Problem 109 - Chemistry: The Central Science

Question

The value of \(K_{\text {sp }}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14}\). (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) ? (b) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}{\underline{\phantom{xx}}}^{2-}\left(K_{f}=5 \times 10^{3}\right)\). If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a \(\mathrm{NaBr}\) solution, what is the initial concentration of \(\mathrm{NaBr}\) needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?\)

Step-by-Step Solution

Answer

The molar solubility of Cd(OH)₂ is approximately \(1.85\times 10^{-5} \mathrm{~mol} / \mathrm{L}\). The initial concentration of NaBr required to achieve the target molar solubility is approximately \(0.026 \mathrm{~mol} / \mathrm{L}\).

Step 1: Calculate the solubility of Cd(OH)₂

For the given exercises, which can be represented as: \(Cd(OH)₂(s) ⇌ Cd²⁺(aq) + 2OH⁻(aq)\) The Ksp for this reaction is given as \(2.5\times 10^{-14}\) Let the solubility of Cd(OH)₂ be x. Then, the concentration of Cd²⁺(aq) will also be x and the concentration of OH⁻(aq) will be 2x. Now with the given value of Ksp, let's calculate the solubility. \(K_{sp} = [Cd²⁺][OH⁻]²\)

Step 2: Set up an equation and Solve for x

Since \([Cd²⁺]=x\) and \([OH⁻]=2x\), we can substitute these values into the Ksp equation: \(2.5\times 10^{-14} = (x)(2x)²\) Now, we can solve for x: \(2.5\times 10^{-14} = 4x³\) \(x³ = \dfrac{2.5\times 10^{-14}}{4}\) \(x³ = 6.25\times 10^{-15}\) \(x = \sqrt[3]{6.25\times 10^{-15}}\) \(x = 1.85\times 10^{-5}\) The molar solubility of Cd(OH)₂ is \(1.85\times 10^{-5} \mathrm{~mol} / \mathrm{L}\).

Step 3: Calculate the concentration of NaBr to increase the molar solubility of Cd(OH)₂ to 1.0 x 10⁻³ mol/L

Given that the formation of the complex ion \(CdBr₄^{2-}\) can increase the solubility of Cd(OH)₂ and \(K_f = 5\times 10^3\). The equation for the formation of the complex is: \(Cd²⁺(aq) + 4Br⁻(aq) ⇌ CdBr₄^{2-}(aq)\) The target molar solubility is \(1.0\times 10^{-3} \mathrm{~mol} / \mathrm{L}\) Let the initial concentration of NaBr be y. Since: \([Br⁻]_i = y\) \([Br⁻]_f = y - 4(1.0 *\times 10^{-3})\) Now we have \(K_f = [CdBr₄^{2-}] \div ([Cd²⁺][Br⁻]^4)\)

Step 4: Set up the equation and solve for y

Substitute the values and the target solubility in the Kf equation: \(5\times 10^3 = \dfrac{(1.0\times 10^{-3})}{((1.0\times 10^{-3})([Br⁻]_f)^4)}\) Now substitute the final concentration of Br⁻: \(5\times 10^3 = \dfrac{(1.0\times 10^{-3})}{((1.0\times 10^{-3})(y - 4(1.0\times 10^{-3}))^4)}\) Divide both sides by \(1.0\times 10^{-3}\): \(5\times 10^3 = \dfrac{1}{(y - 4(1.0\times 10^{-3}))^4}\) Now solve for y: \((y - 4(1.0\times 10^{-3}))^4 = \dfrac{1}{5\times 10^3}\) \(y - 4(1.0\times 10^{-3}) = \sqrt[4]{\dfrac{1}{5\times 10^3}}\) \(y = 4(1.0\times 10^{-3}) + \sqrt[4]{\dfrac{1}{5\times 10^3}}\) After calculating, we get: \(y = 0.026\) The initial concentration of NaBr required to achieve the target molar solubility is \(0.026 \mathrm{~mol} / \mathrm{L}\).

Key Concepts

Ksp (Solubility Product Constant)

The Solubility Product Constant, abbreviated as Ksp, is a crucial concept in chemistry. It helps us understand the solubility of ionic compounds. It represents the level at which a compound dissolves in water. When a sparingly soluble compound, like \( \mathrm{Cd(OH)_2} \), is in equilibrium with its ions in a solution, the Ksp provides the maximum concentrations of these ions that can coexist in solution without forming a precipitate.
Understanding Ksp is fundamental because it allows us to predict whether a precipitate will form when solutions are mixed. For example, if the ion product (actual concentrations in solution) exceeds the Ksp, the compound will precipitate out. This is key when dealing with reactions in solutions.

• Ksp is specific to a given temperature. A change in temperature can affect solubility.
• The units of Ksp depend on the dissolution reaction. These can vary because they are derived from the concentrations of ions in mol/L.

This concept also helps in calculating the solubility of a compound, as seen with the problem of \( \mathrm{Cd(OH)_2} \), where we can find the concentration of dissolved ions using its Ksp value.

Molar Solubility

Molar solubility refers to the number of moles of a compound that dissolve to form 1 liter of a saturated solution. For the compound \( \mathrm{Cd(OH)_2} \), the molar solubility is found using the Ksp expression.
Each sparingly soluble salt dissociates in a particular ratio into its ions. For \( \mathrm{Cd(OH)_2} \), the dissociation is:
\[ \mathrm{Cd(OH)_2 (s) \rightleftharpoons Cd^{2+} (aq) + 2OH^{-} (aq) } \]
From the stoichiometry of this dissolution reaction, for every 1 mole of \( \mathrm{Cd(OH)_2} \) that dissolves, 1 mole of \( \mathrm{Cd^{2+}} \) and 2 moles of \( \mathrm{OH^{-}} \) are formed. Therefore, if the solubility is \( x \) mol/L, \([\mathrm{Cd^{2+}}] = x\) and \([\mathrm{OH^{-}}] = 2x\).
The Ksp expression \( [\mathrm{Cd^{2+}}][\mathrm{OH^{-}}]^2 \) gives us a key equation to solve for \( x \), demonstrating how we can compute molar solubility. This was applied to find that the solubility of \( \mathrm{Cd(OH)_2} \) is \( 1.85 \times 10^{-5} \ \mathrm{mol/L} \).
Calculating molar solubility is fundamental in processes such as predicting the extent of reaction and understanding solubility equilibrium.

Complex Ion Formation

Complex ion formation significantly affects the solubility of compounds. It occurs when ligands (usually negative ions or molecules) bond with a cation to form a more stable complex ion. Understanding this concept is essential when working with solutions, as it can alter the expected solubility behavior.
In the given problem, \( \mathrm{CdBr}_4^{2-} \) is a complex ion formed by the cadmium ion \( \mathrm{Cd^{2+}} \) in the presence of bromide ions. The stability of this complex is given by the formation constant, \( K_f \). This constant shows how readily the complex forms in solution and is essential in calculating the molar solubility of \( \mathrm{Cd(OH)_2} \) in the presence of additional bromide ions.
For \( \mathrm{Cd^{2+}} \) forming \( \mathrm{CdBr}_4^{2-} \), the equation is:
\[ \mathrm{Cd^{2+} (aq) + 4Br^- (aq) \rightleftharpoons CdBr_4^{2-} (aq) } \]
Due to this complex formation, additional \( \mathrm{Cd(OH)_2} \) can dissolve because \( \mathrm{Cd^{2+}} \) is being removed from the solution to form \( \mathrm{CdBr}_4^{2-} \), shifting the equilibrium forward.
Complex ion formation helps explain increased solubility in various chemical environments and is crucial for processes like selective precipitation and metal recovery.