Calculate the e.m.f. of the cell
`Pt|H_(2)(1.0atm)|CH_(3)COOH (0.1M)||NH_(3)(aq,0.01M)|H_(2)(1.0atm)|Pt`
`K_(a) (CH_(3)COOH) =1.8 xx 10^(-5), K_(b),(NH_(3)) = 1.8 xx 10^(-5)`

Neglecting the liquid-liquid junction potential, calculate the emf of the following cell at 25^(@)C H_(2) (1 atm)|0.5 M HCOOH|| 1M CH_(3)COOH | (1 atm) H_(2) K_(a) " for " HCOOH and CH_(3) COOH are 1.77 xx 10^(-4) and 1.8 xx 10^(-5) respectively.

Neglecting the liquid-liquid junction potential, calculate the emf of the following cell at 25^(@)C H_(2) (1 atm)|0.5 M HCOOH|| 1M CH_(3)COOH | (1 atm) H_(2) K_(a) " for " HCOOH and CH_(3) COOH are 1.77 xx 10^(-4) and 1.8 xx 10^(-5) respectively.

% 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:

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))]