In an experiment to measure \(K_{\mathrm{sp}}\) of \(\mathrm{CaSO}_{4}\) [D. Masterman, J. Chem. Educ., 64, 409 (1987)], a saturated solution of \(\mathrm{CaSO}_{4}(\mathrm{aq})\) is poured into the ion-exchange column pictured (and described in Chapter 21 ). As the solution passes through the column, \(\mathrm{Ca}^{2+}\) is retained by the ion-exchange medium and \(\mathrm{H}_{3} \mathrm{O}^{+}\) is released; two \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions appear in the effluent solution for every \(\mathrm{Ca}^{2+}\) ion. As the drawing suggests, a \(25.00 \mathrm{mL}\) sample is added to the column, and the effluent is collected and diluted to \(100.0 \mathrm{mL}\) in a volumetric flask. A \(10.00 \mathrm{mL}\) portion of the diluted solution requires \(8.25 \mathrm{mL}\) of \(0.0105 \mathrm{M} \mathrm{NaOH}\) for its titration. Use these data to obtain a value of \(K_{\mathrm{sp}}\) for \(\mathrm{CaSO}_{4}\).

Step by step solution

01

Determine concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions

From the titration process, we know that it took 8.25 mL of 0.0105 M \(\mathrm{NaOH}\) to neutralize a 10.00 mL sample of the solution. We use the molar relationship between \(\mathrm{NaOH}\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\) (since \(\mathrm{NaOH}\) reacts with \(\mathrm{H}_{3} \mathrm{O}^{+}\) to form \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{Na}^{+}\) in a 1:1 ratio) to find the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) in the sample solution. This gives that 0.000086625 mol of \(\mathrm{NaOH}\) have been used. Thus, 0.000086625 mol of the \(\mathrm{H}_{3} \mathrm{O}^{+}\) were in the 10 mL of dilute effluent. We can calculate the original concentration in the 25 mL aliquot before the dilution.

02

Calculate concentration of \(\mathrm{Ca}^{2+}\) ions

As two \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions appear for every \(\mathrm{Ca}^{2+}\) ion, the concentration of \(\mathrm{Ca}^{2+}\) ions is therefore half the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions. Halving the concentration from step 1 will give concentration of \(\mathrm{Ca}^{2+}\) ions.

03

Calculate \(K_{\mathrm{sp}}\) of \(\mathrm{CaSO}_{4}\)

Since the reaction of \(\mathrm{CaSO}_{4}\) breaking down into its ions is \(\mathrm{CaSO}_{4} \longleftrightarrow \mathrm{Ca}^{2+} + \mathrm{SO}_{4}^{2-}\), and the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) ions are the same, we can directly substitute the value obtained from step 2 into the \(K_{\mathrm{sp}}\) expression. That is, \(K_{\mathrm{sp}} = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]\). On solving, we will get the required \(K_{\mathrm{sp}}\) value.

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