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Question

0.01 M Solution of H2A has pH=4 . If Ka1 for the acid is 4.45×107 , the concentration of HA ion in the solution would be :
  1. 1.0×102 M
  2. 4.45×105 M
  3. 8.0×105 M
  4. Unpredictable


Solution

The correct option is B 4.45×105 M
Given, [H2A]=0.01, pH=4, Ka1=4.45×107
The dissociation reaction for the weak dibasic acid is :
H2AH++HAKa=[HA][H+][H2A]
Now by rearranging,  the value of [HA] can be obtained.

[HA]=Ka×[H2A][H+][HA]=4.45×107×1×1021×104[HA]=4.45×105 M

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