Question

The value of $K_{sp}$ 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 ${\mathrm{CdBr}}_4^{2-}(K_f=5\times {10}^3).$ 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 $1.0\times {10}^{-3}\mathrm{mol}/\mathrm{L}?$

Short 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-by-step solution

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: $$Cd(OH{)}_2\stackrel{dissolution}{\to }C{d}^{2+}+2O{H}^{-}$$ Now, write the expression for the solubility product constant, Ksp, of Cd(OH)2: $$K_{sp}=[C{d}^{2+}][O{H}^{-}{]}^2$$ 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 $K_{sp}=(s)(2s{)}^2$

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.

Determine the relationship between Ksp and Kf

The complex formation reaction is: $$C{d}^{2+}+4B{r}^{-}\stackrel{formation}{\to }CdB{r}_4^{2-}$$ 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.

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-]: $1.25\times {10}^{-10}=\frac{[(s)(1.0\times {10}^{-3})]}{(4s)([B{r}^{-}]{)}^4}$ Rearrange for [Br-]: $[[B{r}^{-}]=\sqrt[4]{\frac{(s)(1.0\times {10}^{-3})}{(4s)(1.25\times {10}^{-10})}}$ Substituting the value of s from step 2: $[[B{r}^{-}]=\sqrt[4]{\frac{(6.48\times {10}^{-6})(1.0\times {10}^{-3})}{(4)(6.48\times {10}^{-6})(1.25\times {10}^{-10})}}$ Calculate the value of [Br-]: $[B{r}^{-}]\approx 0.063\frac{mol}{L}$ 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.