Question

Calcium nitrate is added to a sodium sulfate solution that is \(0.0150 \mathrm{M}\) (a) At what concentration of \(\mathrm{Ca}^{2+}\) does a precipitate first start to form? (b) Enough \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is added to make \(\left[\mathrm{Ca}^{2+}\right]=0.0075 \mathrm{M}\). What percentage of the original sulfate ion has precipitated?

Short answer

Answer: The precipitate of calcium sulfate starts to form at a concentration of approximately 3.29 × 10^(-6) M for Ca2+ ion. When the Ca2+ concentration is 0.0075 M, approximately 99.95% of the original sulfate ion has precipitated.

Step-by-step solution

Write the balanced chemical equation for the precipitation reaction.

The balanced chemical equation for the reaction between calcium nitrate (Ca(NO3)2) and sodium sulfate (Na2SO4) is: Ca(NO3)2(aq) + Na2SO4(aq) → 2 NaNO3(aq) + CaSO4(s) The precipitate is calcium sulfate (CaSO4).

Find the Ksp value of CaSO4.

From a solubility product constant table, we find that the Ksp value of CaSO4 is 4.93 × 10^(-5).

Determine the concentration of Ca2+ at which the precipitate starts to form.

To find the concentration of Ca2+ at which the precipitate starts to form, we will use the Ksp equation for CaSO4: Ksp = [Ca2+][SO4^(2-)] Since the sodium sulfate solution is 0.0150 M, the initial concentration of SO4^(2-) is also 0.0150 M. Let x be the concentration of Ca2+ when the precipitate starts to form. Therefore, the ion product (Q) equals Ksp when precipitation starts: Q = x × 0.0150 When Q = Ksp: 4.93 × 10^(-5) = x × 0.0150 Now, solve for x: x = (4.93 × 10^(-5)) / (0.0150) x ≈ 3.29 × 10^(-6) M At a concentration of approximately 3.29 × 10^(-6) M, the precipitate of CaSO4 starts to form.

Calculate the percentage of the original sulfate ion that has precipitated.

Given that Ca2+ concentration is increased to 0.0075 M, we can now calculate the new concentration of SO4^(2-) that has not precipitated: Q = 0.0075 × [SO4^(2-)] 4.93 × 10^(-5) = 0.0075 × [SO4^(2-)] Now, solve for [SO4^(2-)]: [SO4^(2-)] = (4.93 × 10^(-5)) / (0.0075) [SO4^(2-)] ≈ 6.57 × 10^(-6) M To find the percentage of the original sulfate ion that has precipitated, first calculate the amount of sulfate that has precipitated: 0.0150 M (initial) - 6.57 × 10^(-6) M (final) ≈ 0.014993 M (precipitated) Then, calculate the percentage: (0.014993 M / 0.0150 M) × 100 ≈ 99.95% Approximately 99.95% of the original sulfate ion has precipitated when the concentration of Ca2+ is 0.0075 M.