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Question

A 25 mL solution of 0.2 M NH4OH is titrated against 0.2 M HCl. Find the pH of the solution when 25 mL of HCl is added.
pKb(NH4OH)=4.74

A
8.37
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B
5.13
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C
4.74
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D
3.55
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Solution

The correct option is B 5.13
mmol of NH4OH=25×0.2=5
mmol of HCl=25×0.2=5
NH4OH (aq)+HCl (aq)NH4Cl (aq)+H2O (l)Initially: 5 5 0 0Final: 0 0 5 5
Since only salt of weak base and strong acid i.e. NH4Cl is left, it will undergo hydrolysis.
Concentration=MolesTotal Volume

[NH4Cl]=[NH+4]=550=0.1 M

NH+4 (aq)+H2O (l)NH4OH (aq)+H+(aq)
pH=712(pKb+logC)pH=712(4.74+log(0.1))pH=712(4.741)pH=71.87pH=5.13
Theory:
Generally, Methyl Orange is used for weak base and strong acid titrations.


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