[" 63."hydrolysis of "0.1MCH_(3)COONH_(4)" ,when "],[K_(3)=K_(6)=1.8times10^(-5)" is "]

% hydrolysis of 0.1M CH_(3)COONH_(4), when K_(a)(CH_(3)COOH)=K_(b)(NH_(4)OH)=1.8xx10^(-5) is:

% hydrolysis of 0.1M CH_(3)COONH_(4), when K_(a)(CH_(3)COOH)=K_(b)(NH_(4)OH)=1.8xx10^(-5) is:

% hydrolysis of 0.1M CH_(3)COONH_(4), when K_(a)(CH_(3)COOH)=K_(b)(NH_(4)OH)=1.8xx10^(-5) is: (a) 0.55 (b) 7.63 (c) 0.55xx10^(-2) (d) 7.63xx10^(-3)

Calculate the extent of hydrolysis & the pH of 0.02 M CH_3COONH_4 . [K_(b) (NH_(3))=1.8 xx 10^(-5), K_(a) (CH_(3)COOH)=1.8 xx 10^(-5)]

Calculate the percentage hydrolysis & the pH of 0.02M CH_(3)COONH_4 . K_(b)(NH_(3))-1.6xx10^(-5),K_(a)(CH_(3)COOH)=1.6xx10^(-5)

Calculate the extent of hydrolysis and the pH of 0.02 M CH_3COONH_4. [K_b (NH_3)=1.8×10^(−5) , K_a (CH 3COOH)=1.8×10^(−5) ]

pH of a solution of 0.1 M [CH_3COONH_4(aq)] is [given: K_a(CH_3COOH) = K_b(NH_4OH) = 1.8 x 10^-5)]