Ka for CH3COOH is 1.8 × 10-5 and Kb for NH4OH is 1.8 × 10-5. The pH of ammonium acetate will be

Question

Ka for CH3COOH is 1.8 × 10-5 and Kb for NH4OH is 1.8 × 10-5. The pH of ammonium acetate will be (A) 7.005 (B) 4.75 (C) 7.0 (D) between 6 and 7

Tardigrade Admin · Accepted Answer

CH3COONH4 is a salt of weak acid (CH3COOH) and weak base (NH4OH). pH=7+(1/2)(pKa-pKb) pKa=-log Ka =-log(1.8×10-5)=4.74 pKb = - log Kb = - log (1.8 × 10-5) = 4.74 pH=7+(1/2)(4.74-4.74)=7.0

Q. $K_{a}$ for $C H_{3} COO H$ is $1.8 \times 1 0^{- 5}$ and $K_{b}$ for $N H_{4} O H$ is $1.8 \times 1 0^{- 5}$ . The $p H$ of ammonium acetate will be

$7.005$

$4.75$

$7.0$

between $6$ and $7$

Q. Ka​ for CH3​COOH is 1.8×10−5 and Kb​ for NH4​OH is 1.8×10−5. The pH of ammonium acetate will be

7.005

4.75

7.0

between 6 and 7

CH3​COONH4​ is a salt of weak acid (CH3​COOH) and weak base (NH4​OH). pH=7+21​(pKa​−pKb​) pKa​=−logKa​ =−log(1.8×10−5)=4.74 pKb​=−logKb​ =−log(1.8×10−5)=4.74 pH=7+21​(4.74−4.74)=7.0

Q. $K_{a}$ for $C H_{3} COO H$ is $1.8 \times 1 0^{- 5}$ and $K_{b}$ for $N H_{4} O H$ is $1.8 \times 1 0^{- 5}$ . The $p H$ of ammonium acetate will be

$7.005$

$4.75$

$7.0$

between $6$ and $7$