20 mL 0.1( N ) acetic acid is mixed with 10 mL 0.1( N ) solution of NaOH. The pH of the resulting solution is ( p Ka. of acetic acid is 4.74 )

Question

20 mL 0.1( N ) acetic acid is mixed with 10 mL 0.1( N ) solution of NaOH. The pH of the resulting solution is ( p Ka. of acetic acid is 4.74 ) (A) 3.74 (B) 4.74 (C) 5.74 (D) 6.74

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/5cad1210271d3f8f4963b737b0f94189-.png" /> From Henderson's equation, pH = p Ka+ log ([ CH 3 COONa ]/[ CH 3 COOH ]) =4.74+ log (1/1)=4.74

Q. $20 m L 0.1 (N)$ acetic acid is mixed with $10 m L$ $0.1 (N)$ solution of $N a O H$ . The $p H$ of the resulting solution is $(p K_{a}$ of acetic acid is $4.74$ )

3.74

4.74

5.74

6.74

From Henderson's equation,
$p H = p K_{a} + lo g \frac{[ C H _{3} COON a ]}{[ C H _{3} COO H ]}$
$= 4.74 + lo g \frac{1}{1} = 4.74$

Q. 20mL0.1(N) acetic acid is mixed with 10mL 0.1(N) solution of NaOH. The pH of the resulting solution is (pKa​ of acetic acid is 4.74 )