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Question

The pH of the 0.10 M hydrocyanic acid solutions is 5.2. What will be the value of Ka for hydrocyanic acid ?

A
2.0×105
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B
4.0×105
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C
2.0×1010
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D
4.0×1010
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Solution

The correct option is D 4.0×1010
Given,
pH of hydrocyanic acid is 5.2
pH=log H+=5.2
log[H+]=5.2
[H+]=Antilog(5.2)=6.31×106 M
HCN H+ + CN
0.16.31×106 6.31×106M 6.31×106 M
=0.1 M (6.31×106 is too smaller than 0.1, so we can neglect it)
Ka=[H+][CN][HCN]
=(6.31×106M)×(6.31×106)M0.1 M
=3.9816×1010=4.0×1010

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