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Question

0.02 M solution of H2A has pH=4. If Ka1 for the acid is 5×107 M, the concentration of HA ion in the solution would be :

A
2×107 M
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B
1×106 M
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C
3×105 M
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D
1×104 M
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Solution

The correct option is D 1×104 M
Given, [H2A]=0.02, pH=4,Ka1=5×107 M
pH=4log[H+]=4[H+]=104 M

The dissociation reaction for finding Ka1 of H2A is :

H2A (aq)H+ (aq)+HA (aq)Ka1=[HA][H+][H2A]
Now by rearranging, the value of [HA] can be obtained.

[HA]=Ka1×[H2A][H+][HA]=5×107×0.02104[HA]=104 M

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