulfide	$PbS$	$9.04 \times 10^{-29}$
magnesium carbonate	$MgCO_3$	$6.82 \times 10^{-6}$
magnesium hydroxide	$Mg(OH)_2$	$2.06 \times 10^{-13}$
silver chloride	$AgCl$	$1.77 \times 10^{-10}$
silver chromate	$Ag_2CrO_4$	$1.12 \times 10^{-12}$
silver iodide	$AgI$	$8.51 \times 10^{-17}$

Question 15.9: Calculate the molar solubility of BaSO4....

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