AP EAMCET · Chemistry · Ionic Equilibrium
The \(\mathrm{pH}\) of \(0.01 \mathrm{M} \mathrm{BOH}\) solution is 10 . What is its degree of dissociation?
(Given \(\mathrm{K}_{\mathrm{b}}\) of \(\mathrm{BOH}\) is \(1 \times 10^{-6}\) )
- A \(10 \%\)
- B \(5 \%\)
- C \(2 \%\)
- D \(1 \%\)
Answer & Solution
Correct Answer
(A) \(10 \%\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{B}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{BOH}]}=1 \times 10^{-6} \\ & \mathrm{pH}=10 \Rightarrow \mathrm{pOH}=14-10=4 \\ &…
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