Phosphoric acid is a triprotic acid with pK_A1 = 2.12, pK_A2 = 7.2, pK_A3 = 12.3
ID: 482474 • Letter: P
Question
Phosphoric acid is a triprotic acid with pK_A1 = 2.12, pK_A2 = 7.2, pK_A3 = 12.3. Assume that you have a solution containing 1 mM phosphoric acid. Graph the concentrations of each component vs. pH. Calculate the pH of a 1 mM phosphoric acid solution. Write out the charge balance of a 1 mM phosphoric acid solution that also contains 3 mM NaOH and 2 mM HCl. What is the pH of the solution in part c? Now imagine that you have a 1 mM phosphoric acid along with a 3 mM carbonic acid solutions (pK_A1 = 6.3, pK_A2 = 10.3) and 2 mM NaOH. Write out the charge balance. Solve for the pH of the solution in part e.Explanation / Answer
2. Phosphoric acid
- lets say we have 20 ml of 0.001 M H3PO4
titrated with 0.001 M NaOH solution
NaOH (ml) pH
1 0.84
5 1.64
10 2.12
20 4.66
30 7.2
40 9.75
50 12.3
60 12.3
The graph is plotted as NaOH (ml) on x-axis and pH on y-axis
- pH of 0.001 M H3PO4
H3PO4 <===> H+ + H2PO4-
Ka = [H+][H2PO4-]/[H3PO4]
7.58 x 10^-3 = x^2/(0.001 - x)
x^2 + 7.58 x 10^-3x - 7.58 x 10^-6 = 0
x = [H+] = 8.95 x 10^-4 M
pH = -log[H+] = 3.05
- Charge balance
[0.001 M H2PO4-] = [0.001 M Na+]
- pH of solution
pH = 1/2(pKa1 + pKa2)
= 1/2(2.12 + 7.2)
= 4.66
- Charge balance
[0.001 M H2PO4-] + [0.001 M HCO3-] = [0.002 M Na+]
- pH of solution
pH of 0.001 M H2PO4- = 4.66
pH of HCO3-
= 6.3 + log(0.001/0.002) = 6.0
Overall pH = 10.6