TS EAMCET · Chemistry · Ionic Equilibrium
The \(\mathrm{pH}\) of a buffer solution made by mixing \(25 \mathrm{~mL}\) of \(0.02 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}\) and \(25 \mathrm{~mL}\) of \(0.2 \mathrm{M}\) \(\mathrm{NH}_4 \mathrm{Cl}\) at \(25^{\circ}\) is \(\left(\mathrm{pK}\right.\) of \(\mathrm{NH}_4 \mathrm{OH}=4.8\) )
- A \(5.8\)
- B \(8.2\)
- C \(4.8\)
- D \(3.8\)
Answer & Solution
Correct Answer
(B) \(8.2\)
Step-by-step Solution
Detailed explanation
As a mixture of \(\mathrm{NH}_4 \mathrm{OH}\) and \(\mathrm{NH}_4 \mathrm{Cl}\) acts as a basic buffer, so its pH must be basic, (i.e., greater than 7), hence the answer must be 2 nd. It can also be find by calculations :…
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