A 10.00-mL sample of 0.1000 M KH2PO4 was titrated with 0.1000 M HCl
Ka for phosphoric acid (H3PO4): Ka1= 7.50x10-3
; Ka2=6.20x10-8
; Ka3= 4.20x10-10
Calculate the pH after addition of 10.00 mL of HCl to the KH2PO4 solution?
4 answers
Use the Henderson-Hasselbalch equation.
which ka value would we use and would the concentrations of the acid and base be the same?
First, forget what I told you. You are titrating KH2PO4 with HCl.
KH2PO4 + HCl --> H3PO4 + KCl
You notice (which is what I didn't notice when I first read the problem) that 10 mL of 0.1M KH2PO4 is exactly neutralized (neutralized may be the wrong word) by 10 mL of 0.1M HCl so you have 0.05M H3PO4 at the equivalence point. Therefore, the pH will be determined as if you had a pure solution of 0.05M H3PO4 and for that you use k1.
..........H3PO4--> H^+ + H2PO4^-
I.........0.05.....0......0
C..........-x......x......x
E........0.05-x....x......x
Substitute the E line into k1 expression and solve for x = (H^+) and convert to pH. I expect you will need to solve the quadratic equation you will get (meaning that you can not assume 0.05-x = 0.05)
KH2PO4 + HCl --> H3PO4 + KCl
You notice (which is what I didn't notice when I first read the problem) that 10 mL of 0.1M KH2PO4 is exactly neutralized (neutralized may be the wrong word) by 10 mL of 0.1M HCl so you have 0.05M H3PO4 at the equivalence point. Therefore, the pH will be determined as if you had a pure solution of 0.05M H3PO4 and for that you use k1.
..........H3PO4--> H^+ + H2PO4^-
I.........0.05.....0......0
C..........-x......x......x
E........0.05-x....x......x
Substitute the E line into k1 expression and solve for x = (H^+) and convert to pH. I expect you will need to solve the quadratic equation you will get (meaning that you can not assume 0.05-x = 0.05)
Why the ph will be determinated as if you have 0,05M H3PO4???
0 answers