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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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