Asked by help
A 20ml sample of .125M diprotic acid solution was titrated with .1019M KOH.the constants for the acid are k1=5.2*10 ^-5 and K2=3.4*10 ^-10. What added volume of base does each equivalence point occur?
i got .0254L for the first one and .0508 for the second. Can anyone check if i did it rightÉ
i got .0254L for the first one and .0508 for the second. Can anyone check if i did it rightÉ
Answers
Answered by
DrBob222
First equivalence point is
M x L = moles
moles/0.1019 = L. Close to 0.0254 but I get 0.0245 L = 24.5 mL. I wonder if that 20 mL sample really is 20.00 an the 0.125 M is really 0.1250 in which case more s.f. are available and the volume for the first equivalence point is 24.53 mL.
For the second one, it is
40.00 x 0.1250/0.1019 = 49.07
How do we use the constants in this problem? The fact that they are 10^5 apart (approximately) means the two equivalence points are distinct and can be titrated separately.
M x L = moles
moles/0.1019 = L. Close to 0.0254 but I get 0.0245 L = 24.5 mL. I wonder if that 20 mL sample really is 20.00 an the 0.125 M is really 0.1250 in which case more s.f. are available and the volume for the first equivalence point is 24.53 mL.
For the second one, it is
40.00 x 0.1250/0.1019 = 49.07
How do we use the constants in this problem? The fact that they are 10^5 apart (approximately) means the two equivalence points are distinct and can be titrated separately.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.