Complete Question
The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).
(a) before addition of any HCl (b) after addition of 25.0 mL of HCl
Answer:
a The value is 
b 
Explanation:
From the question we are told that
The first pKb value for B is 
The second pKb value for B is 
The volume is 
The concentration of B is ![[B] = 0.60 M](https://tex.z-dn.net/?f=%5BB%5D%20%20%3D%20%200.60%20M)
The concentration of 
Generally the reaction equation showing the first dissociation of B is

Here the ionic constant for B is mathematically represented as
![K_i = \frac{[BH^+] [OH^-]}{[B]}](https://tex.z-dn.net/?f=K_i%20%20%3D%20%20%5Cfrac%7B%5BBH%5E%2B%5D%20%5BOH%5E-%5D%7D%7B%5BB%5D%7D)
Let denot the concentration of [BH^+] as z and since
then
is also z
So [B] = 0.60 - z
Here
is ionic constant for the first reaction of a dibasic base B and the value is

So

=> 
using quadratic formula to solve this equation

Hence the concentration of
is ![[OH^-] =0.0651](https://tex.z-dn.net/?f=%5BOH%5E-%5D%20%3D0.0651)
Generally ![pOH = -log [OH^-]](https://tex.z-dn.net/?f=pOH%20%3D%20%20-log%20%5BOH%5E-%5D)
=> 
=> 
Generally the pH is mathematically represented as


Generally the volume of
at the second dissociation of the base B is 
The volume of the
half way to the first dissociation of the base is 25mL
Now the pOH at half way to the first dissociation of the base is

=> 
=> 
Generally the pH after addition of 25.0 mL of HCl is
\
=> 