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Question

Chemistry Question on Equilibrium

KaK_a for CH3COOHCH_3COOH is 1.8×1051.8 \times 10^{-5} and KbK_b for NH4OHNH_4OH is 1.8×1051.8 \times 10^{-5}. The pHpH of ammonium acetate will be

A

7.0057.005

B

4.754.75

C

7.07.0

D

between 66 and 77

Answer

7.07.0

Explanation

Solution

CH3COONH4CH_3COONH_4 is a salt of weak acid (CH3COOH)(CH_3COOH) and weak base (NH4OH)(NH_4OH). pH=7+12(pKapKb)pH=7+\frac{1}{2}\left(pK_{a}-pK_{b}\right) pKa=logKapK_{a}=-log\,K_{a} =log(1.8×105)=4.74=-log\left(1.8\times10^{-5}\right)=4.74 pKb=logKbpK_b = - log \,K_b =log(1.8×105)=4.74= - log (1.8 \times 10^{-5}) = 4.74 pH=7+12(4.744.74)=7.0pH=7+\frac{1}{2}\left(4.74-4.74\right)=7.0