andieje
asked on
basic ph calculation
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
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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