Question

Calculate the ph and hydrolysis in a 1m nacn solution

Answer 1

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