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Chemistry · Ch 8 — Ionic Equilibrium

Solubility Product

8.9

Solubility Product

Precipitation reactions in qualitative inorganic analysis -- such as dilute HClHCl precipitating Pb2+Pb^{2+} as sparingly soluble PbCl2PbCl_2, or calcium oxalate precipitating in kidney stones -- are understood through the solubility equilibrium between an undissociated sparingly soluble salt and its constituent ions. For a general salt XmYn(s)⇌mXn+(aq)+nYm−(aq)X_mY_n(s) \rightleftharpoons mX^{n+}(aq)+nY^{m-}(aq), the equilibrium constant is K=[Xn+]m[Ym−]n[XmYn]K=\dfrac{[X^{n+}]^m[Y^{m-}]^n}{[X_mY_n]}; since the solid's concentration is itself constant in a heterogeneous equilibrium, it is absorbed to give the solubility product, Ksp=[Xn+]m[Ym−]nK_{sp}=[X^{n+}]^m[Y^{m-}]^n -- the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the balanced equation. The solubility product decides whether a precipitate forms when solutions containing the constituent ions are mixed: computing the same expression using the actual (possibly non-equilibrium) concentrations present gives the ionic product, and comparing the two tells the outcome -- if the ionic product exceeds KspK_{sp}, the solution is supersaturated and precipitation occurs; if it is less than KspK_{sp}, the solution is unsaturated and no precipitation occurs; if the two are equal, the solution is exactly saturated and at equilibrium. …

Misc example-8.9Example 8.9 – will lead chloride precipitate?

Worked out. 1 mL of 0.1M Pb(NO3)2Pb(NO_3)_2 is mixed with 0.5 mL of 0.2M NaCl; Ksp(PbCl2)=1.2×10−5K_{sp}(PbCl_2)=1.2\times10^{-5}. Total volume = 1.5 mL. Moles Pb2+=0.1×1×10−3=10−4Pb^{2+}=0.1\times1\times10^{-3}=10^{-4}, so [Pb2+]=10−4/(1.5×10−3)=6.7×10−2[Pb^{2+}]=10^{-4}/(1.5\times10^{-3})=6.7\times10^{-2} M. Moles Cl−=0.2×0.5×10−3=10−4Cl^-=0.2\times0.5\times10^{-3}=10^{-4}, so [Cl−]=10−4/(1.5×10−3)=6.7×10−2[Cl^-]=10^{-4}/(1.5\times10^{-3})=6.7\times10^{-2} M. Ionic product =[Pb2+][Cl−]2=(6.7×10−2)(6.7×10−2)2=3.01×10−4=[Pb^{2+}][Cl^-]^2=(6.7\times10^{-2})(6.7\times10^{-2})^2=3.01\times10^{-4}. Since 3.01×10−4>Ksp=1.2×10−53.01\times10^{-4} > K_{sp}=1.2\times10^{-5}, …