Problem 109 - Chemistry: The Central Science in SI Units, Global Edition

Question

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14} .(\mathbf{a})\) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?(\mathbf{b})\) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of 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 initial concentration of NaBr needed to increase the molar solubility of Cd(OH)₂ to \(1.0 \times 10^{-3} \frac{mol}{L}\) is approximately 0.063 mol/L.

Step 1: Find the molar solubility of Cd(OH)2 from its Ksp value

To find the molar solubility of Cd(OH)2, first, we start by writing a balanced dissociation equation: \[Cd(OH)_2 \xrightarrow{dissolution} Cd^{2+} + 2 OH^-\] Now, write the expression for the solubility product constant, Ksp, of Cd(OH)2: \[K_{sp} = [Cd^{2+}][OH^-]^2\] We are given the value of Ksp, and we can solve for the molar solubility, s, of Cd(OH)2. Let the molar solubility be denoted by s, thus the [Cd2+] = s and [OH-] = 2s \(K_{sp} = (s)(2s)^2\)

Step 2: Calculate the value of s

Now, insert the given Ksp value and solve for s: \(2.5 \times 10^{-14} = (s)(2s)^2\) \(2.5 \times 10^{-14} = 4s^3\) Divide both sides by 4: \(\frac{2.5 \times 10^{-14}}{4} = s^3\) Take the cube root of both sides: \(s = \sqrt[3]{\frac{2.5 \times 10^{-14}}{4}}\) Finally, calculate the value of s: \(s \approx 6.48 \times 10^{-6} \frac{mol}{L}\) Now, we move to the second part of the problem.

Step 3: Determine the relationship between Ksp and Kf

The complex formation reaction is: \[Cd^{2+} + 4Br^- \xrightarrow{formation} CdBr_4^{2-}\] The equilibrium constant for this reaction, Kf, is given as 5 x 10^3. The relationship between Ksp and Kf is derived as follows: \[K = K_{sp}K_{f} = (s)(2s)^2 \times \frac{[(s)(1.0 \times 10^{-3})]}{(4s)(1.0 \times 10^{-3})^4}\] We are given the value of K and solve for s.

Step 4: Calculate the initial concentration of NaBr

First, calculate the value of the equilibrium constant K: \(K = K_{sp} K_{f}\) \(K = (2.5 \times 10^{-14})(5 \times 10^3)\) \(K = 1.25 \times 10^{-10}\) Now, insert the given molar solubility of 1.0 x 10^-3 mol/L and solve for the initial concentration [Br-]: \(1.25 \times 10^{-10} = \frac{[(s)(1.0 \times 10^{-3})]}{(4s)([Br^-])^4}\) Rearrange for [Br-]: \([[Br^-] = \sqrt[4]{\frac{(s)(1.0 \times 10^{-3})}{(4s)(1.25 \times 10^{-10})}}\) Substituting the value of s from step 2: \([[Br^-] = \sqrt[4]{\frac{(6.48 \times 10^{-6})(1.0 \times 10^{-3})}{(4)(6.48 \times 10^{-6})(1.25 \times 10^{-10})}}\) Calculate the value of [Br-]: \([Br^-] \approx 0.063 \frac{mol}{L}\) Therefore, the initial concentration of NaBr needed to increase the molar solubility of Cd(OH)2 to 1.0 x 10^-3 mol/L is 0.063 mol/L.

Key Concepts

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solvent, reaching a saturated solution at equilibrium.
This concept is essential in determining how much of a particular compound can be dissolved before the solution is saturated.
In the case of \(\text{Cd(OH)}_2\), we find its molar solubility by using the solubility product constant, \(K_{sp}\).
When \(\text{Cd(OH)}_2\) dissolves, it dissociates into \(\text{Cd}^{2+}\) and \(\text{OH}^-\) ions: \[\text{Cd(OH)}_2 \rightleftharpoons \text{Cd}^{2+} + 2 \text{OH}^-\]The solubility product, \(K_{sp}\), is given by the equation: \[K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2\]Here, \([\text{Cd}^{2+}] = s\) and \([\text{OH}^-] = 2s\), where \(s\) is the molar solubility of the compound.
Solving for \(s\), you can determine how much \(\text{Cd(OH)}_2\) is soluble in water under the given conditions.
You simply need to find the value of \(s\) by substituting the given \(K_{sp}\) value and performing basic algebraic operations.

Complex Ion Formation

Complex ion formation significantly increases a compound's solubility.
When certain ions in solution form a complex ion, it can change the dynamics of solubility equilibrium.
In this exercise, \(\text{Cd(OH)}_2\) can form a complex ion with bromide ions, \(\text{CdBr}_4^{2-}\).
The equilibrium expression for this process is characterized by the formation constant, \(K_f\).
The formation of this complex ion ultimately shifts the equilibrium and decreases the concentration of free \(\text{Cd}^{2+}\) ions.
This effect leads to an increase in the solubility of \(\text{Cd(OH)}_2\).
Mathematically, the relationship in the presence of a complex-ion is expressed by: \[K = K_{sp} \times K_{f}\]The increase in solubility depends on both the original solubility product \(K_{sp}\) and the formation constant \(K_f\).Lowering the concentration of the ions forming \(\text{Cd(OH)}_2\) helps maintain a higher concentration of the complex ion.
Thus, careful manipulation of ion concentrations can effectively maximize solubility.

Solubility Product Constant (Ksp)

The solubility product constant, or \(K_{sp}\), is a specific type of equilibrium constant.
It applies to the dissolution equilibrium of sparingly soluble compounds in water.
\(K_{sp}\) helps predict whether a compound will precipitate or remain dissolved under certain conditions.
The value of \(K_{sp}\) indicates how much of the compound can dissolve in water.
A small \(K_{sp}\) means the compound is not very soluble.
For \(\text{Cd(OH)}_2\), the dissolution can be described by: \[\text{Cd(OH)}_2 \rightleftharpoons \text{Cd}^{2+} + 2 \text{OH}^-\]The constant \(K_{sp}\) is then: \[K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2\]Knowing \(K_{sp}\) is fundamental to calculating the maximum possible concentration of ions in the solution before precipitate starts forming.In attracting equal measures of ions back into a solid state, \(K_{sp}\) plays a critical role in equilibrium calculations and is essential in complex scenarios like predicting the results of mixing solutions.
Understanding \(K_{sp}\) provides a foundation for leveraging alterations in concentrations to control and predict solubility in various settings.