# Assessment Question Set #1 – Solubility Equilibria

Q1. Write the balanced reaction for the dissolution of iron (III) hydroxide in water.

Q1 is a knowledge-based question for quantitative analysis students. Chemistry students at this level should be able to write the chemical formula of an ionic compound and express the dissociation of an ionic compound in water, which is general chemistry material.

$\ce{Fe(OH)3 (s) ⇌ Fe^3+ (aq) + 3 OH- (aq)}\nonumber$

Q2. Write the equilibrium constant expression for the solubility of iron (III) hydroxide.

Q2 is a comprehension-based question for quantitative analysis students. If a chemistry student can recall the definition of an equilibrium constant expression, then that student can modify the definition of an equilibrium constant expression for any solubility equilibrium.

$K_{sp}= [Fe^{3+}] [OH^- ]^3\nonumber$

Q3. Calculate the molar solubility of iron (III) hydroxide in pure water.

Q3 is an application-based question for determining the solubility of an ionic compound in water.

In pure water, pH = pOH = 7.00. Therefore, in pure water, [OH-] = 1.0 x 10-7 M :

$\begin{array}{lccccc} \ce{&Fe(OH)3 (s) &\rightleftharpoons &Fe^3+ (aq) &+ &3 OH- (aq)}\\ \textrm{Initial} &\textrm{-----} & &0 & &1.0 \times 10^{-7}\\ \textrm{Change} &\ce{-X & &+X & &+3X}\\ \textrm{Equilibrium} &\textrm{-----} & &\ce{X} & &\mathrm{1.0 \times 10^{-7} + 3X} \end{array}\nonumber$

Assume that X represents the molar solubility of iron (III)hydroxide

$K_{sp}= [Fe^{3+} ] [OH^- ]^3\nonumber$

$1.6 \times 10^{-39}= (X) (1.0 \times 10^{-7}+3X)^3\nonumber$

assuming $$3X≪1.0 \times 10^{-7}\: M$$

$1.6 \times 10^{-39}= (X) (1.0 \times 10^{-7} )^3\nonumber$

$1.6 \times 10^{-18}\: M= X\nonumber$

using 5% rule, $$3X<(0.05)(1.0 \times 10^{-7}\: M)$$

$$3(1.6 \times 10^{-18}\: M)<5.0 \times 10^{-9}\: M$$, assumption holds

Q4. Calculate the molar solubility of iron (III) hydroxide at pH 6.4, and compare it to the calculated molar solubility of iron (III) hydroxide in Q3.

Q4 is an analysis-based question because it requires that a student differentiate between the solubility of iron hydroxide in a solvent where one of the ion concentrations (i.e. hydroxide) is held constant (i.e. a buffer) and where the concentration is not held constant.

In a pH buffer, the hydronium ion concentration, and therefore the hydroxide ion concentration, is constant. In a buffered solution with a pH of 6.4:

$\mathrm{pH + pOH = 14}\nonumber$

$\mathrm{pOH = 14 - pH = 14 - 6.4 = 7.6}\nonumber$

$\mathrm{[OH^-] = 10^{-pOH} = 10^{-7.6} = 2.5 \times 10^{-8}\: M}\nonumber$

$K_{sp}= [Fe^{3+} ] [OH^- ]^3\nonumber$

$1.6 \times 10^{-39}= (X) (2.5 \times 10^{-8}\: M)^3\nonumber$

$1.0 \times 10^{-16}\: M= X\nonumber$

since $$1.0 \times 10^{-16}\: M > 4.8 \times 10^{-18}\: M$$,

the solubility of Fe(OH)3 increases as the solution becomes more acidic

Q5. Calculate the molar solubility of iron (III) hydroxide when the ionic strength of the pH 6.4 buffer is 0.10 mol L-1.

Q5 is a synthesis-based question because it requires a student to understand how ionic strength controls the activity coefficient of an analyte, to understand how this activity coefficient changes the magnitude of the equilibrium constant, and to apply these concepts in the determination of the molar solubility of the ionic compound. In this problem, a student needs to:

• Use the Debye-Hückel equation to calculate the activity coefficient of each ionic species in the iron (III) hydroxide solubility equilibrium

$\mathrm{\log γ_x= \dfrac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}\nonumber$

$\mathrm{γ_{Fe^{3+}} = 10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}= 10^{\Large\frac{(-0.51)(+3^2 ) \sqrt{0.10}}{1+3.3(0.9\: nm) \sqrt{0.10}}}=0.1784}\nonumber$

$\mathrm{γ_{OH^-}= 10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}= 10^{\Large\frac{(-0.51)(-1^2 ) \sqrt{0.10}}{1+3.3(0.35\: nm) \sqrt{0.10}}}=0.7619}\nonumber$

• Calculate the concentration-based equilibrium constant using the thermodynamically-based equilibrium constant and the activity coefficients calculated in the previous step

$K_{sp}^{'}= \dfrac{K_{sp}}{γ_{Fe^{3+} } (γ_{OH^-} )^3 }=\dfrac{1.6 \times 10^{-39}}{(0.1784) (0.7619)^3 }=2.0 \times 10^{-38}\nonumber$

• Calculate the molar solubility of the ionic compound using the concentration-based equilibrium constant

$K_{sp}^{'}= [Fe^{3+} ] [OH^- ]^3\nonumber$

$2.0 \times 10^{-38}= (X) (2.5 \times 10^{-8}\: M)^3\nonumber$

$1.3 \times 10^{-15}\: M= X\nonumber$

Q6. To prevent precipitation of iron (III) hydroxide in household water systems, hydrogen peroxide is often added as an oxidizing agent to municipal water sources containing iron (II) ion concentrations above 0.3 mg L‑1. Compare the effectiveness of this treatment at pH 6.4 versus pH 2.0 at a dissolved iron (i.e. Fe2+) concentration of 0.3 mg L-1.

Q6 is an evaluation-based question as it asks the student to use equilibrium chemistry to compare the effectiveness of iron removal during water purification under two sets of conditions.

Since hydrogen peroxide is used as an oxidizing agent to remove iron, the first step for the student is to recognize that there is a large difference in solubility constants between iron (II) hydroxide and iron (III) hydroxide, and that this difference may lead to marked differences in solubility between the two salts. The second step is to use each solubility equilibrium constant expression to determine the molar solubility of each salt at each pH and compare it to the analyte concentration of interest (0.3 mg L-1) to evaluate the effectiveness of hydrogen peroxide.

The critical molarity of dissolved iron in water for this problem is:

$\left(\dfrac{0.3\: mg\: Fe}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Fe}{55.85\: g\: Fe}\right)=5.4 \times 10^{-6}\: M\: Fe\nonumber$

At pH 6.4:

Iron (III) hydroxide

$K_{sp}= [Fe^{3+} ] [OH^- ]^3\nonumber$

$1.6 \times 10^{-39}= (X) (2.5 \times 10^{-8}\: M)^3\nonumber$

$1.0 \times 10^{-16}\: M= X\nonumber$

Iron (II) hydroxide

$K_{sp}= [Fe^{3+} ] [OH^- ]^2\nonumber$

$8.0 \times 10^{-16}= (X) (2.5 \times 10^{-8}\: M)^2\nonumber$

$1.3\: M= X\nonumber$

When the solution pH is slightly acidic, the oxidation of ferrous ion by the addition of hydrogen peroxide results in the formation of an iron (III) hydroxide salt which is substantially less soluble than the iron (II) hydroxide salt. Under these conditions, hydrogen peroxide is effective at decreasing the dissolved iron concentration below the 0.3 mg L-1 threshold.

At pH 2.0:

Iron (III) hydroxide

$K_{sp}= [Fe^{3+} ] [OH^- ]^3\nonumber$

$1.6 \times 10^{-39}= (X) (1.0 \times 10^{-12}\: M)^3\nonumber$

$1.6 \times 10^{-3}\: M= X\nonumber$

Iron (II) hydroxide

$K_{sp}= [Fe^{3+} ] [OH^- ]^2\nonumber$

$8.0 \times 10^{-16}= (X) (1.0 \times 10^{-12}\: M)^2\nonumber$

$$8 \times 10^8\: M= X$$ (unrealistically large-interpret as freely soluble)

Under highly acidic conditions, both iron hydroxide salts are more soluble than the 0.3 mg L-1 threshold for dissolved iron. This example highlights the significance of sample acidification below pH 2.0 as a method of sample preservation for metals analysis.