A 205 mg sample of a diprotic acid is dissolved in enough water to make 250.0 ml of solution. The pH of this solution is 2.15. A saturated solution of calcium hydroxide (Ksp= 1.3 x 10-6) is prepared by adding excess calcium hydroxide to water and then removing the undissolved solid by filtration. Enough of the calcium hydroxide solution is added to the solution of the acid to reach the second equivalence point. The pH at the second equivalence is 9.40. The first dissociation constant for the acid is 1.70 x 10-2. Assume that the volumes of solutions are additive and that the first K for the acid is at least 1000 times greater than the second K.
a- Calculate the molar mass of the acid
b- Calculate the second dissociation constant for the acid.
8 answers
I've calculated the molar mass as 82g/mol, but I can't figure out how to get the dissociation constant.
I think you can do this two ways.
1:
For a diprotic acid, (HA^-) = (H^+); therefore, (A^2-) = k2 and you calculate (A^2-) from the titration with Ca(OH)2.
2:
You need (A^2-), but it will hydrolyze at the second equivalence point to give yu
A^2- + HOH ==> HA + OH^-
Kb = (Kw/k2) = (OH^-)(HA)/(A^2-)
The H gives you OH and HA, you know A from the titration data and you know Kw. You should get the same answer either way although I've never tried to see if I can.
Did I see a post in the last day or so from you about the Intrepid Hero from Michigan State who messed up the titration with an unknown acid. I took a shot at that one and I was curious if I got close to the pKa.
1:
For a diprotic acid, (HA^-) = (H^+); therefore, (A^2-) = k2 and you calculate (A^2-) from the titration with Ca(OH)2.
2:
You need (A^2-), but it will hydrolyze at the second equivalence point to give yu
A^2- + HOH ==> HA + OH^-
Kb = (Kw/k2) = (OH^-)(HA)/(A^2-)
The H gives you OH and HA, you know A from the titration data and you know Kw. You should get the same answer either way although I've never tried to see if I can.
Did I see a post in the last day or so from you about the Intrepid Hero from Michigan State who messed up the titration with an unknown acid. I took a shot at that one and I was curious if I got close to the pKa.
Yes, the intrepid hero post was mine- these are both extra credit problems from our class and I'm having some trouble with this unit. For that one the Ka I came up with was 2.22*10^-5.
As for this problem, your logic makes sense but I'm unsure how to calculate (A^2-) from the titration.
As for this problem, your logic makes sense but I'm unsure how to calculate (A^2-) from the titration.
Just a question, how did you calculate 82g/mol for the first question? I am trying to figure this out as well, but I get 41.5. I just want to check.
...H2A->...H+..+..HA-
...x.......0......0
-.00708..+.00708..+.00708
x-.00708...00708...00708
K1=.017=(.00708)^2/x-.00708
x=.01
The acid is .01M.
There are 250 mL.
There are .205g of acid.
.205g=.0025mol
.205/.0025=82
...x.......0......0
-.00708..+.00708..+.00708
x-.00708...00708...00708
K1=.017=(.00708)^2/x-.00708
x=.01
The acid is .01M.
There are 250 mL.
There are .205g of acid.
.205g=.0025mol
.205/.0025=82
We have (0.205/82) = 0.0025 mols H2A.
For Ca(OH)2 ==> Ca^2+ + 2OH^-
.....solid.......x.......2x
Ksp = 1.36E-6 = 4x^3
and x == [Ca(OH)2] = 0.00688 mols/L = M
Ca(OH)2 + H2A ==> CaA + 2H2O
Since this is a 1:1 ratio, it will take 0.0025 mol Ca(OH)2 to neutralize the acid.
M = mols/L or L = mols/M = 0.0025/0.00688 = about 363 mL.
The H2A was in 250 mL; therefore, total volume is 250 + 363 (you can clean up the significant figures and any decimal places) = about 613 mL. So (A^2-) = 0.0025 mols/0.613L = ?M
I don't trust the first method for k2; stick to the second method. I get about 6.4E-8.
For Ca(OH)2 ==> Ca^2+ + 2OH^-
.....solid.......x.......2x
Ksp = 1.36E-6 = 4x^3
and x == [Ca(OH)2] = 0.00688 mols/L = M
Ca(OH)2 + H2A ==> CaA + 2H2O
Since this is a 1:1 ratio, it will take 0.0025 mol Ca(OH)2 to neutralize the acid.
M = mols/L or L = mols/M = 0.0025/0.00688 = about 363 mL.
The H2A was in 250 mL; therefore, total volume is 250 + 363 (you can clean up the significant figures and any decimal places) = about 613 mL. So (A^2-) = 0.0025 mols/0.613L = ?M
I don't trust the first method for k2; stick to the second method. I get about 6.4E-8.
Okay, I made an accounting error. I calculated a Ka2 value different from yours and am not sure that it is correct. I would repost this question if Dr. Bob222 doesn't answer this soon.
The calculated Ka in one of my previous comments is from a different question- it's not related to this problem.
1 answer