# Calculating Solubility Product Quotient Q

This is part of the HSC Chemistry course under the topic Solution Equilibria

### HSC Chemistry Syllabus

• Derive equilibrium expressions for saturated solutions in terms of Ksp and calculate the solubility of an ionic substance from its Ksp value

• Predict the formation of a precipitate given the standard reference values for Ksp

### How to calculate solubility product quotient Q

This video will explore what the solubility product quotient Q is, and look at how we calculate its value.

• Similar to equilibrium quotient Q, the solubility quotient is the value assigned to a dissociation reaction when it is not at equilibrium.
• When Q < Ksp, the dissociation system has not reached the point of saturation, it is unsaturated. This means more ions will dissolve and become hydrated by solvent molecules e.g. water.
• When Q > Ksp, the dissociation system has exceeded the point of saturation, it is saturated. This means precipitate will form as no more ions can be hydrated by solvent molecules.
• Important: high Ksp indicates more ions can dissolve before saturation (equilibrium) is reached. This means high solubility. In contrast, low Ksp indicates less ions can dissolve before saturation is reached, low solubility.

## Practice Questions

Determining formation of precipitate

Question 1

0.100 g of BaSO4 is added to 500.0 mL of water at 25 ºC. Calculate the Qsp of this barium sulfate solution and thus determine whether a precipitate will form.

Solution:

$$BaSO_4(s) \rightleftharpoons Ba^2+ (aq) + SO_4^{2–} (aq)$$

$$n(BaSO_4) = \frac{0.1}{137.3+32.07+16 \times 4} = 4.29 \times 10^{–6}$$

$$Q_{sp} = [Ba^{2+}][SO_4^{2–}]$$

$$[Ba^{2+}] = [SO_4^{2-}] = [BaSO_4] = \frac{n}{V} = \frac{4.29 \times 10^{-6}}{0.5\, \text{L}} = 8.57 \times 10^{–8}$$

$$Q_{sp} = (8.57 \times 10^{–8})^2 = 7.34 \times 10^{–7}$$

$$K_{sp} (BaSO_4) = 1.08 \times 10^{–10}$$

Since Q_{sp} > K_{sp} a precipitate will form

Question 2

Will lead(II) chloride precipitate when 50 mL of 0.10 M Pb(NO3)2 solution is mixed with 50 mL of 0.10 M NaCl solution? Support your answer with a balanced chemical equation and calculations.

Solution:

$$Pb(NO_3)_2(aq) + 2NaCl(aq) \rightarrow PbCl_2(s) + 2NaNO_3(aq)$$

$$n(Pb(NO_3)_2) = 0.050 \times 0.1 = 0.05 \text{ } mol$$

$$n(NaCl) = 0.050 \times 0.1 = 0.05 \text{ } mol$$

$$\because n(PbNO_3)_2 : nNaCl = 1:2 \text{ }$$

$$NaCl \text{ is a limiting reagent}$$

$$n(PbCl_2) = \frac{0.05}{2} = 0.025$$

$$PbCl_2 \rightleftharpoons Pb^{2+}(aq) + 2Cl^–(aq)$$

$$\text{After mixing the solutions }V = 0.05 \text{ } L + 0.05 \text{ } L = 0.1 \text{ } L$$

$$[Pb^{2+}] = \frac{0.0025}{0.1} = 0.025$$

$$[Cl^–] = \frac{0.0025 \times 2}{0.1} = 0.05$$

$$Q_{sp} = [Pb^{2+}][Cl^–]^2 = [0.025][0.05]^2 = 6.25 \times 10^{–5}$$

$$\text{from data sheet } K_{sp} (PbCl_2) = 1.7 \times 10^{–5}$$

$$\because Q_{sp} > K_{sp} \text{ a precipitate will form}$$

Question 3

Equal volumes of 0.25 mol L–1 solutions of silver nitrate and calcium chloride are mixed at 25 ºC. Predict whether a precipitate will form. Support your answer with calculations.

Solution:

$$2AgNO_3 + CaCl_2 \rightarrow 2AgCl_2(s) + Ca(NO_3)_2(aq)$$

$$\because n(AgNO_3) : n(CaCl_2) = 2:1$$

$$AgNO_3 \text{ is the limiting reagent}$$

$$n(AgCl) 0.25 \times V = 0.25 \text{ }V$$

$$\text{New volume after adding solutions } = V + V = 2V$$

$$[AgCl] = \frac{0.25 \text{ }V}{2V} = 0.125$$

$$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^–(aq)$$

$$[Ag^+] = [Cl^–] = [AgCl] = 0.125$$

$$Q_{sp} = [Ag^+][Cl^–] = [0.125]^2 = 1.56 \times 10^{–2} \text{ 3 s.f.}$$

$$\text{from data sheet } K_{sp} (AgCl) = 1.77 \times 10^{–10}$$

$$\because Q_{sp} > K_{sp} \text{ a precipitate will form}$$

Previous Section: Calculating Solubility and Solubility Product

Next Section: Common Ion Effect