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 K_sp and calculate the solubility of an ionic substance from its K_sp value

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

How to Calculate Ionic Product 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 < K_sp, 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 > K_sp, 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 K_sp indicates more ions can dissolve before saturation (equilibrium) is reached. This means high solubility. In contrast, low K_sp 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 Q_sp 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

  

BACK TO MODULE 5: EQUILIBRIUM AND ACID REACTIONS