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Question

The pH of 0.1 M monobasic acid is 4.50. The acidity constant (Ka) of the monobasic acid is:
  1. 1.0×108
  2. 1.0×107
  3. 1.0×105
  4. 1.0×104

A
1.0×105
B
1.0×107
C
1.0×104
D
1.0×108
Solution
Verified by Toppr

pH=4.5[H+]=104.5=3.16×105

Since the acid is monobasic, it will dissociate in the aqueous solution as

HAH++A

Since, 0.1>>3.16×105, [HA]=0.1 mol L1

Also [H+]=[A]

Ka=[H+]2[HA]

Ka=(3.16×105)20.1=1×108

Hence, option D is the correct answer.

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