Ncert Exercises & Solutions

$K_{a}$ for ${CH}_{3} COOH$ is $1.8 \times 10^{- 5}$ and K_b for ${NH}_{4} OH$ is $1.8 \times 10^{- 5}$ . The pH of ammonium acetate will be:

1. 7.005

2. 4.75

3. 7.0

4. Between 6 and 7

Hint: Here, the calculation of pH occurs in terms of pK_a and pK_b

Explanation:

Step 1:

The following expression is used to calculate the pH of ammonium acetate

pH = 7 + \frac{{pK}_{a} - {pK}_{b}}{2}

Given that,

K_{a}

for

{CH}_{3} COOH = 1.8 \times 10^{- 5}

K_{b}

for

{NH}_{4} OH = 1.8 \times 10^{- 5}

Step 2:

Ammonium acetate is a salt of a weak acid and weak acid. For such salts

pH = 7 + \frac{{pK}_{a} - {pK}_{b}}{2}

= 7 + [- \log

1.8 \times 10^{- 5}] - [- \log

1.8 \times 10^{- 5}] 2

= 7 + \frac{4.74 - 4.74}{2} = 7.00