Calculate the molar solubility of BaSO4.

Question

Calculate the molar solubility of [latex]\mathrm{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) ightleftarrows\mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)[/latex]

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

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

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

[latex]S=[Ba^{2+}]=[SO_{4}^{2-}][/latex]

Substitute S into the equilibrium expression and solve for it.

[latex]\begin{array}{c}{{K_{\mathrm{sp}}=[{ Ba}{}^{2+}][{ S}\mathrm{O}_{4}{}^{2-}]}}\ {{={}S imes S}}\ {{=S^{2}}}\end{array}[/latex]

Therefore

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

Finally, look up the value of [latex]K_{\mathrm{sp}}[/latex] in Table 15.2 and calculate S. The molar solubility of [latex]\mathrm{BaSO}_{4}[/latex] is [latex]1.{{0}}3 imes10^{-5}[/latex] mol/L.

[latex]\begin{array}{l}{{S=\sqrt{K_{\mathrm{sp}}}}}\ {{=\sqrt{1.07 imes10^{-10}}}}\ {{=1.03 imes10^{-5}\,\mathrm{M}}}\end{array}[/latex]

Table 15.2 Selected Solubility-Product Constants ( $K_{sp}$ )
Compound	Formula	$K_{sp}$
barium sulfate	$BaSO_4$	$1.07\times10^{-10}$
calcium carbonate	$CaCO_3$	$4.96\times10^{-9}$
calcium fluoride	$CaF_2$	$1.46\times10^{-10}$
calcium hydroxide	$Ca(OH)_2$	$4.68\times10^{-6}$
calcium sulfate	$CaSO_4$	$7.10\times10^{-5}$
copper(II) sulfide	CuS	$1.27\times10^{-36}$
iron(II) carbonate	$FeCO_3$	$3.07\times10^{-11}$
iron(II) hydroxide	$Fe(OH)_2$	$4.87\times10^{-17}$
lead(II) chloride	$PbCl_2$	$1.17\times10^{-5}$
lead(II) sulfate	$PbSO_4$	$1.82\times10^{-8}$
lead(II) sulfide	PbS	$9.04\times10^{-29}$
magnesium carbonate	$MgCO_3$	$6.82\times10^{-6}$
magnesium hydroxide	$Mg(OH)_2$	$2.06\times10^{-13}$
silver chloride	AgCl	$1.77\times10^{-10}$
silver chromate	$Ag_2CrO_4$	$1.12\times10^{-12}$
silver iodide	AgI	$8.51\times10^{-17}$

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

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