Question

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 \(\mathrm{CaSO}_{4}\).

Short answer

The calculated Solubility Product Constant, \(K_{\mathrm{sp}}\), of \(\mathrm{CaSO}_{4}\) is calculated using the steps as described.

Step-by-step solution

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.

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.

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.

Key concepts

Ion Exchange Column

An ion exchange column is a piece of laboratory equipment used to separate ions in a solution. This process involves the exchange of ions between a liquid phase and a solid phase. In the experiment involving calcium sulfate (\(\text{CaSO}_4\)), the solution is passed through the ion exchange column which contains a resin that catches calcium ions (\(\text{Ca}^{2+}\)) while releasing hydronium ions (\(\text{H}_3\text{O}^+\)) into the solution. This exchange process is important as it alters the composition of the solution and helps to determine how much calcium is in the original sample through further analysis.
The key points about ion exchange columns are:

• They trap specific ions from a solution.
• In this experiment, two \(\text{H}_3\text{O}^+\) ions are released for every calcium ion captured.
• This helps in deducing the concentration of \(\text{Ca}^{2+}\) in a given solution after a known procedure such as titration.

Titration

Titration is a common technique used to determine the concentration of a solute in a solution. In this particular experiment, titration helps quantify the amount of \(\text{H}_3\text{O}^+\) ions that were released after the ion exchange process. By reacting these ions with a known concentration of sodium hydroxide (\(\text{NaOH}\)), students are able to figure out the original concentration of the ions.
Here's how titration works in this context:

• A diluted solution of the effluent is titrated.
• The transformation follows a 1:1 ratio where \(\text{NaOH}\) neutralizes \(\text{H}_3\text{O}^+\) to form water and salt.
• By measuring the volume of \(\text{NaOH}\) needed to reach the endpoint, you can calculate the amount of \(\text{H}_3\text{O}^+\) ions in the solution.

This method allows precise determination of the ion concentration, which is vital for calculating the solubility product constant, \(K_{sp}\).

Calcium Sulfate Solubility

Calcium sulfate's solubility is a key topic in chemical equilibrium. The Solubility Product Constant (\(K_{sp}\)) is crucial as it indicates how much calcium sulfate can dissolve in water before it reaches saturation and starts forming precipitate.
Here’s how \(K_{sp}\) works:

• Calcium sulfate dissociates in water into calcium ions (\(\text{Ca}^{2+}\)) and sulfate ions (\(\text{SO}_4^{2−}\)).
• The \(K_{sp}\) expression for \(\text{CaSO}_4\) is \([\text{Ca}^{2+}][\text{SO}_4^{2-}]\).
• Since both ion concentrations are equal at equilibrium, \(K_{sp}\) can be calculated using the concentration of \(\text{Ca}^{2+}\) derived through titration.

Understanding \(K_{sp}\) assists in predicting the behavior of slightly soluble salts in various solutions and environmental conditions.