A sample of 45.0 mL of a 0.0325 M HCN solution is titrated with a 0.0200 M NaOH solution.
Ka of HCN given: 6.2*10^-10

A. What volume of NaOH is used in the titration in order to reach the equivalence point?
       For this part, I used stoichiometry and got 73.125 mL as the volume
B. what is the ph of the solution at the equivalence point?
      I do not understand how to calculate this and why the ph is not 7 even though it is at equivalence point.

C. What is the molar concentration of CN- at the equivalence point?
How do you find this from the answer in part b?

Thank you!

Sirisha sorry for the late reply
I am sure first part you calculated by using M1V1=M2V2
for the second part -
HCN is a weak acid and when it is neutralized by NaOH it forms NaCN and H2O
now you know that a weak acid forms a strong conjugate base
here also CN- is a strong base and hence the pH will be more than 7(soln will be basic)
CN-+H2O ---> HCN+ OH-
Now you need to know pH of this soln
there are many ways of doing it
HCN+ H2O ----> H3O+ +CN-
now you can make an ice table for the above reaction
Initial CN- conc is not zero .This you are going to calculate from the amount of NaCN formed during neutralization reaction with NaOH
Ka value will be given
take out [H3O+] and this will give pH of the soln also
b) From the ice table which you have made above the equilibrium conc of CN- is the molar conc of the CN- at the equivalence point
I hope this is clear

these questions are important and you must do practice on them