MHT CET · Chemistry · Ionic Equilibrium
Calculate the pH of 0.02 M monobasic acid having \(2 \%\) dissociation.
- A 3.4
- B 4.5
- C 5.1
- D 5.8
Answer & Solution
Correct Answer
(A) 3.4
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{HA} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-} \\ & {[\mathrm{H}]^{+}=\alpha \mathrm{C}=\frac{2}{100} \times 0.02 \mathrm{~m}=0.0004 \mathrm{~m}} \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log (0.0004)=3.4\end{aligned}\)
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