Calculate the pH at the following points:?

Question

when 50.0 mL of 1.0 M methylamine (CH3NH2) (Kb = 4.4 x 10^-4) is titrated with 0.50M HCl.

a. after 50.0 mL of 0.50 mL HCl has been added

b. at the stoichiometric point.

Answer 1

a.after 50.0 mL of 0.50 mL HCl has been added, you are at 1/2 reaction, where pOH = pKb

Kb = 4.4 x 10^-4  ..... so pOh = pKb = 3.36

your answer  is : pH = 10.64

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b.at the stoichiometric point, the 50 ml of  1.0 M methylamine (CH3NH2) has been neutralized by 100 ml of 0.5M HCl,... diluting the product "(CH3NH3)+"  1.0Molar (50ml /150ml) into 0.33 Molar

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(CH3NH3)+ --> (CH3NH2)  & H+

0.33 molar --->   x   ,,,   ...  ... & x

Ka = Kw / Kb = 1e-14 / 4.4e-4 = 2.27e-11

2.27e-11 = [ (CH3NH2)] [H+] / [(CH3NH3)+]

2.27e-11 = [x] [x] / [0.33]

x2= 7.5e-10

x= [H+] = 2.7e-5

your last answer is:  pH = 4.56