In qualitative analysis, Bi³⁺ is detected by appearance of precipitate of BiO(OH). Calculate pH when the following equilibrium exists at 298K:

BiO(OH)(s) ⇋ BiO⁺(aq)+OH⁻(aq)

K =4×10⁻¹⁰, log2=0.3010

For the dissolution equilibrium

$BiO(OH)(s)\; \rightleftharpoons \; BiO+(aq)\; +\; OH-(aq)$

the solid does not appear in the equilibrium expression, so the thermodynamic constant is

$K\; =\; [BiO+][OH-]$

Because one mole of solid produces one mole of each ion, the concentrations of the two ions are equal at equilibrium. Let

$[OH-]\; =\; [BiO+]\; =\; x$

Then

$K\; =\; x^2$

Given $K\; =\; 4\; \times \; 10-10$, solve for $x$:

$x\; =\; \surd K\; =\; \surd (4\; \times \; 10-10)\; =\; \surd 4\; \times \; \surd (10-10)\; =\; 2\; \times \; 10-5 M$

Now calculate the pOH:

$pOH\; =\; -log$₁₀[OH⁻] = −log(2 × 10⁻⁵)

Use the logarithm property $log(ab)\; =\; log\; a\; +\; log\; b$ and the given value $log 2\; =\; 0.3010$:

$log(2\; \times \; 10-5)\; =\; log 2\; +\; log 10-5=\; 0.3010\; -\; 5\; =\; -4.6990$

Therefore

$pOH\; =\; -(-4.6990)\; =\; 4.699$

Finally, obtain the pH from the relation $pH\; +\; pOH\; =\; 14$ (at 298 K):

$pH\; =\; 14\; -\; pOH\; =\; 14\; -\; 4.699\; =\; 9.301$

The calculated pH is 9.301.