Problem 51 - Chemical Principles: The Quest for Insight

Question

Determine the \(K_{p}\) for the following sparingly soluble substances, given their molar solubilitics: (a) \(\mathrm{AgBr}, 8.8 \times 10^{-7} \mathrm{~mol} \cdot \mathrm{L}^{-1}\); (b) \(\mathrm{PbCrO}_{4}, 1.3 \times\) \(10^{-7} \mathrm{~mol} \cdot \mathrm{L}^{-1}\); (c) \(\mathrm{Ba}(\mathrm{OH})_{2}, 0.11 \mathrm{~mol} \cdot \mathrm{L}^{-1}\); (d) \(\mathrm{MgF}_{2}\) \(1.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\).

Step-by-Step Solution

Answer

The solubility products (\(K_{sp}\)) are (a) \(7.744 \times 10^{-13}\) for AgBr, (b) \(1.69 \times 10^{-14}\) for PbCrO4, (c) \(0.05324\) for Ba(OH)2, (d) \(2.7648 \times 10^{-8}\) for MgF2.

Step 1: Write the balanced chemical equation for AgBr dissolution

Solubility product, denoted as \(K_{sp}\), is determined by the molar solubility of the substance. The dissolution of silver bromide (AgBr) in water is represented by the following equation: \[\mathrm{AgBr(s)} \rightleftharpoons \mathrm{Ag^{+}(aq)} + \mathrm{Br^{-}(aq)}\] For every mole of \(AgBr\) that dissolves, one mole of \(Ag^{+}\) and one mole of \(Br^{-}\) ions are formed.

Step 2: Calculate the \(K_{sp}\) for AgBr

The \(K_{sp}\) for a salt is calculated by multiplying the concentrations of the ions produced upon dissolution. For \(AgBr\): \[K_{sp} = [Ag^{+}][Br^{-}]\] Given that the molar solubility of \(AgBr\) is \(8.8 \times 10^{-7} \mathrm{mol/L}\), both \(Ag^{+}\) and \(Br^{-}\) will have the same molar concentration: \[K_{sp}(AgBr) = ( 8.8 \times 10^{-7} )( 8.8 \times 10^{-7} )\] \[K_{sp}(AgBr) = 7.744 \times 10^{-13}\]

Step 3: Write the balanced chemical equation for PbCrO4 dissolution

For lead(II) chromate (PbCrO4), the dissolution process is represented by: \[\mathrm{PbCrO4(s)} \rightleftharpoons \mathrm{Pb^{2+}(aq)} + \mathrm{CrO4^{2-}(aq)}\] One mole of \(Pb^{2+}\) and one mole of \(CrO4^{2-}\) are produced for every mole of \(PbCrO4\) that dissolves.

Step 4: Calculate the \(K_{sp}\) for PbCrO4

Using the molar solubility given for \(PbCrO4\), \(1.3 \times 10^{-7} \mathrm{mol/L}\), the \(K_{sp}\) is calculated as follows: \[K_{sp} = [Pb^{2+}][CrO4^{2-}]\] \[K_{sp}(PbCrO4) = (1.3 \times 10^{-7})(1.3 \times 10^{-7})\] \[K_{sp}(PbCrO4) = 1.69 \times 10^{-14}\]

Step 5: Write the balanced chemical equation for Ba(OH)2 dissolution

Barium hydroxide (Ba(OH)2) dissolves in water as follows: \[\mathrm{Ba(OH)2(s)} \rightlefharpoons \mathrm{Ba^{2+}(aq)} + 2\mathrm{OH^{-}(aq)}\] For every mole of \(Ba(OH)2\), one mole of \(Ba^{2+}\) and two moles of \(OH^{-}\) are formed.

Step 6: Calculate the \(K_{sp}\) for Ba(OH)2

For \(Ba(OH)2\): \[K_{sp} = [Ba^{2+}][OH^{-}]^2\] Given the molar solubility of \(Ba(OH)2\) is \(0.11 \mathrm{mol/L}\): \[K_{sp}(Ba(OH)2) = (0.11)(2 \times 0.11)^2\] \[K_{sp}(Ba(OH)2) = 0.11 \times 0.484\] \[K_{sp}(Ba(OH)2) = 0.05324\]

Step 7: Write the balanced chemical equation for MgF2 dissolution

For magnesium fluoride (MgF2), the dissolution process is given by: \[\mathrm{MgF2(s)} \rightlefharpoons \mathrm{Mg^{2+}(aq)} + 2\mathrm{F^{-}(aq)}\] One mole of \(Mg^{2+}\) and two moles of \(F^{-}\) are formed upon the dissolution of one mole of \(MgF2\).

Step 8: Calculate the \(K_{sp}\) for MgF2

Using the given molar solubility for \(MgF2\), \(1.2 \times 10^{-3} \mathrm{mol/L}\): \[K_{sp} = [Mg^{2+}][F^{-}]^2\] \[K_{sp}(MgF2) = (1.2 \times 10^{-3})(2 \times 1.2 \times 10^{-3})^2\] \[K_{sp}(MgF2) = (1.2 \times 10^{-3})(4.8 \times 10^{-3})^2\] \[K_{sp}(MgF2) = (1.2 \times 10^{-3})(2.304 \times 10^{-5})\] \[K_{sp}(MgF2) = 2.7648 \times 10^{-8}\]

Key Concepts

Molar Solubility

Molar solubility is a term used to describe the amount of a substance that can dissolve in a given volume of solvent at a specific temperature to form a saturated solution. It is expressed in moles per liter ()). Molar solubility is particularly important for sparingly soluble salts, which do not fully dissolve in water but reach an equilibrium between the solid and dissolved ions. Understanding molar solubility is crucial for predicting the extent of a substance's solubility in a solvent, which in turn helps in various applications, including pharmaceuticals, chemical reactions, and water treatment.

Balanced Chemical Equation

A balanced chemical equation is essential in chemistry as it represents the conservation of mass in a chemical reaction. Every chemical reaction abides by the law of conservation of mass, meaning the number of atoms for each element in the reactants must equal the number in the products. In the context of dissolution, a balanced chemical equation helps us understand the molar ratios of the ions formed when a compound dissolves. These ratios are used when calculating the solubility product constant ()) as shown in the example solutions, where the stoichiometry of the dissolution process dictates the number of ions produced.

Ksp Calculation

The solubility product constant ()), is a value that represents the extent to which a compound can dissolve in water. Calculating ) involves writing the balanced chemical equation for the dissolution of the compound, determining the molar concentration of each ion at equilibrium, and applying these concentrations to the expression for ). The product of these concentrations raised to the power of their coefficients in the balanced equation gives us ). For instance, in the exercise, the ) for )) is calculated by squaring the concentration of the silver and bromide ions because the ratio of their formation is 1:1.

Dissolution Process

The dissolution process involves the breaking down of a solid into its constituent ions when it is placed in a solvent like water. This process is essential for understanding solubility and ) calculation. It's important to note that for slightly soluble ionic compounds, the dissolution process reaches a state of dynamic equilibrium. At this point, the rate of dissolution equals the rate of precipitation, which results in a constant concentration of ions in the solution. This is what we measure when we discuss molar solubility. The balanced chemical equations used in ) calculations stem directly from understanding the dissolution process of each particular compound.