MHT CET · Chemistry · Ionic Equilibrium
Ammonium acetate which is \(0.01 \quad \mathrm{M}\), is hydrolysed to \(0.001 \mathrm{M}\) concentration. Calculate the change in \(\mathrm{pH}\) in \(0.001 \mathrm{M}\) solution, if initially \(\mathrm{pH}=\mathrm{pK}_{a}\)
- A 5
- B 10
- C 100
- D 1
Answer & Solution
Correct Answer
(D) 1
Step-by-step Solution
Detailed explanation
\(\mathrm{CH}_{3} \mathrm{COONH}_{4}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}~+\) \(\mathrm{NH}_{4} \mathrm{OH} \)
\( 0.01 \mathrm{M} 0.001 \mathrm{M} \)
\( \mathrm{pH}=\mathrm{p} K_{a}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONH}_{4}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \)
\( =\mathrm{p} K_{a}+\log \left[\frac{0.01}{0.001}\right] \)
\( =\mathrm{p} K_{a}+\log 10 \)
\( =\mathrm{p} K_{a}+1 \)
\( \therefore \text { Change in } \mathrm{pH}=1\)
\( 0.01 \mathrm{M} 0.001 \mathrm{M} \)
\( \mathrm{pH}=\mathrm{p} K_{a}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONH}_{4}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \)
\( =\mathrm{p} K_{a}+\log \left[\frac{0.01}{0.001}\right] \)
\( =\mathrm{p} K_{a}+\log 10 \)
\( =\mathrm{p} K_{a}+1 \)
\( \therefore \text { Change in } \mathrm{pH}=1\)
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