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 \(: \log 2=0.3010\) )

(D) 9.301
The equilibrium reaction is given by:
\(BiO(OH)(s) \rightleftharpoons BiO^{+}(aq)+OH^{-}(aq)\)
The equilibrium constant expression is:
\(K=\left[BiO^{+}\right]\left[OH^{-}\right]\)
Let the solubility of \(BiO ( OH )\) be \(s\). Then \(\left[ BiO ^{+}\right]=s\) and \(\left[ OH ^{-}\right]=s\).
Substituting the values into the equilibrium expression:
\(s^2=4 \times 10^{-10}\)
\(s=2 \times 10^{-5} M\)
Therefore, the concentration of hydroxide ions is:
\(\left[OH^{-}\right]=2 \times 10^{-5} M\)
Calculating the pOH :
\(pOH=-\log \left[OH^{-}\right]=-\log \left(2 \times 10^{-5}\right)=5-\log 2\)
Given \(\log 2=0.3010:\)
\(pOH=5-0.3010=4.699\)
The pH of the solution at 298 K is:
\(pH=14-pOH=14-4.699=9.301\)