|Dec19-04, 01:56 PM||#1|
According to theory, which amounts of the following acid and its conjugate base would you need to prepare 150 ml of a 0.3 M phosphate buffer at pH 7.9? Use the pKa table below to help you in your answer.
H3PO4 <–> H2PO4- + H+ pKa = 2.3
H2PO4- <–> HPO4-2 + H+ pKa = 7.2
HPO4-2 <–> PO4-3 + H+ pKa = 12.1
Molecular weights: PO4-3 = 95, Na = 23, H = 1
a) 7.5 ml of 0.3M NaH2PO4 and 37.5 ml of 0.3M Na2HPO4 plus water
b) 37.5 ml of 0.3M NaH2PO4 and 7.5 ml of 0.3M Na2HPO4 plus water
answer:--> c) 25 ml of 0.3M NaH2PO4 and 125 ml of 0.3M Na2HPO4
d) 125 ml of 0.3M NaH2PO4 and 25 ml of 0.3M Na2HPO4
e) 0.05 ml of 0.3M NaH2PO4 and 0.25 ml of 0.3M Na2HPO4 plus water
f) 0.25 ml of 0.3M NaH2PO4 and 0.05 ml of 0.3M Na2HPO4 plus water
I don't understand why c) would be the correct answer. Please help!
|Dec19-04, 03:22 PM||#2|
Can you should show some of your reasoning people might tell you what is wrong and point you toward the right direction
|Dec19-04, 04:18 PM||#3|
Well, I am not sure how to get c as the answer because that is the answer that was given. This question appeared on a test.
I thought the answer should be b because after using the equation pH=pKa+log(A/HA) I got the results of 37.5 and 7.5, but then in theory water is not added to the solution to create a buffer unless I believe a titration is made. I may be mixing this information up with something else. In theory no titration is made and the Henderson-Hasselbach equation is used. That is why I do not understand how the values for C are correct. Thank you.
|Jul15-08, 04:08 PM||#4|
I don't understand why c) would be the correct answer.
can you give me the lows please
|Jul15-08, 04:29 PM||#5|
This question looks hard, but in fact is pretty obvious.
First of all - out of 6 answers listed, 4 are wrong, because they don't give 0.3M buffer solution. That leaves only two answers to select from. pKa2 is 7.2 - so to have buffer with pH lower than that you need a solution which contains more acid than base. That leaves only one answer.
Sure, that requires assumption that one of the answers given is correct
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|Jul22-10, 04:17 PM||#6|
V=150 ml= 0.15 L
MW Na2HPO4=(23 x 2) + 1+ 95 = 142
MW NaH2PO4= 23 + (1 x 2) + 95= 120
H2PO4- <–> HPO4-2 + H+
pH = pKa +log conjugated base/ acid
7.9=7.2 + log Na2HPO4/ NaH2PO4
0.7= log Na2HPO4/ NaH2PO4
1) Na2HPO4/ NaH2PO4 = 5.011
2) Na2HPO4 + NaH2PO4= 0.3 M
From 1 & 2 : Na2HPO4= 0.245 NaH2PO4= 0.049
G (wight) [gr]= C(concentration) [M] x mw (molecular weight) X v (volume) [L]
Na2HPO4 G= 0.245 x 0.15 x 142= 5.218 gr in 0.15 L water
NaH2PO4 G= 0.049 x 0.15 x 120= 0.882 gr in 0.15 L water
V= g/C x MW
Na2HPO4 V= 5.218/ 0.3 x 142 = 0.122 L = 122 ml of 0.3M Na2HPO4
NaH2PO4 V = 0.882 / 0.3 x 120= 0.0245 = 24.5 ml of 0.3M NaH2PO4
These answers are close enought to C)
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