20 ml of 0.1 M NaOH is added to 30 text ml 0.1 text M NaOH 30 text ml 0.2 ( text M CH)3 textCOOH (( textpK) texta = 4.74) , calculate the textpH of the resulting solution? [given: log 2=0.3 ]

Question

20 ml of 0.1 M NaOH is added to 30 text ml 0.1 text M NaOH 30 text ml 0.2 ( text M CH)3 textCOOH (( textpK) texta = 4.74) , calculate the textpH of the resulting solution? [given: log 2=0.3 ] (A) 3.44 (B) 4.01 (C) 4.44 (D) 4.71

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/d05df2a4f794fcf533c3056879896981-.png" /> The resulting solution is an acidic buffer. so textpH = textpK texta + log ([ textSalt]/[ textAcid]) =4.71+log (2/4) =4.44

Q. $20 m l$ of $0.1 M N a O H$ is added to $30 ml 0.1 M NaOH 30 ml 0.2 (M CH)_{3} COOH ((pK)_{a} = 4.74)$ , calculate the $pH$ of the resulting solution? [given: $l o g 2 = 0.3$ ]

3.44

4.01

4.44

4.71

The resulting solution is an acidic buffer. so $pH = pK_{a} + l o g \frac{[ Salt ]}{[ Acid ]}$

$= 4.71 + l o g \frac{2}{4}$

$= 4.44$

Q. 20ml of 0.1MNaOH is added to 30 ml0.1 M NaOH 30 ml0.2( M CH)3​COOH ((pK)a​=4.74) , calculate the pH of the resulting solution? [given: log2=0.3 ]