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Question

A solution of 0.1M weak base B is titrated with 0.1M 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

Step 1: The Henderson-Hasselbalch equation for the buffer solution is given below-

pOH=pKb+log[salt][base]

pOH=14-pH

salt=concentration of salt

base=concentration of base

Step 2: Calculation of pOH-

In the graph, at the equivalence point (V=6mL)- complete neutralisation takes place.

At half neutralisation point (V=3mL)- buffer formation occurs (pH=11).

pOH=14-11

pOH=3

Step 3: Calculation of salt and base -

B+HABH++A-AtpH=11[0.1×3mmol]0.1×3mmol[0.3mmol]totalvolume=6mL

salt=0.36 mol/L

base=0.36mol/L

Step 3: Calculation of pKb for the given reaction -

pOH=pKb+logsaltbase

3=pKb+log0.360.36

3=pKb+log1

3=pKb

So, the pKb of the base is 3.


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