20 mL of 0.2 M NaOH is added to 50 mL of 0.2 M acetic acid. The pH of this solution after mixing is (Ka = 1.8 × 10-5)

Question

20 mL of 0.2 M NaOH is added to 50 mL of 0.2 M acetic acid. The pH of this solution after mixing is (Ka = 1.8 × 10-5) (A) 4.5 (B) 2.3 (C) 3.8 (D) 4

Tardigrade Admin · Accepted Answer

NaOH + CH3COOH -> CH3COONa + H2O <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/60f86269f3ae08891bd48962e0738737-.png" /> pH =- log (1.8 × 10-5)+ log (4 / 70/6 / 70) =4.56

Q. $20 m L$ of $0.2 M N a O H$ is added to $50 m L$ of $0.2 M$ acetic acid. The pH of this solution after mixing is $(K a = 1.8 \times 1 0^{- 5})$

4.5

2.3

3.8

4

$NaOH + CH X_{3} COOH CH X_{3} COONa + H X_{2} O$

$p H = - lo g (1.8 \times 1 0^{- 5}) + lo g \frac{4/70}{6/70}$
$= 4.56$

Q. 20mL of 0.2MNaOH is added to 50mL of 0.2M acetic acid. The pH of this solution after mixing is (Ka=1.8×10−5)

4.5

2.3

3.8

4

NaOH+CHX3​COOH​CHX3​COONa+HX2​O pH=−log(1.8×10−5)+log6/704/70​ =4.56

Q. $20 m L$ of $0.2 M N a O H$ is added to $50 m L$ of $0.2 M$ acetic acid. The pH of this solution after mixing is $(K a = 1.8 \times 1 0^{- 5})$

$NaOH + CH X_{3} COOH CH X_{3} COONa + H X_{2} O$

$p H = - lo g (1.8 \times 1 0^{- 5}) + lo g \frac{4/70}{6/70}$
$= 4.56$