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The pH of the solution containing $\mathrm{50 \mathrm{~mL}}$ each of $\mathrm{ 0.10 \mathrm{M}}$ sodium acetate and $\mathrm{ 0.01 \mathrm{M}}$ acetic acid is [Given $\mathrm{pK}_{a} \mathrm{of} \: \mathrm{CH}_{3} \mathrm{COOH}={4.57}$ ]

Option: 1
$3.57$

Option: 2
$4.57$

Option: 3
$2.57$

Option: 4
$5.57$

Answers (1)

best_answer

Given salt ( sodium acetate ), $\mathrm{M=0.10,V=50mL}$ Weak acid

$\mathrm{\left ( CH_{3}COOH \right ),pKa=4.57}$ and $\mathrm{M=0.01}$ So, weak acid and salt form an

acidic buffer

Then, $\mathrm{p{H} =p K_{a}+\log \frac{[\text { salt] }]}{[\text { Acid] }}} \\$

$\mathrm{p{H} =4.57+\log \frac{(0.1)}{(0.01)}} \\$

$\mathrm{p{H} =5.57 }$

Trick: acetate ion is a base and if we add base in acid the $\mathrm{p{H} }$ of solution will

increase. So, $\mathrm{p{H} }$ of solution will be greater than $4.57$

Hence the correct option is 4.

Posted by

vinayak

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