Question: $K_{a}$for $CH_{3}COOH$ is $1.8 \times 10^{- 5}$and $K_{b}$ for $NH_{4}OH$ is \(1.8 \times...

Question

$K_{a}$ 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.

Answer 1

: $CH_{3}COONH_{4}$ is a salt of weak acid $(CH_{3}COOH)$ and weak base $(NH_{4}OH).$

$pH = 7 + \frac{1}{2}(pK_{a} - pK_{b})$

$pK_{a} = - \log{}K_{a} = - \log{}(1.8 \times 10^{- 5}) = 4.74$

$pK_{b} = - \log{}K_{b} = - \log{}(1.8 \times 10^{- 5}) = 4.74$

$pH = 7 + \frac{1}{2}(4.74 - 4.74) = 7.0$

Question: \(K_{a}\)for \(CH_{3}COOH\) is \(1.8 \times 10^{- 5}\)and \(K_{b}\) for \(NH_{4}OH\) is \(1.8 \times...