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

Both conc. Being 1 M, the factor by which the percent hydrolysis of CH_3COO^- is greater in CH_3COONH_4 than in CH_3COONa is : [K_a(CH_3COOH)=K_b(NH_4OH)=1.8xx10^(-5)]

Carefully observe the given figure and using data provided find the EMF of shown Galvenic cell in volt: Solution A is 0.1M each in NH_(4)OH and NH_(4)CI and solution B is 0.1M each in CH_(3)COOH and CH_(3)COONa^(+) . [Given: K_(a)(CH_(3)COOH) = 10^(-5), K_(b) (NH_(4)OH) = 10^(-5) and (2.303RT)/(F) = 0.06 volt]

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)]

What is the concentration of CH_(3)COOH(aq.) in a solution prepared by dissolving 0.01 mole of NH_(4)^(+)CH_(3)COO^(-) in 1 L H_(2)O ? [K_(a(CH_(3)COOH))=1.8xx10^(-5)),K_(b(NH_(4)OH)=1.8xx10^(-5))]

Which of the following cell can produce maximum electrical work? ["Given" : P_(H_(2))=1 atm, K_(a)(CH_(3)COOH)=K_(b)(NH_(4)OH)=10^(-5)]

In a solution containing 0.02M CH_(3)COOH and 0.01M C_(6)H_(5)COOH . If K_(CH_(3)COOH) and K_(C_(6)H_(5)COOH) are 1.8xx10^(-5) and 6.4xx10^(-5) respectively, then calculate the pH of solution.

Calculate the pH of a solution containing 0.1 M CH_(3)COOH and 0.15 M CH_(3)COO^(-) . (K_(a) "of" CH_(3)COOH=1.8xx10^(-5))