CBSE Class 11-science Answered
Concentration of HCN and NaCN in a solution is 0.01M each.Calculate [H30+]& [OH-].Ka(HCN)=0.00000000072
Asked by nilay pandey | 12 Oct, 2013, 10:36: PM
Expert Answer
(i) HCN 
H+ + CN-
(0.01 - X)M XM XM
(ii) NaCN Na+ + CN-
Initial 0.01 M 0M 0M
Final 0.0M 0.01M 0.01M
So at equilibrium, [H+] = XM
[CN-]Total = [CN-]Acid + [CN-]Salt
= XM + 0.01M = 1.01M
[HCN] = 0.01M XM = 0.01M
= -log[H+] = log 7.2 X 10-10
Answered by Hanisha Vyas | 14 Oct, 2013, 01:37: PM
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