basic ph calculation

andieje
andieje used Ask the Experts™
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Hi

I seem to be getting this pH calculation wrong and I can't see where:

What is pH of 25cm3 0.1M NH3 after 24cm3 0.1M Hcl added. pkb NH3 = 4.75

If you react 24cm3 of HCL with ammonia you will use up 24cm3 of ammonia leaving 1cm3.
This contains 1*10^-4 moles of ammonia in a volume of 49cm3 = 2*10^-3 M
 You will generate 0.0024 mole of NH4+ from the reaction. This has a molarity of 0.49M

Kb = [OH-][NH4+]  /  [NH3]
1.778 * 10^-5 = [oh-] * [0.49]   /  [2*10^-3]
pOH = 7.14

I can't see what I am doing wrong. Apparently answer is:

[H+] = 1.3501211260092E-08
[OH-] = 7.4067428524424E-07
pH = 7.8696272670623
pOH = 6.1303727329377

many thanks
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You are correct that you have 0.0024 mol NH4 generated in the reaction

You are ~correct in the concentration of NH3 that you have left over, 0.0001mol / 49ml = 2.0408*10^-3M

You are incorrect in the concentration of NH4 in the reaction, when stating "has a molarity of 0.49M"
The fact that 0.49M is almost 5x greater than any of the starting concentrations you should have spotted the error in this assertion earlier.

The correct concentration of 0.0489M will produce the "apparently answer is" values.


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