Q.

The PH of 0.1 M monobasic acid is 4.5. The Ka and PKa values, respectively, are

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a

1.0×10-6, 6

b

1.0×10-7, 7

c

1.0×10-8, 8

d

none of these

answer is C.

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Detailed Solution

PH=-logH+H+=Antilog -PH         =Antilog (-4.5)         =Antilog 5¯.5=3.16×10-5H+=A-=3.16×10-5HA=C-Cα=C-H+         =0.1-3.16×10-5=0.1Ka=H+A-HA=3.16×10-520.1=1.0×10-8PKa=-log Ka=-log 10-8=+8
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