MHT CET · Chemistry · Ionic Equilibrium
A buffer solution is prepared by mixing \(0.2 \mathrm{M} \mathrm{NH} \mathrm{O}_4 \mathrm{OH}\) and \(1 \mathrm{M} \mathrm{NH}_4 \mathrm{Cl}\). What is the pH value of buffer solution? (Give \(\mathrm{pK}_{\mathrm{b}}=7.744\) )
- A 5.56
- B 6.99
- C 8.44
- D 9.56
Answer & Solution
Correct Answer
(A) 5.56
Step-by-step Solution
Detailed explanation
For a basic buffer solution,
\(\begin{aligned}
\mathrm{pOH} & =\mathrm{pK}_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]} \\
& =7.744+\log _{10} \frac{1}{0.2} \\
& =7.744+0.698 \\
\therefore \quad \mathrm{pOH} & =8.44 \\
\therefore \quad \mathrm{Now} & , \mathrm{pH}+\mathrm{pOH}=14 \\
\therefore \quad \mathrm{pH} & =14-\mathrm{pOH}=14-8.44=5.56
\end{aligned}\)
\(\begin{aligned}
\mathrm{pOH} & =\mathrm{pK}_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]} \\
& =7.744+\log _{10} \frac{1}{0.2} \\
& =7.744+0.698 \\
\therefore \quad \mathrm{pOH} & =8.44 \\
\therefore \quad \mathrm{Now} & , \mathrm{pH}+\mathrm{pOH}=14 \\
\therefore \quad \mathrm{pH} & =14-\mathrm{pOH}=14-8.44=5.56
\end{aligned}\)
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