Question

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) is adjusted to \(8.0 ?\) (b) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when \(100 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(5.0 \times 10^{-2} \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution?

Short answer

(a) Upon adjusting the pH of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) to \(8.0\), \(\mathrm{Ca}(\mathrm{OH})_{2}\) will not precipitate since the ion product, Q = \(2.5 \times 10^{-13}\), is less than the solubility product constant, Ksp = \(4.68 \times 10^{-11}\). (b) When \(100 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(5.0 \times 10^{-2} \mathrm{M} \mathrm{Na}_{2}\mathrm{SO}_{4}\) solution, \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) will not precipitate since the ion product, Q = \(9.3 \times 10^{-6}\), is less than the solubility product constant, Ksp = \(1.2 \times 10^{-5}\).

Step-by-step solution

Calculate the concentration of hydroxide ions

Given the pH of the solution is \(8.0\), we can find the hydroxide ion concentration (\([\mathrm{OH}^{-}]\)) using the following relationship: \(pH + pOH = 14\) Since \(pH = 8.0\), \(pOH = 14 - pH = 6\) \( [\mathrm{OH}^{-}] = 10^{-pOH} = 10^{-6} \, \mathrm{M} \)

Calculate the ion-product for Calcium Hydroxide

The ion-product for Calcium Hydroxide can be calculated as follows: Q = [Ca²⁺][OH⁻]² Since the calcium ion concentration is given as \(0.050 \, \mathrm{M}\), we can plug the values in the equation: Q = (0.050)(10^{-6})² Q = \(2.5 \times 10^{-13}\)

Compare Q to Ksp for Calcium Hydroxide

The solubility product constant (Ksp) for \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(4.68 \times 10^{-11}\). Since Q (2.5 x 10^(-13)) is less than Ksp (4.68 x 10^(-11)): Q < Ksp Therefore, \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitation will not occur in this solution. (b)

Calculate the concentrations of ions in the mixed solution

In this case, we have to find the \(\mathrm{Ag}^+\) and \(\mathrm{SO}_4^{2-}\) ion concentrations in the mixed solution. Volume of the mixed solution = \(100\,\mathrm{mL} + 10\,\mathrm{mL} = 110\,\mathrm{mL} \) Silver ion concentration: C1 = 0.050 M (initial concentration in \(\mathrm{AgNO}_{3}\) solution) V1 = 100 mL (volume of \(\mathrm{AgNO}_{3}\) solution) V2 = 110 mL (total volume) C2 = (C1 × V1) / V2 C2 = (0.050 × 100) / 110 C2 = 0.0455 M (concentration of \(\mathrm{Ag}^+\) ions in the mixed solution) Sulfate ion concentration: C1 = \(5.0 \times 10^{-2} \mathrm{M}\) (initial concentration in \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution) V1 = 10 mL (volume of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution) V2 = 110 mL (total volume) C2 = (C1 × V1) / V2 C2 = (\(5.0 \times 10^{-2}\) × 10) / 110 C2 = \(4.55 \times 10^{-3} \,\mathrm{M}\) (concentration of \(\mathrm{SO}_4^{2-}\) ions in the mixed solution)

Calculate the ion-product for Silver Sulfate

The ion-product for Silver Sulfate can be calculated as follows: Q = [Ag⁺]²[SO₄²⁻] Plugging in the values from the calculation above: Q = (0.0455)²(\(4.55 \times 10^{-3}\)) Q = \(9.3 \times 10^{-6}\)

Compare Q to Ksp for Silver Sulfate

The solubility product constant (Ksp) for \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) is \(1.2 \times 10^{-5}\). Since Q (9.3 x 10^(-6)) is less than Ksp (1.2 x 10^(-5)): Q < Ksp Therefore, \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) precipitation will not occur when the solutions are mixed.