Phosphoric acid, H3PO4, ionizes according to the equations, H3PO4(aq) ⇌ H+(aq) + H2PO4 –(aq) K1 = 7. — Acid-Base Equilibrium / Buffers Chemistry Question
Problem Context
Phosphoric acid, H3PO4, ionizes according to the equations,
H3PO4(aq) ⇌ H+(aq) + H2PO4 –(aq) K1 = 7.1 × 10-3
H2PO4 –(aq) ⇌ H+(aq) + HPO4 2-(aq) K2 = 6.2 × 10-8
HPO4 2-(aq) ⇌ H+(aq) + PO4 3-(aq) K3 = 4.5 × 10-13
Write the equilibrium expression for the ionization of H3PO4 and find the pH of a 1.5 M solution of H3PO4.
Model Answer
for a 1.5 M solution, assuming no initial concentration of reactants and that the amount of phosphoric acid that reacts is small compared to the original volume,
Solving for x, x = 1.03 × 10-1 M. Use successive approximations to check that the amount of reacting phosphoric acid doesn’t change the answer… plug in 1.397 (1.5 - 0.103 = 1.397) rather than 1.5,
x = 1.00 × 10-1 M the change is small enough to accept this answer.
A student is asked to prepare a phosphate buffer with a pH of 7.00. Identify the species that should be used in this solution and calculate their ratio.
Model Answer
To obtain a pH of 7 the Ka should be close to 7. In this case it would be Ka2. Thus the species that would be present in this buffer should be H2PO4–(aq) and HPO42-(aq).
The ratio can be found by the equation,
and the ratio is,
Assume that 50.0 mL of the buffer solution in b. are available in which the more abundant buffer species has a concentration of 0.10 M. Determine the [H+] in this solution after 2.0 x 10-3 mol of NaOH are added.
Model Answer
H2PO4–(aq) is the more abundant species in the buffer from part b.
0.050 L × 0.10 mol·L-1 = 0.0050 mol H2PO4– initially
0.0020 mol of OH– is added, so the amount of H2PO4–(aq) left is (1 to 1 stoichiometry)
0.0050 – 0.0020 = 0.0030 mol H2PO4–(aq)
If the concentration of H2PO4–(aq) is 0.10 M the ratio calculated earlier indicates that the initial concentration of HPO42-(aq) must be 0.62 × 0.10 = 0.062 M, so
0.050 L × 0.062 mol·L-1 = 0.0031 mol HPO42- initially
and 0.0020 mol are formed in the reaction so there is 0.0051 mol HPO42-(aq) present.
Calculating the amount of hydrogen ion present,
solving for [H+] gives, 3.65 × 10-8 M
Determine the [H+] in a 0.20 M solution of Na3PO4.
Model Answer
PO4 3– + H2O ⇌ HPO4 2– + OH– and
Kb = Kw / Ka3 = 1.0 × 10-14 / 4.5 × 10-13 = 0.0222
Obtain concentration of hydroxide, initially assume no reaction of phosphate and let x = [OH–],
0.0222 = x^2 / 0.20 so x = [OH–] = 6.67 × 10-2 M now for phosphate, 0.20 – 0.0667 = 0.1333 M
0.0222 = x^2 / 0.1333 so x = [OH–] = 5.44 × 10-2 M now for phosphate, 0.20 - 0.0544 = 0.1456 M
0.0222 = x^2 / 0.1456 so x = [OH–] = 5.70 × 10-2 M now for phosphate, 0.20 - 0.0570 = 0.143 M
0.0222 = x^2 / 0.143 so x = [OH–] = 5.63 × 10-2 M which is acceptably close to the previous iteration, now use Kw to calculate [H+],
[H+] = 1.0 × 10-14 / 5.63 × 10-2 = 1.77 × 10-13 M