Problem 35 - Chemistry

Question

Calculate the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=5.9 \times 10^{-15}.\)

Step-by-Step Solution

Answer

The molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) is approximately \(1.44\times10^{-5}\) mol/L.

Step 1: Write the balanced dissolution reaction and corresponding equilibrium expression

The dissolution reaction for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is given by: \[ \mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(aq) + 2\mathrm{OH}^-(aq). \] The corresponding equilibrium expression for Ksp is given by: \[ K_{sp}=[\mathrm{Cd}^{2+}][\mathrm{OH}^-]^2, \] where \([\mathrm{Cd}^{2+}]\) and \([\mathrm{OH}^-]\) represent the equilibrium concentrations of cadmium ions and hydroxide ions, respectively.

Step 2: Set up a reaction table

To determine the equilibrium concentrations, we need to set up a reaction table: | | \(\mathrm{Cd}^{2+}\) | \(\mathrm{OH}^-\) | |-------------|-------------|-----------| | Initial | 0 | 0 | | Change | +s | +2s | | Equilibrium | s | 2s | where s represents the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\). Now, substitute the equilibrium concentrations into the equilibrium expression: \[ K_{sp}=(s)(2s)^{2}. \]

Step 3: Solve for the molar solubility

Plug the given Ksp value into the equation: \(5.9\times10^{-15}=(s)(2s)^{2}\). Solving for s: \[ 5.9\times10^{-15}=s(4s^2)\Rightarrow s^3=\frac{5.9\times10^{-15}}{4}, \] so: \[ s=\sqrt[3]{\frac{5.9\times10^{-15}}{4}} \approx 1.44\times10^{-5}. \] The molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) is approximately \(1.44\times10^{-5}\) mol/L.

Key Concepts

Equilibrium Expression

In chemistry, the equilibrium expression is a mathematical representation of the concentrations of reactants and products in a chemical equilibrium. For our specific case of the dissolution of \(\mathrm{Cd}(\mathrm{OH})_{2}\), the chemical equation is formulated as \(\mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(aq) + 2\mathrm{OH}^-(aq)\). Here, the substances are in equilibrium, which means there is a constant ratio of the products' concentrations to the reactants'.

The expression for the solubility product constant \(K_{sp}\) is derived by taking the concentrations of the ions formed. In our example:

• \([\mathrm{Cd}^{2+}]\) is the concentration of cadmium ions in equilibrium.
• \([\mathrm{OH}^-]^2\) is the squared concentration of hydroxide ions, since two hydroxide ions are produced for each cadmium ion.

The equilibrium expression for this reaction is: \(K_{sp}=[\mathrm{Cd}^{2+}][\mathrm{OH}^-]^{2}\). This helps us predict how many of the solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) can dissolve in water until the system reaches equilibrium.

Ksp (solubility product constant)

The solubility product constant, \(K_{sp}\), is a crucial concept in understanding solubility in chemistry. It provides insight into how soluble a compound will be in water. Specifically, \(K_{sp}\) is the product of the molar concentrations of the ions, each raised to the power of its coefficient in the balanced equation.

For \(\mathrm{Cd}(\mathrm{OH})_{2}\), the \(K_{sp}\) value is given as \(5.9 \times 10^{-15}\). This low \(K_{sp}\) value suggests that \(\mathrm{Cd}(\mathrm{OH})_{2}\) is not very soluble in water, as only a small amount will dissolve before reaching saturation and establishing equilibrium.

• \(K_{sp}\) is specific to temperature; the same substance may have different solubility at different temperatures.
• This constant allows chemists to predict the point at which a solution will become saturated and precipitation will begin.

Understanding \(K_{sp}\) helps in contexts like predicting mineral formation in natural waters or designing pharmaceuticals to ensure they dissolve properly in the body.

Reaction Table

A reaction table is a simple yet effective tool used in chemistry to keep track of changes in concentrations of species as a reaction progresses towards equilibrium. It breaks down the stages from initial concentrations, through changes, to equilibrium concentrations. In our example of \(\mathrm{Cd}(\mathrm{OH})_{2}\), the reaction table helps us determine the extent to which the compound will dissolve.

The table is set up as follows:

• Initial concentrations: Start with zero concentrations of \(\mathrm{Cd}^{2+}\) and \(\mathrm{OH}^-\) because no reaction has occurred initially.
• Change (using variable \(s\)): The concentration of \(\mathrm{Cd}^{2+}\) increases by \(s\), and \(\mathrm{OH}^-\) increases by \(2s\) as each molecule of \(\mathrm{Cd}(\mathrm{OH})_{2}\) yields two hydroxide ions.
• Equilibrium concentrations: We use \(s\) and \(2s\) for \(\mathrm{Cd}^{2+}\) and \(\mathrm{OH}^-\) respectively, obtaining these by substituting back into the \(K_{sp}\) equation.

This systematic approach using a reaction table ensures clarity when solving equilibrium problems, helping us pinpoint how much of a substance can be expected to dissolve under given conditions.