`%` 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)`

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)]

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)

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]

The hydrolysis constant of 0.1M aqueous solution of sodium acetate if K_(a) of CH_(3)COOH = 1.8 xx 10^(-5) is

Calculate pH of (a) 10^(-1)MCH_(3)COOH (b) 10^(-3)CH_(3)COOH (C )10^(-6)M cH_(3)COOH Take K_(a)=2xx10^(-5) at 25^(@)C .

The degree of dissociation of 0.1N CH_(3)COOH is (K_(a)= 1xx10^(-5))

The degree of hydrolysis of 0.01 M NH_(4)CI is (K_(h) = 2.5 xx 10^(-9))

Total number of solutions from the following which has pHlt7 at 25^(@)C . (K_(a)(CH_(3)COOH)=1.8xx10^(-5),K_(b)(NH_(4)OH)=1.8xx10^(-5)) (i) 10^(-8) MHCl (ii) 0.01 M solution of NH_(4)Cl (iii) 0.01 M solution of CH_(3)COONa (iv) 0.01M solution is B(OH)_(3) (v) 0.01M solution of CH_(3)COONH_(4)

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))

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))]