112) Ionisation constant of CH3COOH is 1.7 * 10^ -5 and the conc. of H+ ions is 3.4 * 10^ -4. Then find out the initial conc. of CH3COOH molecules?

113) Solution of 0.1N NH4OH and 0.1 N NH4Cl has pH9.25 . Find pKb

1. We know that ionization constant of an acid(HA) :

K_a = [H⁺] [A^-]/ [HA]

So, in case of acetic acid CH₃COOH,  we have K_a = 1.7 x 10^-5 and concentration of H⁺ ion [H⁺] = 3.4 x 10^-4 

Then the initial concentration of CH₃COOH could be obtained as:

K_a = [H⁺] [ CH₃COO^-]/ [ CH₃COOH]

We know that, during a reaction: 

CH₃COOH <=> CH₃COO^- + H⁺ 

let Initial amount, 1..........0,,....... 0

change ; (1-x²) .................x........ x

So, the concentration of both the ions will be same.

Hence, Concentration of  CH₃COOH ,[CH₃COOH] =  [H⁺] [ CH₃COO^-]/K_a 

Or,  [ CH₃COOH] = 3.4 x 10^-4 x  3.4 x 10^-4 / 1.7 x 10^-5 = 6.8 x 10^-3 

2. We have, The solution of NH₄OH and NH₄Cl are basic buffer solutions like (BOH + BA)

Hence, pOH = pK_b + log [salt]/ [base] --eq.1.

Now as , pH + pOH = 14 , so from the question , pOH  = 14- pH = 14- 9.25 =4.75.

Hence, from equation 1

pK_b = pOH- log [salt]/[base] = 4.75 - log [0.1 ]/ [ 0.1] = 4.75

Hence, pK_b = 4.75