Question

Calculate the pH at the equivalence point for the titration of 0.300 M HCN with 0.300M NaOH. (Ka of HCN is 4.9 x 10-10).

Answer 1

Solution :-

At the equivalence point all the HCN is converted to the CN- which is the conjugate base

since the molarities are same for the HCN and NaOH threfore the volume required to reach the equivalence point are same

then molarity of the CN- is halved so the concnetration of the CN- = 0.300 M /2 = 0.150 M

now lets write the equation of the CN- with water

CN-   +H2O ----- > HCN + OH-

0.150                      0          0

-x                         +x           +x

0.150-x               x              x

Kb of CN- = Kw/ Ka

                 =1*10^-14 / 4.9*10^-10

                 = 2.04*10^-5

Kb= [HCN][OH-]/[CN-]

2.04*10^-5 =[x][x]/[0.150-x]

since kb is small then we can neglect the x from dem=nominator then we get

2.04*10^-5 =[x][x]/[0.150]

2.04*10^-5 * 0.150 = x^2

3.06*10^-6 =x^2

taking square root of both sides we get

1.75*10^-3 = x =[OH-]

now lets calculate the pOH

pOH= -log[ OH-]

       = -log [1.75*10^-3]

      = 2.76

pH+ pOH = 14

pH= 14 - pOH

    = 14 - 2.76

    = 11.24

so at the equivalence point pH = 11.24