Q67 / 180 · NEET 2026 (Cancelled) · Chemistry

In a qualitative analysis $\text{Bi}^{3+}$ is detected by appearance of precipitate of $\text{BiO(OH)(s)}$. Calculate pH when the following equilibrium exists at $298$ K :
$\text{BiO(OH)(s)} \rightleftharpoons \text{BiO}^{+}\text{(aq)} + \text{OH}^{-}\text{(aq)}$,
$K = 4 \times 10^{-10}$
(Given : $\log 2 = 0.3010$)

  • A
    $8.714$
  • B
    $4.699$
  • C
    $5.286$
  • Correct answer
    D
    $9.301$

Solution

The equilibrium reaction is given by:

$\text{BiO(OH)(s)} \rightleftharpoons \text{BiO}^{+}\text{(aq)} + \text{OH}^{-}\text{(aq)}$

The equilibrium constant expression is:

$K = [\text{BiO}^{+}][\text{OH}^{-}]$

Let the solubility of $\text{BiO(OH)}$ be $s$. Then $[\text{BiO}^{+}] = s$ and $[\text{OH}^{-}] = s$.

Substituting the values into the equilibrium expression:

$s^2 = 4 \times 10^{-10}$

$s = 2 \times 10^{-5} \text{ M}$

Therefore, the concentration of hydroxide ions is:

$[\text{OH}^{-}] = 2 \times 10^{-5} \text{ M}$

Calculating the pOH:

$\text{pOH} = -\log[\text{OH}^{-}] = -\log(2 \times 10^{-5}) = 5 - \log 2$

Given $\log 2 = 0.3010$:

$\text{pOH} = 5 - 0.3010 = 4.699$

The pH of the solution at $298 \text{ K}$ is:

$\text{pH} = 14 - \text{pOH} = 14 - 4.699 = 9.301$

Answer: $9.301$

Hence, the correct answer is option: 'D'.

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