Question
In a qualitative analysis, $Bi^{3+}$ is detected by appearance of precipitate of $BiO(OH)(s)$. Calculate pH when the following equilibrium exists at 298 K :
$$BiO(OH) (s) \rightleftharpoons BiO^+ (aq) + OH^- (aq)$$
$K = 4 \times 10^{-10}$
(Given : $\log 2 = 0.3010$)
This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
9.301
Equilibrium Calculation for $BiO(OH)$ Dissolution
The dissolution of $BiO(OH)(s)$ establishes the following equilibrium:
$BiO(OH) (s) \rightleftharpoons BiO^+ (aq) + OH^- (aq)$The equilibrium constant ($K$) expression is given by:
$K = [BiO^+][OH^-]$We are given $K = 4 \times 10^{-10}$.
Determining Hydroxide Ion Concentration
Since the solid $BiO(OH)$ does not appear in the equilibrium expression, and assuming initial concentrations of $BiO^+$ and $OH^-$ are zero before dissolution:
- Let $[OH^-] = x$.
- From the stoichiometry, $[BiO^+] = [OH^-] = x$.
- Substituting into the $K$ expression: $K = x \times x = x^2$.
Therefore:
$x^2 = 4 \times 10^{-10}$ $x = \sqrt{4 \times 10^{-10}} = 2 \times 10^{-5} \, \text{M}$So, the hydroxide ion concentration is $[OH^-] = 2 \times 10^{-5}$ M.
Calculating pH from pOH
First, calculate the pOH:
$\text{pOH} = -\log_{10}[OH^-]$ $\text{pOH} = -\log_{10}(2 \times 10^{-5})$ $\text{pOH} = -(\log_{10} 2 + \log_{10} 10^{-5})$ $\text{pOH} = -(\log_{10} 2 - 5)$Using the given value $\log_{10} 2 = 0.3010$:
$\text{pOH} = -(0.3010 - 5)$ $\text{pOH} = 5 - 0.3010$ $\text{pOH} = 4.699$At 298 K, the relationship between pH and pOH is:
$\text{pH} + \text{pOH} = 14$Now, calculate the pH:
$\text{pH} = 14 - \text{pOH}$ $\text{pH} = 14 - 4.699$ $\text{pH} = 9.301$Was this answer helpful?
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