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Question

The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.1 M NaOH will be: (pKa for CH3COOH = 4.74 )

A
4.74
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B
8.87
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C
9.1
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D
8.57
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Solution

The correct option is C 8.87
0.2 M CH3COOH 0.2 M NaOH
Both react as NaOH is strong base
0.22MCH3COO is left in sals
=01MCH3COO is left in sals
CH3COO+H2OCH3COOH+OHK=KwKa
αnkc
=1074Ka×0.1
107×10Ka
Ka104.74 (PKa=4.74)
α=10710104.74
=107×10+2.67
=104.33<0.1
[OH]=Cα
=101×104.33
=105.33
pOH=5.33
pH=14p01+
=8.67


1057643_1070123_ans_5d2455195fa44a10a1067137269a7a02.png

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