[AmyJ]

I am given a solution of NH4NO3 and HCN.

With the NH4NO3 I got the Ka using the Kb of NH3.  The I did x²/[NH4NO3] = Ka, to get the [H⁺].

For the HCN I did the same thing but used Ka of HCN.  Then I added the [H⁺] concentrations and did -log[H⁺] to find the pH.

I am getting the wrong answer.
Any ideas?

[enahs]

How much of the two weak acids do you have? Is this an addition of two solutions, and so are you making sure to use the total volume of solutions?


Try and show all your work and calculations, please, so we can see what went wrong where.

[AmyJ]

Calculate pH of the following solution: .265 M NH₄NO₃ and .105 M HCN.

I did:

NH₄ + H₂0 -> NH₃ + H₃O

Kb = NH₃ = 1.8*10^-5
So Ka = 10^-14/1.8*10^-5 = 5.56*10^-10

x²/.265 = 5.56*10^-10
[H⁺] = 1.21*10^-5

HCN -> H⁺ + CH^-

Ka HCN = 6.2*10^-10
x²/.105 = 6.2*10^-10
[H⁺] = 8.06*10^-6

I added the [H⁺] concentrations (1.21*10^-5
+ 8.06*10^-6 = 2.01*10^-5

pH = -log[H⁺] = 4.69

Apparently my answer is slightly off...

[enahs]

When you have such weak acids you must take into account the acid dissociation of water its self.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/weaka.php#verywk

I did not read that site very throughly, hope it explains well, just the first result.

If that does not fix "slightly off" problem just say so.