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Question

A solution of 0.1 M weak base B is titrated with 0.1 M of a strong acid (HA). The variation of pH of the solution with the volume of HA added is shown in the figure below.

What is the pkb of the base? The neutralization reaction is given by

B+HABH++A


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Solution

B+HABH+A
From the figure, it is clear that there is a sharp change in the curve when 6 ml of HA has been added. So, this is the equivalence point. As the concentration of B and HA are same, volume of B = volume of HA = 6 ml
So, total volume =12 ml
At equivalence point, mili mol of B = mili mol of HA
=molarity×volume(ml)=0.1×6=0.6 m mol
Hence, concentration of A=0.612=120 M
As pH=6,[H]+=106 M
But, [H]+=kwkb×c
or,106=1014×0.6kb×12
kb=5×104
pkb=4log5=40.7=3.3

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