Chapter 18:  Solubility Equilibrium

 

Section 18-1: Solubility Equilibrium Reactions and Ksp Expressions

Section 18-2: Molar Solubility and Ksp Calculations

Section 18-3: Predicting Precipitation Using Qsp

Section 18-4: Experiment - Determining Ksp Using pH

Section 18-5:  The Common-Ion Effect on Solubility

Section 18-6:  Fractional Precipitation

Chapter 18 Practice Exercises and Review Quizzes

 

 

 

 

 

Section 18-1:  Solubility Equilibrium Reactions and Ksp Expressions

Ionic compounds that are described as “insoluble” in water are actually soluble to a certain extent.  We can represent the dissolving of these slightly soluble ionic compounds in water using a solubility equilibrium reaction that shows the ionic solid on the left and the separated aqueous cations and anions on the right.  For example, the solubility equilibrium reaction for Ag₂CO₃ is written as follows:

 

Ag₂CO₃ (s) ⇌ 2 Ag⁺ (aq) + CO₃^2- (aq)

 

The equilibrium constant K_c for solubility equilibrium reactions is typically expressed as K_sp, known as the solubility product constant.  Since solubility equilibrium reactions are heterogeneous, the ionic solid is omitted from the K_sp expression.  Thus, the K_sp expression for Ag₂CO₃ is written as follows:

 

K_sp = [Ag⁺]²[CO₃^2-]

 

Sample Exercise 18A:

 

Write the solubility equilibrium reaction and the K_sp expression for magnesium phosphate.

 

Solution:

 

Mg₃(PO₄)₂ (s) ⇌ 3 Mg²⁺ (aq) + 2 PO₄^3- (aq)          K_sp = [Mg²⁺]³[PO₄^3-]²

 

 

Section 18-2:  Molar Solubility and Ksp Calculations

The extent to which an ionic compound dissolves in water can be expressed by indicating the molar solubility (s) in M.  If we know the value of K_sp for an ionic compound, we can calculate the molar solubility using a RICE chart.  The ionic solid is omitted from the left side of the RICE chart, and the relative molarities of the separated aqueous cations and anions formed on the right side of the RICE chart are found using the coefficients from the solubility equilibrium reaction as follows:

 

Sample Exercise 18B:

 

Calculate the molar solubility of magnesium phosphate (K_sp = 1.0 x 10^-25).

 

Solution:

 

K_sp = [Mg²⁺]³[PO₄^3-]²

1.0 x 10^-25 = (3s)³(2s)²

s = 3.9 x 10^-6 M

 

If we know the molar solubility of an ionic compound, we can calculate the value of K_sp using a RICE chart as follows:

 

Sample Exercise 18C:

 

The molar solubility of Ag₂CO₃ is 1.3 x 10^-4 M.  Calculate the value of K_sp for Ag₂CO₃.

 

Solution:

 

K_sp = [Ag⁺]²[CO₃^2-] = (2s)²(s) = 4s³ = 4(1.3 x 10^-4)³ = 8.8 x 10^-12

 

 

Section 18-3:  Predicting Precipitation Using Qsp

 

To predict if a solid precipitate will form when two ionic compound solutions are mixed, we will employ the following steps:

 

1. Using solubility rules, identify any combination of cation + anion that is unlikely to form an insoluble precipitate.  These will be spectator ions.

 

2. For any combination of cation + anion that may form an insoluble precipitate, write a solubility equilibrium reaction with the formula of the solid precipitate on the left and the separate aqueous cations and anions on the right.

 

3. Calculate the initial molarities of the cations and anions on the right by taking into account the dilution that occurs when the two solutions are mixed to increase the total volume.

 

4. Calculate the reaction quotient Q_sp using the calculated initial molarities of the cations and anions, and then compare Q_sp to K_sp (given in problem or on K_sp data table) as follows:

 

Q_sp > K_sp:  solubility equilibrium reaction goes to left = solid precipitate forms

Q_sp = K_sp:  solution is saturated, no precipitation occurs

Q_sp < K_sp:  solution is unsaturated, no precipitation occurs

 

 Sample Exercise 18D:

 

Predict if precipitation will occur when 33 mL of 0.0016 M Pb(NO₃)₂ is mixed with 11 mL of 0.0072 M KI.  (K_sp = 8.5 x 10^-9 for PbI₂)

 

Solution:

1. K⁺ and NO₃^- = spectator ions

 

2. PbI₂ (s) ⇌ Pb²⁺ (aq) + 2 I^- (aq)

 

3. total volume after mixing = 33 mL + 11 mL = 44 mL

 

[Pb²⁺]_i = 0.0016 M(33 mL/44 mL) = 0.0012 M

 

[I^-]_i = 0.0072 M(11 mL/44 mL) = 0.0018 M

 

4. Q_sp = (0.0012)(0.0018)² = 3.9 x 10^-9 < K_sp , no precipitate

 

 

Section 18-4:  Experiment – Determining Ksp Using pH

 

For the dissolving of a metal hydroxide in water, the molarity of hydroxide ions at equilibrium on the RICE chart can be determined by measuring the pH of a saturated solution.  The molarity of hydroxide ions can be used to determine the molar solubility of the metal hydroxide, which can then be used to calculate the value of K_sp as follows:

 

Sample Exercise 18E:

 

A metal hydroxide with the formula M(OH)₂ was mixed with water and stirred until a saturated solution was created.  The pH of the solution was found to be 9.53.  Calculate the value of K_sp for the metal hydroxide.

 

Solution:

 

pOH = 14.00 – 9.53 = 4.47

[OH^-] = 10^-4.47 = 3.4 x 10^-5 M = 2s

s = 1.7 x 10^-5 M

K_sp = [M²⁺][OH^-]² = (s)(2s)² = 4s³ = 4(1.7 x 10^-5)³ = 2.0 x 10^-14

 

 

Section 18-5:  The Common-Ion Effect on Solubility

The solubility will decrease if an ionic compound is dissolved in an aqueous solution that already contains one of the ions produced in the solubility equilibrium reaction.  This is an example of the common-ion effect.  As shown in Sample Exercise 18B above, the molar solubility of magnesium phosphate in pure water is 3.9 x 10^-6 M.  We will now demonstrate the common-ion effect by showing that the molar solubility decreases when magnesium phosphate is dissolved in an aqueous solution where the initial concentration of magnesium ions is not 0 M:        

 

Sample Exercise 18F:

 

Calculate the molar solubility of magnesium phosphate (K_sp = 1.0 x 10^-25) in 0.15 M Mg(NO₃)₂.

 

Solution:

NO₃^- = spectator ion, [Mg²⁺]_i = 0.15 M:

 

 

K_sp = [Mg²⁺]³[PO₄^3-]²

1.0 x 10^-25 = (0.15 + 3s)³(2s)²

s = 2.7 x 10^-12 M

 

 

Section 18-6:  Fractional Precipitation

A mixture of two aqueous cations or two aqueous anions can be effectively separated by the process of fractional (or selective) precipitation if nearly 100% of the amount of one ion precipitates while the second ion remains in solution as an aqueous reagent ion is added dropwise to the mixture.  The ion from the original mixture that precipitates first can be collected as part of the solid precipitate upon filtration, while the other ion from the original mixture will remain in solution as part of the filtrate that passes through the filter paper.  The problem below demonstrates how to determine which ion from the original mixture will precipitate first and if the two ions can be effectively separated by fractional precipitation:

 

Sample Exercise 18G:

 

An aqueous solution of AgNO₃ is added dropwise to an aqueous mixture containing 0.10 M CO₃^2- and 0.15 M I^-.

 

a.  Calculate the minimum molarity of Ag⁺ that must be reached to initiate precipitation of CO₃^2- (K_sp for Ag₂CO₃ = 8.8 x 10^-12) and the minimum molarity of Ag⁺ that must be reached to initiate precipitation of I^- (K_sp for AgI = 8.5 x 10^-17).  Which precipitates first, CO₃^2- or I^-?

 

b. At the point when the second ion from the original mixture begins to precipitate, what percentage of the first ion’s initial molarity still remains unprecipitated in the solution?  Can the CO₃^2- and I^- mixture be effectively separated by fractional precipitation?           

  

Solution:

Write solubility equilibrium reactions and K_sp expressions for both possible precipitates formed between the added reagent ion Ag⁺ (NO₃^- = spectator ion) and each of the anions in the original mixture:

 

Ag₂CO₃ (s) ⇌ 2 Ag⁺ (aq) + CO₃^2- (aq)     K_sp = [Ag⁺]²[CO₃^2-]

 

AgI (s) ⇌ Ag⁺ (aq) + I^- (aq)     K_sp = [Ag⁺][I^-]

 

Into each K_sp expression, substitute the K_sp value and the initial molarity of the appropriate anion from the original mixture.  Solve each expression to calculate the molarity of Ag⁺:

 

8.8 x 10^-12 = [Ag⁺]²(0.10)

[Ag⁺] = 9.4 x 10^-6 M = minimum that must be reached to precipitate CO₃^2-

 

8.5 x 10^-17 = [Ag⁺](0.15)

[Ag⁺] = 5.7 x 10^-16 M = minimum that must be reached to precipitate I^-

 

Since less Ag⁺ must be added to precipitate I^-, I^- precipitates first.

 

b.

Precipitation of CO₃^2- will begin when [Ag⁺] = 9.4 x 10^-6 M.  Therefore, we can substitute this value into the K_sp expression for AgI to calculate the molarity of I^- that still remains unprecipitated in the solution at the point when CO₃^2- begins to precipitate:

 

8.5 x 10^-17 = (9.4 x 10^-6)[I^-]

[I^-] = 9.0 x 10^-12 M

 

Dividing the result by the initial molarity of I^- in the original mixture and then multiplying by 100% yields the percentage of the initial molarity of I^- that still remains unprecipitated in the solution at the point when CO₃^2- begins to precipitate:

 

(9.0 x 10^-12 M/0.15 M) x 100% = 6.0 x 10^-9 %

 

Only a very small percentage of the original amount of I^- remains in solution.  Therefore, since nearly 100% of the original amount of I^- is incorporated into a solid precipitate before any of the CO₃^2- can leave the solution, the CO₃^2- and I^- mixture can be effectively separated by fractional precipitation.

 

 

Chapter 18 Practice Exercises and Review Quizzes:

 

 

 

 

 

 

 

 

 

 

 

 

 

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