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: Write the equilibrium expression.
For: \[ BiO(OH)(s)\rightleftharpoons BiO^+(aq)+OH^-(aq), \] solid is not included in equilibrium expression. So: \[ K=[BiO^+][OH^-]. \]

Step 2: Let solubility be \(s\).
From the reaction: \[ [BiO^+]=s \] and: \[ [OH^-]=s. \] Therefore: \[ K=s^2. \]

Step 3: Substitute the value of \(K\).
\[ s^2=4\times10^{-10}. \] Taking square root: \[ s=2\times10^{-5}. \] So: \[ [OH^-]=2\times10^{-5}. \]

Step 4: Calculate pOH.
\[ pOH=-\log[OH^-]. \] \[ pOH=-\log(2\times10^{-5}). \] \[ pOH=-(\log2+\log10^{-5}). \] \[ pOH=-(0.3010-5). \] \[ pOH=4.699. \]

Step 5: Calculate pH.
At \(298K\): \[ pH+pOH=14. \] \[ pH=14-4.699. \] \[ pH=9.301. \] Therefore, the pH is: \[ 9.301. \]