Chemistry Labs

Problem 1

To determine the solubility product of copper(II) iodate, Cu(IOX3)X2\ce{Cu(IO3)2}, a saturated aqueous solution of Cu(IOX3)X2\ce{Cu(IO3)2} at 25 °C is analysed by iodometric titration in acidic solution: an excess of iodide is added and the liberated iodine is titrated with sodium thiosulphate. 30.00 cm330.00\ \mathrm{cm^3} of a 0.100 M sodium thiosulphate solution are needed for 20.00 cm320.00\ \mathrm{cm^3} of the saturated solution. (1.1) Write the sequence of balanced equations for the reactions described. (1.2) Calculate the initial concentration of CuX2+\ce{Cu^{2+}} and the solubility product of copper(II) iodate. Activity coefficients can be neglected.
Step 3 of 3: Concentration and solubility product
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[CuX2+]=2.31×10−40.02000=1.15×10−2 mol dm−3;Ksp=[CuX2+][IOX3X−]2=4[CuX2+]3=6.1×10−6[\ce{Cu^{2+}}] = \frac{2.31\times10^{-4}}{0.02000} = 1.15\times10^{-2}\ \mathrm{mol\,dm^{-3}};\qquad K_{sp} = [\ce{Cu^{2+}}][\ce{IO3^-}]^2 = 4[\ce{Cu^{2+}}]^3 = 6.1\times10^{-6}
Analysis

In a saturated solution of pure Cu(IOX3)X2\ce{Cu(IO3)2} the stoichiometry gives [IOX3X−]=2[CuX2+][\ce{IO3^-}] = 2[\ce{Cu^{2+}}], so Ksp=[CuX2+] (2[CuX2+])2=4×(1.15×10−2)3=6.1×10−6K_{sp} = [\ce{Cu^{2+}}]\,(2[\ce{Cu^{2+}}])^2 = 4\times(1.15\times10^{-2})^3 = 6.1\times10^{-6}.

Common pitfall. Do not forget the factor 4: with [IOX3X−]=2[CuX2+][\ce{IO3^-}] = 2[\ce{Cu^{2+}}], KspK_{sp} is 4[CuX2+]34[\ce{Cu^{2+}}]^3, not [CuX2+]3[\ce{Cu^{2+}}]^3.