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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: diablo on March 24, 2007, 10:38:43 AM

Title: calculate pH of a solution
Post by: diablo on March 24, 2007, 10:38:43 AM
Hi, i disremember some basic stuff so thats why i got this prob ;D  :

you've got 0,1M solution of ammonium formate and following data:

Ka (formic acid) 1,77x10-4
Kb (ammonium hydroxide) 1,77x 10-5
and Kwater 1x10-14

Now calculate the pH of this solution.

Actually i dont know which equation i should use to calculate the pH. So i wrote down the equation:

NH4COOH <--> HCOOH + NH3

Any ideas what to do know?

Thx & greets

Diabloooooo
Title: Re: calculate pH of a solution
Post by: Borek on March 24, 2007, 10:49:32 AM
http://www.chembuddy.com/?left=pH-calculation&right=toc

http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified
Title: Re: calculate pH of a solution
Post by: diablo on March 24, 2007, 11:16:44 AM
thx for your fast answer!

--> nice site; plz correct me if i'm wrong:

so the result would be 1/2 x (pkw + pka- pkb) --> as stated at the end of the 2nd link

i calculated my pka = 3,75 and my pkb = 4,75 and result would be pH =7,5 --> but i thought that i have to enter the concentration (0,1M) anywhere?

greets, Diablooooooo
Title: Re: calculate pH of a solution
Post by: Borek on March 24, 2007, 02:05:52 PM
Quote from: diablo on March 24, 2007, 11:16:44 AM
so the result would be 1/2 x (pkw + pka- pkb) --> as stated at the end of the 2nd link

i calculated my pka = 3,75 and my pkb = 4,75 and result would be pH =7,5 --> but i thought that i have to enter the concentration (0,1M) anywhere?

In some cases pH is almost independent on the concentration, as shown in the table at the bootom of the page mentioned. Correct pH value is slightly different, but it is very hard to calculate - it requires solving of 4th degree polynomial in this case. If you ever need exact value, you may try pH calculator BATE.
Title: Re: calculate pH of a solution
Post by: diablo on March 24, 2007, 02:46:59 PM
ok, thx alot  for your help  ;D