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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Rutherford on June 16, 2012, 10:46:53 AM

Title: pH of NH4CN solution
Post by: Rutherford on June 16, 2012, 10:46:53 AM
Calculate the pH of a 0.7M solution of NH4CN. KbNH3=1.79*10-5, KaHCN=7.9*10-10.

The reactions are:
NH4++H2O ::equil:: NH3+H3O+ Ka=Kh=Kw/Kb=5.59*10-10
0.7                  x      x
CN-+H2O ::equil:: HCN+OH- Kb=Kw/Ka=1.27*10-5
0.7                y     y

[H+]=x=1.98*10-5
[OH-]=y=2.98*10-3
There are more OH- ions :rarrow: y-x≈2.98*10-3
pOH=2.52
pH=14-2.52=11.48 but in my book it is 9.18.
What did I do wrong?
Title: Re: pH of NH4CN solution
Post by: Borek on June 16, 2012, 11:59:03 AM
You can't assume there are two different, separate reactions taking place. You ended with OH- being neutralized by H+ - once they react, they are removed from the solution, so the equilibrium shifts.

http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified
Title: Re: pH of NH4CN solution
Post by: Rutherford on June 16, 2012, 12:52:02 PM
Understood, thanks.