A 25mL solution of 0.2MNH4OH is titrated against 0.2MHCl. Find the pH of the solution when 25mL 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:5500Final:0055 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=0.1M
NH+4(aq)+H2O(l)⇌NH4OH(aq)+H+(aq) pH=7−12(pKb+logC)pH=7−12(4.74+log(0.1))pH=7−12(4.74−1)pH=7−1.87pH=5.13 Theory: Generally, Methyl Orange is used for weak base and strong acid titrations.