MHT CET · Chemistry · Ionic Equilibrium
The degree of dissociation of a \(0.01 \mathrm{M}\) weak acid is \(10^{-3}\). Its \(\mathrm{pOH}\) is
- A \(\underline{5}\)
- B 3
- C 9
- D 11
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
\(\left[\mathrm{H}^{+}\right]=\sqrt{K_{a} \cdot C}=139\)
Or \(=\sqrt{\alpha^{2} \cdot C^{2}} \quad\left[\therefore K_{a}=\alpha^{2} C\right]\)
\(= \sqrt{\left(10^{-3}\right)^{2} \cdot(0.01)^{2}} \)
\( = 10^{-5} \)
\( \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] =-\log 10^{-5} \)
\( =5\)
\(\mathrm{pOH}=14-5=9\)
Or \(=\sqrt{\alpha^{2} \cdot C^{2}} \quad\left[\therefore K_{a}=\alpha^{2} C\right]\)
\(= \sqrt{\left(10^{-3}\right)^{2} \cdot(0.01)^{2}} \)
\( = 10^{-5} \)
\( \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] =-\log 10^{-5} \)
\( =5\)
\(\mathrm{pOH}=14-5=9\)
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