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Chemistry · Ch 8 — Ionic Equilibrium

Determination of Solubility Product from Molar Solubility

8.9.1

Determination of Solubility Product from Molar Solubility

KspK_{sp} can be calculated from the molar solubility -- the maximum number of moles of solute dissolving per litre. For XmYn(s)⇌mXn+(aq)+nYm−(aq)X_mY_n(s) \rightleftharpoons mX^{n+}(aq)+nY^{m-}(aq), one mole of XmYnX_mY_n furnishes mm moles of Xn+X^{n+} and nn moles of Ym−Y^{m-}; if ss is the molar solubility, [Xn+]=ms[X^{n+}]=ms and [Ym−]=ns[Y^{m-}]=ns, so Ksp=[Xn+]m[Ym−]n=(ms)m(ns)n=mm nn sm+nK_{sp}=[X^{n+}]^m[Y^{m-}]^n=(ms)^m(ns)^n=m^m\,n^n\,s^{m+n} -- a single formula relating KspK_{sp} and ss for any salt stoichiometry once mm and nn are known. …

Misc example-8.10Example 8.10 – Ksp-solubility relation for BaSO4 and Ag2CrO4

Worked out. Establish the KspK_{sp}-solubility relation for (a) BaSO4BaSO_4 and (b) Ag2CrO4Ag_2CrO_4. (a) BaSO4(s)⇌Ba2+(aq)+SO42−(aq)BaSO_4(s) \rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq): with molar solubility ss, [Ba2+]=s[Ba^{2+}]=s and [SO42−]=s[SO_4^{2-}]=s, so Ksp=(s)(s)=s2K_{sp}=(s)(s)=s^2. (b) Ag2CrO4(s)⇌2Ag+(aq)+CrO42−(aq)Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq)+CrO_4^{2-}(aq): [Ag+]=2s[Ag^+]=2s and [CrO42−]=s[CrO_4^{2-}]=s, so Ksp=(2s)2(s)=4s3K_{sp}=(2s)^2(s)=4s^3. …