A) \[5+\log 2-\log 3\]
B) \[5-\log 2\]
C) \[5-\log 3\]
D) \[5-\log 6\]
Correct Answer: B
Solution :
\[{{K}_{a}}={{10}^{-5}}\Rightarrow p{{K}_{a}}=-{{\operatorname{logK}}_{a}}=-\log {{10}^{-5}}=5\] \[\text{Initial}\,\underset{\text{1}\,\text{mole}}{\mathop{\text{HA}}}\,+\underset{0}{\mathop{\text{NaOH}}}\,\xrightarrow{{}}\underset{0}{\mathop{\text{NaA}}}\,+\underset{0}{\mathop{{{\text{H}}_{\text{2}}}\text{O}}}\,\] Final \[\left( 1-\frac{1}{3} \right)\,\text{mol}\] \[\frac{1}{3}\text{mol}\] \[\frac{\text{1}}{\text{3}}\,\text{mol}\] \[=\frac{2}{3}\,\text{mol}\] \[\frac{1}{3}\,\text{mol}\] (Assumed weak acid to be monoprotic, since only one dissociation constant value is provided.) Final solution acts as an acidic buffer. \[pH=p{{K}_{a}}+\log \frac{[salt]}{[acid]}\] or \[pH=5+\log \frac{\frac{1}{3}}{\frac{2}{3}}\] \[=5+\log \frac{1}{2}\] \[\therefore \] \[pH=5-\log 2\]
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