need help with this chem problem please?

Question

Find the volume (mL) and the pH of 0.135 M HCl needed to reach the equivalence point(s) in titrations of the following. (You need to find the pH at the equivalence point, not the initial pH of the solution.)

(a) 53.0 mL of 0.216 M NH3 (NH3 Kb = 1.5*10^-5)

(b) 20.0 mL of 1.28 M CH3NH2 (CH3NH2 Kb= 4.4*10^-4

Answer 1

In both cases you have a monoprotic acid reacting with a monoprotic base, so the stoichiometry of the reaction is 1:1 and at the equivalence point mole acid =mole base =>
Ma*Va=Mb*Vb =>
Va=(Mb/Ma)*Vb

For (a) Va =(0.216/0.135)*53.0 =84.8 mL
for (b) Va =(1.28/0.135)*20.0 = 189.6 mL

If I write NH3 as H-NH2 you can see that both bases can be written as R-NH2 where R=H for NH3 and R=CH3 for CH3NH2
So we can get a general scheme for both cases
At the equivalence point all of the base has turned to R-NH3(+) so if the concentration of R-NH3(+) is C we have

.. .. .. .. .. .. R-NH3(+) <=> R-NH2 + H(+)
Initial .. .. .. .. C
Dissociate. .. x
Produce .. ... .. .. .. .. .. .. .. .. x. . .. .. .. x
At Equilbr. .. C-x .. .. .. .. .. .. x .. .. .. .. x

Ka= [R-NH3][H+]/[R-NH3(+)] = x^2/(C-x)
and Ka=Kw/Kb so

Kw/Kb=x^2/(C-x)
and for C>>x we assume C-x=C so Kw/Kb=x^2/C =>

x=Squareroot(C*Kw/Kb)

and pH=-logx

How much is C?
The moles of R-NH3(+) are equal to the moles of the respective base (1:1 stoichiometry) but the volume has increased (V=Va+Vb). So

C=Mb*Vb/(Va+Vb)

For (a) C =0.216*53/(84.8+53.0) = 0.083
and x =SQRT( 0.083*(10^-14)/ (1.5*10^-5) )= 7.44*10^-6 << 0.083 so our assumption is OK and pH=5.13

For (b) C =1.28*20.0/(189.6+20.0) = 0.122
and x =SQRT( 0.122*(10^-14)/ (4.4*10^-4) ) = 1.66*10^-6<< 0.083 so our assumption is OK and pH=5.78