The problem is too long to work; however, if you will tell me what you know about it and what you don't understand about it I can help you through it.
The process involves three stages.
1. at the beginning, 0 mL KOH added.
2. hydrolysis of the salts at each equivalence point.
3. the pH of all points between equivalence point.
4. After the last equivalence point.
Calculate the pH for each of the following points in the titration of 50.0 mL of a 2.7 M H3PO3(aq) with 2.7 M KOH(aq).
pKa1 = 1.3 and pKa2 = 6.7
a) before addition of any KOH
b) after addition of 25.0 mL of KOH
c) after addition of 50.0 mL of KOH
d) after addition of 75.0 mL of KOH
e) after addition of 100.0 mL of KOH
4 answers
I'm not sure how to set up the equations using the volumes, and I'm not even sure what givens I'm supposed to use.
mols H3PO4 = M x L = 2.7 x 0.050 = 0.135.
(1) NaOH + H3PO4 ==> NaH2PO4 + H2O
(2) NaOH + NaH2PO4 ==> Na2HPO4 + H2O
Where are the equivalence points for equn 1 and 2 above?
So it will take 0.135 mol KOH for the first and another 0.135 for the second. Since the M of the KOH is the same as M H3PO4, this means mL will be the same. Therefore, the first equivalence point (the first H is titrated) with 50 mL KOH and the second with another 50 mL KOH. Look at the problem.
a is at the beginning.
c. is the first e.p.
b. is halfway between beginning and e/p/
e. is the 2nd e.p.
e. halfway between 1st and 2nd e.p.
a. at the beginning.
...........H3PO4 ==> H^+ + H2PO4^-
I...........2.7M......0......0
C...........-x.........x......x
E..........2.7-x.......x.......x
kl = (H^+)(H2PO4^-)/(H3PO4)
Substitute and solve for x = (H^+) and convert to pH.
For part c and e you have hydolysis of the salts. c is (H^+)= sqrt(k1k2)
e is (H^+) = sqrt(k2k3)
Parts b and d are solved using the Henderson-Hasselbalch equation.
(1) NaOH + H3PO4 ==> NaH2PO4 + H2O
(2) NaOH + NaH2PO4 ==> Na2HPO4 + H2O
Where are the equivalence points for equn 1 and 2 above?
So it will take 0.135 mol KOH for the first and another 0.135 for the second. Since the M of the KOH is the same as M H3PO4, this means mL will be the same. Therefore, the first equivalence point (the first H is titrated) with 50 mL KOH and the second with another 50 mL KOH. Look at the problem.
a is at the beginning.
c. is the first e.p.
b. is halfway between beginning and e/p/
e. is the 2nd e.p.
e. halfway between 1st and 2nd e.p.
a. at the beginning.
...........H3PO4 ==> H^+ + H2PO4^-
I...........2.7M......0......0
C...........-x.........x......x
E..........2.7-x.......x.......x
kl = (H^+)(H2PO4^-)/(H3PO4)
Substitute and solve for x = (H^+) and convert to pH.
For part c and e you have hydolysis of the salts. c is (H^+)= sqrt(k1k2)
e is (H^+) = sqrt(k2k3)
Parts b and d are solved using the Henderson-Hasselbalch equation.
Thanks!
7 answers