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#### Borek

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« Reply #15 on: March 09, 2008, 01:25:28 PM »
pH = 10.26

That's OK. Could be you have accidentally used correct equation, but 10.26 it is.
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#### Lisa_G

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« Reply #16 on: March 09, 2008, 01:36:29 PM »
thanks for that, i calculated for kb2 as well and i got pH 5.63 using the same method, is it suppose to be below or above 7

#### Borek

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« Reply #17 on: March 09, 2008, 01:54:21 PM »
You can't calculate effects of the second dissociation step the same way. However, second dissociation constant is so low, you can safely ignore it.

More on pH calculation here:

http://www.chembuddy.com/?left=pH-calculation&right=toc

and specific information about calculation of multiprotic acid pH:

http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-simplified

(note that for base you just replace Ka with Kb and H+ with OH-).
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#### Lisa_G

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« Reply #18 on: March 09, 2008, 04:59:03 PM »
thanks for that information, very helpful

i got this question
a buffer solution contains equal volumes of H3PO4 (0.5M) and its conj base H2PO4- (0.45M)
Acid dissociation constant Ka = 7.5 x10 -3

What is the pH

Because the volumes are equal i used the henderson hasslebach equation and got

pH = (-log 7.5 x10-3) + log 0.45
0.5          pH = 2.08

is that right ?

and the 2nd question i got is

10 cm3 of 0.05M H3O+ is added to 1dm3 of the solution above. what is the new pH ?

i got

H3O+ moles = 10
1000    x 0.05 = 5x10-4

AFter addition of H3O+ (in 1dm3) H3PO4 = 0.5 - 5x10-4 = 0.50 moles

[H3PO4] = 1000
1010        x 0.50 = 0.50M

H2PO4- = 0.45 + 5x10-4 = 0.45 moles

[H2PO4-] = 1000
1010        x 0.45 = 0.46M

pH = (- log 7.5 x10-3) + log 0.46
0.50

pH = 2.08

is the method i used correct

#### Borek

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« Reply #19 on: March 09, 2008, 05:33:29 PM »
2.08 twice is OK, but your calculations are hard to follow. Better write it like pH=2.12-log(0.45/0.5), it is much more obvious what you are doing.

0.5 - 5x10-4 = 0.50 is not true. You should never round down intermediate results.
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#### Lisa_G

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