Problem 8 - General Chemistry: Principles and Modern Applications

Question

In the qualitative cation analysis procedure, \(\mathrm{Bi}^{3+}\) is detected by the appearance of a white precipitate of bismuthyl hydroxide, \(\mathrm{BiOOH}(\mathrm{s})\): \(\mathrm{BiOOH}(\mathrm{s}) \rightleftharpoons \mathrm{BiO}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\) \(K_{\mathrm{sp}}=4 \times 10^{-10}\) Calculate the \(\mathrm{pH}\) of a saturated aqueous solution of \(\mathrm{BiOOH}\).

Step-by-Step Solution

Answer

The pH of a saturated aqueous solution of BiOOH is 9.30.

Step 1: Write the equilibrium expression

The equilibrium reaction can be written as: \(\mathrm{BiOOH}(\mathrm{s}) \rightleftharpoons\mathrm{BiO}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\). The solubility product expression becomes: \[ K_{sp} = [\mathrm{BiO}^+][\mathrm{OH}^-] \]

Step 2: Set up the solubility relationship

Let's denote the solubility of BiOOH as \(s\). Then, \([\mathrm{BiO}^+]=s\) and \([\mathrm{OH}^-]=s\). Substituting these into the \(K_{sp}\) expression we get: \( K_{sp}=s^2 = 4 \times 10^{-10}\). Solving for \(s\) we get \(s=\sqrt{4 \times 10^{-10}} = 2 \times 10^{-5}\) M

Step 3: Calculating pH

Now, since \(\mathrm{pOH} = -\log[\mathrm{OH}^-]\), we can substitute \(s = [\mathrm{OH}^-]\) into the formula to calculate the \(\mathrm{pOH} = -\log(2 \times 10^{-5}) \approx 4.70\). Then, because \(\mathrm{pOH} + \mathrm{pH} = 14\) in water at 25 C, we can substitute the \(\mathrm{pOH}\) into the equation to get \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 4.70 = 9.30\).

Step 4: Rounding to correct significant figures

Following the significant figure rules, the final answer can be rounded to two decimal places, giving a \(\mathrm{pH} = 9.30\).

Key Concepts

Solubility Product Constant

The solubility product constant, denoted as \( K_{sp} \), is a special type of equilibrium constant. It helps us understand the solubility of sparingly soluble salts in water. When a salt dissolves in water, it breaks down into its constituent ions.
For example, the dissolution of bismuthyl hydroxide \( \text{BiOOH} \) can be represented by the equation: - \( \text{BiOOH(s)} \rightleftharpoons \text{BiO}^+(\text{aq}) + \text{OH}^-(\text{aq}) \)
This equation shows the formation of bismuthyl ions \( \text{BiO}^+ \) and hydroxide ions \( \text{OH}^- \) in solution. The solubility product constant is then given by: - \( K_{sp} = [\text{BiO}^+][\text{OH}^-] \)
\( K_{sp} \) provides a measure of the extent to which a salt can dissolve in water, with a smaller \( K_{sp} \) value indicating lower solubility.

Bismuthyl Hydroxide

Bismuthyl hydroxide, commonly written as \( \text{BiOOH} \), is a compound that appears as a white precipitate. It is involved in the qualitative analysis of cations, especially in detecting the presence of \( \text{Bi}^{3+} \) ions.
Bismuthyl hydroxide is not highly soluble in water, which is why it forms a precipitate rather than dissolving completely. When added to a solution, it establishes an equilibrium between the solid \( \text{BiOOH} \) and its ions in solution:- \( \text{BiOOH(s)} \rightleftharpoons \text{BiO}^+(\text{aq}) + \text{OH}^-(\text{aq}) \)
This reversible reaction is key to understanding how bismuthyl hydroxide behaves in solutions and how we can use its properties in chemical analysis.

Equilibrium Expression

An equilibrium expression relates the concentrations of reactants and products of a reaction at equilibrium. For bismuthyl hydroxide, the relevant expression based on its dissolution in water is:- \( K_{sp} = [\text{BiO}^+][\text{OH}^-] \)
This expression helps in calculating the solubility of the compound. By setting it equal to the known \( K_{sp} \) value, we can solve for the solubility \( s \) of \( \text{BiOOH} \).
- Let \( [\text{BiO}^+] = s \)- Let \( [\text{OH}^-] = s \)
Substitute these into the equilibrium expression, we have \( s^2 = 4 \times 10^{-10} \). Solving this gives \( s = 2 \times 10^{-5} \) M, meaning \( \text{BiOOH} \) dissolves at this solubility in water.

pH Calculation

Calculating the pH of a solution is an important skill in chemistry. pH is a measure of how acidic or basic a solution is.
When we have \( \text{BiOOH} \) in a saturated solution, we first calculate the concentration of hydroxide ions \( [\text{OH}^-] \), which equals the solubility \( s = 2 \times 10^{-5} \) M.
- To find pOH, use the formula: \( \text{pOH} = -\log[\text{OH}^-] \)- Substitute the value: \( \text{pOH} = -\log(2 \times 10^{-5}) \approx 4.70 \)
Now, pH and pOH are related by the equation \( \text{pOH} + \text{pH} = 14 \). Solving for pH:- \( \text{pH} = 14 - \text{pOH} = 14 - 4.70 = 9.30 \)
This calculation reveals that a saturated solution of \( \text{BiOOH} \) is mildly basic.