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).

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

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  

     The concentration of

Generally the reaction equation showing the first dissociation of B is  

Here the ionic  constant for B is mathematically represented as

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  

Generally  

=>    

=>    

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  

    \

=>  

b

explanation:

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