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Topic: Can someone please help (phys chem) (Read 15782 times)
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Borek
Mr. pH
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Re: Can someone please help (phys chem)
«
Reply #15 on:
March 09, 2008, 01:25:28 PM »
Quote from: Lisa_G on March 09, 2008, 12:54:15 PM
pH = 10.26
That's OK. Could be you have accidentally used correct equation, but 10.26 it is.
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ChemBuddy
chemical calculators - stoichiometry, pH, concentration, buffer preparation,
titrations.info
Lisa_G
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Re: Can someone please help (phys chem)
«
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
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Borek
Mr. pH
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Posts: 27779
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Re: Can someone please help (phys chem)
«
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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ChemBuddy
chemical calculators - stoichiometry, pH, concentration, buffer preparation,
titrations.info
Lisa_G
Regular Member
Posts: 22
Mole Snacks: +1/-2
Re: Can someone please help (phys chem)
«
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 H
3
PO
4
(0.5M) and its conj base H
2
PO
4
-
(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) H
3
PO
4
= 0.5 - 5x10
-4
= 0.50 moles
[H
3
PO
4
] =
1000
1010 x 0.50 = 0.50M
H
2
PO
4
-
= 0.45 + 5x10
-4
= 0.45 moles
[H
2
PO
4
-
] =
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
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Borek
Mr. pH
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Posts: 27779
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Gender:
I am known to be occasionally wrong.
Re: Can someone please help (phys chem)
«
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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ChemBuddy
chemical calculators - stoichiometry, pH, concentration, buffer preparation,
titrations.info
Lisa_G
Regular Member
Posts: 22
Mole Snacks: +1/-2
Re: Can someone please help (phys chem)
«
Reply #20 on:
March 19, 2008, 05:30:42 PM »
yeh i listened to what you said, i got 2.08 and 2.07, thanks for all your help, i really appreciate it
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