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

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09-03-2018
Chemistry

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

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ChemistryHelper2024 ChemistryHelper2024 · Answer 1 · 2018-03-18T10:03:49+01:00

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

AsiaWoerner AsiaWoerner · Answer 2 · 2019-02-25T15:49:34+01:00

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

[tex]CN^{-}+H_{2}O-->HCN+OH^{-}[/tex]

For this the equilibrium constant will be "Kb"

[tex]Kb=\frac{[CN^{-}][OH^{-}]}{[CN^{-}]}[/tex]

The relation between Kb and Ka is

[tex]Kb=\frac{Kw}{Ka}=\frac{10^{-14} }{4.9X10^{-10} }[/tex]

Let the amount of salt hydrolyzed is "x"

Therefore [HCN] = [OH-]=x

[CN-]=0.135-x

Putting values

[tex]Kb=\frac{10^{-14} }{4.9X10^{-10} }=2.04X10^{-5}[/tex]

[tex]2.04X10^{-5}=\frac{x^{2}}{(0.135-x)}[/tex]

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