Question

Water from a well is found to contain \(3.0 \mathrm{mg}\) of calcium ion per liter. If \(0.50 \mathrm{mg}\) of sodium sulfate is added to one liter of the well water without changing its volume, will a precipitate form? What should \(\left[\mathrm{SO}_{4}^{2-}\right]\) be to just start precipitation?

Short answer

What concentration of sulfate ions is needed to start precipitation? Answer: No, a precipitate will not form when 0.50 mg of sodium sulfate is added to the well water. The concentration of sulfate ions needed to start precipitation is 3.20 × 10⁻⁴ M.

Step-by-step solution

Calculate the concentration of added sulfate ions

Since 0.50 mg of sodium sulfate (Na2SO4) is added to one liter of the well water, we need to find the concentration of sulfate ions added. First, convert the mass of sodium sulfate to moles: \(0.50 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}} \times \frac{1 \mathrm{mol}}{142.04 \mathrm{g/mol}} = 3.52 \times 10^{-6} \mathrm{mol}\) Since the volume of water is one liter, the concentration of added sulfate ions is: \(\frac{3.52 \times 10^{-6}\mathrm{mol}}{1\mathrm{L}} = 3.52 \times 10^{-6} \mathrm{M}\)

Calculate the initial concentration of calcium ions

Given that the well water contains 3.0 mg of calcium ions per liter, we should convert this mass into concentration as well: \(3.0\mathrm{mg} \times \frac{1\mathrm{g}}{1000\mathrm{mg}} \times \frac{1\mathrm{mol}}{40.08\mathrm{g/mol}} = 7.49\times 10^{-5}\mathrm{mol}\) Thus, the initial concentration of calcium ions is: \(\frac{7.49 \times 10^{-5}\mathrm{mol}}{1\mathrm{L}}=7.49 \times 10^{-5}\mathrm{M}\)

Use solubility product constant (Ksp) for calcium sulfate

The solubility product constant (Ksp) for CaSO4 is 2.4 × 10⁻⁵. The reaction for the formation of calcium sulfate from its ions can be written as: Ca²⁺(aq) + SO₄²⁻(aq) ⇌ CaSO₄(s) According to the Ksp expression, we have: \(K_{sp} = [Ca^{2+}][SO_4^{2-}]\)

Calculate the ion product (Q) and compare with Ksp

Ion product (Q) is given by: \(Q = [Ca^{2+}_{initial}][SO_4^{2-}_{added}]\) Using the given concentrations, we get: \(Q = (7.49 \times 10^{-5}\mathrm{M})(3.52 \times 10^{-6}\mathrm{M}) = 2.636 \times 10^{-10}\) Now, compare Q with Ksp: \(Q = 2.636 \times 10^{-10} < K_{sp} = 2.4 \times 10^{-5}\) Since Q < Ksp, a precipitate will not form.

Find the required concentration of sulfate ions to start precipitation

To just begin precipitating, the ion product (Q) must equal Ksp. \(K_{sp} = [Ca^{2+}][SO_4^{2-}]\) Solve for the required sulfate ion concentration: \(\left[ SO_4^{2-} \right]_{required} = \frac{K_{sp}}{[Ca^{2+}]}\) Using the known values: \(\left[ SO_4^{2-} \right]_{required} = \frac{2.4 \times 10^{-5}}{7.49 \times 10^{-5}\mathrm{M}} = 3.20 \times 10^{-4}\mathrm{M}\) So, the required concentration of sulfate ions to just start precipitation is 3.20 × 10⁻⁴ M.