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Question

50 mL of 0.2 M HCl is titrated against the 0.1 M NaOH solution. What will be the pH of the solution when 20 mL NaOH is added ?

A
0.95
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B
4.2
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C
1.25
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D
2.69
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Solution

The correct option is A 0.95
mmol of HCl=50×0.2=10
mmol of NaOH added=20×0.1=2.0

HCl (aq)+NaOH (aq)NaCl (aq)+H2O (l)Initial: 10 2.0 0 0Final: 8 0 2.0 2.0

After titration the remaining HCl=8 mmol .
Final Concentration =Moles of HCl remaining Total Volume
Total Volume =50+20=70 mL
[HCl]=870 M=0.11 MpH=log[H+]pH=log[0.11]pH=2log(11)=0.95

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