MHT CET · Chemistry · Ionic Equilibrium
A buffer solution is prepared by mixing equimolar acetic acid and sodium acetate. If ' \(\mathrm{K}_{\mathrm{a}}\) ' of acetic acid is \(1.78 \times 10^{-5}\), find the \(\mathrm{pH}\) of buffer solution.
- A 4.75
- B 8.9
- C 9.4
- D 2.6
Answer & Solution
Correct Answer
(A) 4.75
Step-by-step Solution
Detailed explanation
For acidic buffer,
\(\mathrm{pH} =\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]}=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{1}{1}=\mathrm{pK}_{\mathrm{a}} \)
\( \mathrm{pK}_{\mathrm{a}} =-\log _{10} \mathrm{~K}_{\mathrm{a}}=-\log _{10}\left(1.78 \times 10^{-5}\right) \)
\( =-\left(\log _{10} 1.78+\log _{10} 10^{-5}\right) \)
\( =-(0.25-5)=4.75\)
Therefore, \(\mathrm{pH}\) of buffer solution \(=4.75\)
\(\mathrm{pH} =\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]}=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{1}{1}=\mathrm{pK}_{\mathrm{a}} \)
\( \mathrm{pK}_{\mathrm{a}} =-\log _{10} \mathrm{~K}_{\mathrm{a}}=-\log _{10}\left(1.78 \times 10^{-5}\right) \)
\( =-\left(\log _{10} 1.78+\log _{10} 10^{-5}\right) \)
\( =-(0.25-5)=4.75\)
Therefore, \(\mathrm{pH}\) of buffer solution \(=4.75\)
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