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Question

100 mL of 0.1 M NaOH solution is titrated with 100 mL of 0.05 M H2SO4 solution. The pH of the resulting solution is : (for H2SO4,Ka1=,Ka2=102)

A
7
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B
7.2
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C
7.4
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D
6.8
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Solution

The correct option is B 7.2
no of milimoles of NaOH=100×0.1=10
no of milimoles of H2SO4=100×0.05=5
now 2NaOH+H2SO4Na2SO4+2H2O
Since 1 mole H2SO4 rects with 2 mole of NaOH
the above reaction will go till complete neutralization and after that [SO4]2=5200M and this will go anion hydrolysis
SO24+H2OHSO4+OH
Where HSO4 is conjugate acid with Ka=102
pH=12[pKw+pKa+logC]
pH=12[14+21.6]=7.2


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