Calculate the pH at the equivalence point for the titration of 0.22 M HCN with 0.22 M NaOH.?

Question

(Ka= 4.9 x 10^-10 for HCN)

Answer 1

At the equivalence point you have only the salt NaCN which will hydrolyse (since HCN is a weak acid) releasing OH-.

HCN + NaOH -> NaCN +H2O
According to the stoichiometry you will have as many moles NaCN forming as the HCN that reacted. Assume that the volume of HCN solution was V.
Then moles NaCN= moles HCN= M*V =0.22*V
Since the concentrations of the acid and base are equal, you need equal volumes of the two for the titration.
So the final volume will be Vfinal= 2*V and the concentration of NaCN, C=mole/Vfinal = 0.22*V/2*V =0.11

.. .. .. .. .. .. .. CN(-) + H2O <=> HCN + OH(-)
Initial .. .. .. .. 0.11
React .. .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. x.. .. .. x
At Equilb. .. 0.11-x .. .. .. .. .. .. .. .. x.. .. .. x

Kb= [HCN][OH-] / [CN-] = x^2/(0.11-x) =>
=> x^2 + Kbx -Kb*0.11=0
but Kb=Kw/Ka =(10^-14)/(4.9*10^-10) =2.04*10^-5
and
x^2 + (2.04*10^-5)x+ (2.04*10^-5)*0.11=0 =>
x^2 + (2.04*10^-5)x+ 2.244*10^-6 =0

x=0.5*( -(2.04*10^-5) +Squareroot( (2.04*10^-5)^2 + 4*2.244*10^-6)
=> x= 1.49*10^-3
pOH=-logx= -log(1.49*10^-3) =2.83
pH= 14- pOH= 11.17

This was the detailed solution. You could have assumed that x<<0.11 and thus 0.11-x= 0.11. Then the equation would be simplified

Kb= x^2/0.11 =2.04*10^-5 =>
x =SquareRoot(0.11*2.04*10^-5) = 1.49*10^-3 and get the same result.
Note that if the calculated x was in the same order of magnitude as 0.11 when you did this simplification you would have to use the detailed solution

Answer 2

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Answer 3

pH = pKa
since the concentration of salt is same as that of base.........
pH = 10 - log(4.9)

Answer 4

pH = 8.65