Question

In a qualitative analysis, \(Bi^{3+}\) is detected by appearance of precipitate of \(BiO(OH)\). Calculate pH when the following equilibrium exists at \(298K\): \[ BiO(OH)(s)\rightleftharpoons BiO^+(aq)+OH^-(aq) \] \[ K=4\times10^{-10} \] \[ \text{Given: }\log2=0.3010 \]

• A. \(4.699\)
• B. \(9.301\)
• C. \(5.286\)
• D. \(8.714\)

Hint:

For sparingly soluble hydroxide-type equilibria, calculate \([OH^-]\), then find \(pOH\), and finally use \(pH=14-pOH\).

Answer 1

The Correct Option is B

Step 1: Understanding the Topic:
This problem focuses on "Ionic Equilibrium," specifically the solubility of sparingly soluble bases and the resulting pH of their saturated solutions. The equilibrium constant $K$ provided is equivalent to the solubility product constant ($K_{sp}$) for Bismuth oxyhydroxide. When this solid dissolves, it produces hydroxide ions, making the solution basic. Our goal is to determine the concentration of these hydroxide ions to find the alkalinity of the system.
Step 2: Key Formulas and Approach:
For a dissolution equilibrium of the type $AB(s) \rightleftharpoons A^+(aq) + B^-(aq)$:
$K = [A^+][B^-]$
If $s$ is the molar solubility, then $[A^+] = s$ and $[B^-] = s$.
$pOH = -\log[OH^-]$
$pH = 14 - pOH$ (at 298 K)

Step 3: Detailed Explanation:

Setup Concentrations: From the stoichiometry of the reaction $BiO(OH) \rightleftharpoons BiO^+ + OH^-$, one mole of solid produces one mole of $BiO^+$ and one mole of $OH^-$. If the solubility is $s$, then $[BiO^+] = s$ and $[OH^-] = s$.
Calculate Solubility: \[ K = s \times s = s^2 = 4 \times 10^{-10} \] \[ s = \sqrt{4 \times 10^{-10}} = 2 \times 10^{-5} \text{ mol/L} \]
Calculate pOH: Since $[OH^-] = 2 \times 10^{-5} \text{ M}$: \[ pOH = -\log(2 \times 10^{-5}) = -(\log 2 + \log 10^{-5}) \] \[ pOH = -(0.3010 - 5) = 4.699 \]
Calculate pH: At standard temperature (298 K), the sum of pH and pOH is 14. \[ pH = 14 - 4.699 = 9.301 \]
Step 4: Final Answer:
The pH of the solution is 9.301, which corresponds to option (D).