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?
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)
Why the ph will be determinated as if you have 0,05M H3PO4???
To calculate the pH after adding 10.00 mL of HCl to the KH2PO4 solution, we can use the concept of acid-base titration and the knowledge of the equilibrium constants (Ka) for the phosphoric acid (H3PO4).
Here's the step-by-step process to calculate the pH:
1. Write the balanced equation for the reaction between KH2PO4 and HCl:
KH2PO4 + HCl -> KCl + H3PO4
2. Determine the initial moles of KH2PO4:
Volume of KH2PO4 solution = 10.00 mL = 0.01000 L
Concentration of KH2PO4 solution = 0.1000 M
Moles of KH2PO4 = Concentration × Volume = 0.1000 M × 0.01000 L = 0.00100 mol
3. Determine the moles of HCl added:
Volume of HCl solution = 10.00 mL = 0.01000 L
Concentration of HCl solution = 0.1000 M
Moles of HCl = Concentration × Volume = 0.1000 M × 0.01000 L = 0.00100 mol
4. As KH2PO4 is a weak acid and HCl is a strong acid, the HCl will completely dissociate in water. Therefore, the moles of H3PO4 formed will be equal to the moles of HCl added.
5. Calculate the moles of H3PO4 formed:
Moles of H3PO4 = Moles of HCl = 0.00100 mol
6. From the balanced equation in step 1, we can see that the ratio of moles of KH2PO4 to moles of H3PO4 is 1:1. Therefore, the moles of KH2PO4 remaining after the reaction will be equal to the initial moles of KH2PO4 minus the moles of H3PO4 formed.
7. Calculate the moles of KH2PO4 remaining:
Moles of KH2PO4 remaining = Moles of KH2PO4 initial - Moles of H3PO4 formed
= 0.00100 mol - 0.00100 mol = 0 mol
8. Calculate the final concentration of KH2PO4:
Volume of KH2PO4 solution remaining = 10.00 mL - 10.00 mL = 0 mL = 0 L
Concentration of KH2PO4 remaining = Moles of KH2PO4 remaining / Volume of KH2PO4 remaining
= 0 mol / 0 L = Undefined
9. Since the concentration of KH2PO4 remaining is zero, we can consider it to be negligible. Therefore, the resulting solution will be a solution of H3PO4.
10. To calculate the pH of the resulting H3PO4 solution, we need to consider the dissociation of the three acidic protons (H+) according to their respective equilibrium constants (Ka) values.
11. For the first dissociation of H3PO4, we can use the Ka1 value:
H3PO4 -> H+ + H2PO4-
The equilibrium expression for Ka1 is:
Ka1 = [H+][H2PO4-] / [H3PO4]
12. Since the concentration of H3PO4 is equal to the moles of H3PO4 divided by the volume of the resulting solution, and the volume of the resulting solution is the sum of the volumes of KH2PO4 and HCl solutions, we can calculate the concentration of H3PO4.
13. Calculate the concentration of H3PO4:
Volume of resulting solution = Volume of KH2PO4 + Volume of HCl
= 10.00 mL + 10.00 mL = 20.00 mL = 0.02000 L
Concentration of H3PO4 = Moles of H3PO4 / Volume of resulting solution
= 0.00100 mol / 0.02000 L = 0.0500 M
14. Now, we can substitute the known values into the equilibrium expression for Ka1 and solve for [H+].