Question

Find the pH of 0.135M NaCN solution. For HCN, Ka=4.9⋅10−10,

Answer 1

The given salt of weak acid and strong base will hydrolyze as

`CN^(-)+H_(2)O-->HCN+OH^(-)`

For this the equilibrium constant will be "Kb"

`Kb=([CN^(-)][OH^(-)])/([CN^(-)])`

The relation between Kb and Ka is

`Kb=(Kw)/(Ka)=(10^(-14) )/(4.9X10^(-10) )`

Let the amount of salt hydrolyzed is "x"

Therefore [HCN] = [OH-]=x

[CN-]=0.135-x

Putting values

`Kb=(10^(-14) )/(4.9X10^(-10) )=2.04X10^(-5)`

`2.04X10^(-5)=(x^(2))/((0.135-x))`

WE may ignore x in denominator as Kb is very low

On solving

x = 1.66X10⁻³ =[OH⁻]

pOH = -log[OH⁻] = -log(1.66X10⁻³)=2.78

Therefore

pH = 14 - pOH = 14-2.78 = 11.22

Answer 2

NaCN is considered as salt of weak acid (HCN) and strong base (NaOH):
we have salt concentration Cs = 0.135 M NaCN and Ka = 4.9 x 10⁻¹⁰
use this formula to calculate pH:
pH = 1/2 pKw + 1/2 pKa - 1/2 pCs
pKw = - log Kw, pKa = - log Ka and pCs = - log Cs
pKw = - log (1 x 10⁻¹⁴) = 14
pKa = - log (4.9 x 10⁻¹⁰) = 9.31
pCs = - log (0.135) = 0.87
pH = (1/2 x 14) + (1/2 x 9.31) - (1/2 x 0.87) = 11.22
Note that it is basic solution (pH more than 7) because it consists of strong base and weak acid