Acids and Bases

Polyprotic Acids: AP Chemistry Rundown

Polyprotic acids are acids that can donate more than one proton (H⁺) per molecule in solution. This means they have multiple ionization steps, each with its own equilibrium constant.

General Concepts

• Definition: Acids that can donate more than one proton (H⁺).
• Examples: Sulfuric acid (H₂SO₄), carbonic acid (H₂CO₃), phosphoric acid (H₃PO₄).
• Ionization Steps: Each proton donation is a separate step with its own equilibrium.

Ionization Steps and Equilibrium Constants

A polyprotic acid with ’n’ protons will have ’n’ ionization steps. Each step has an associated acid dissociation constant (Ka).

For a generic diprotic acid, H₂A:

1. First Ionization: $$ H_2A(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HA^-(aq) \quad K_{a1} = \frac{[H_3O^+][HA^-]}{[H_2A]} $$
2. Second Ionization: $$ HA^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^{2-}(aq) \quad K_{a2} = \frac{[H_3O^+][A^{2-}]}{[HA^-]} $$

• Key Point: Generally, $ K_{a1} > K_{a2} > K_{a3}… $ . This is because it’s easier to remove a proton from a neutral molecule than from a negatively charged ion. Removing a positively charged proton from a negatively charged species requires more energy.

Calculating pH of Polyprotic Acid Solutions

Calculating the pH of a polyprotic acid solution requires considering the multiple ionization steps. However, due to the significant difference in $ K_a $ values, we can often simplify the calculation.

• Simplification: In most cases, the contribution of the second (and subsequent) ionization steps to the [H₃O⁺] concentration is negligible compared to the first ionization step. Therefore, we can often calculate the pH by only considering the first ionization.

• When Simplification Fails: The simplification is not valid when:

  • $ K_{a1} $ and $ K_{a2} $ are close in value (difference is less than 100x).
  • The acid is very dilute.
  • You need to calculate the concentration of the fully deprotonated species (e.g., [A²⁻] in the H₂A example).

Example: Calculating pH of a Diprotic Acid Solution

Let’s say we have a 0.1 M solution of H₂A with $ K_{a1} = 1.0 \times 10^{-3} $ and $ K_{a2} = 1.0 \times 10^{-7} $ .

1. First Ionization: Set up an ICE table for the first ionization:

   	H₂A	H₃O⁺	HA⁻
   Initial	0.1	0	0
   Change	-x	+x	+x
   Equilibrium	0.1-x	x	x

   $ K_{a1} = \frac{[H_3O^+][HA^-]}{[H_2A]} = \frac{x^2}{0.1-x} = 1.0 \times 10^{-3} $

   Since $ K_{a1} $ is small, we can often assume that ‘x’ is negligible compared to 0.1 (0.1-x ≈ 0.1). However, in this case, we should check the 5% rule.

   $ x = \sqrt{(1.0 \times 10^{-3})(0.1)} = \sqrt{1.0 \times 10^{-4}} = 0.01 $

   $ \frac{0.01}{0.1} \times 100% = 10% $ This is greater than 5%, so we must use the quadratic formula:

   $ x^2 + 0.001x - 0.0001 = 0 $

   $ x = \frac{-0.001 \pm \sqrt{0.001^2 - 4(1)(-0.0001)}}{2(1)} = \frac{-0.001 \pm \sqrt{0.000001 + 0.0004}}{2} = \frac{-0.001 \pm \sqrt{0.000401}}{2} $

   $ x = \frac{-0.001 \pm 0.020025}{2} $

   We take the positive root: $ x = \frac{0.019025}{2} = 0.0095125 $

   $ [H_3O^+] \approx 0.0095 M $

   $ pH = -log(0.0095) \approx 2.02 $

2. Second Ionization: Now consider the second ionization:

   	HA⁻	H₃O⁺	A²⁻
   Initial	0.0095	0.0095	0
   Change	-y	+y	+y
   Equilibrium	0.0095-y	0.0095+y	y

   $ K_{a2} = \frac{[H_3O^+][A^{2-}]}{[HA^-]} = \frac{(0.0095+y)(y)}{0.0095-y} = 1.0 \times 10^{-7} $

   Since $ K_{a2} $ is very small, we can assume y is negligible compared to 0.0095 (0.0095+y ≈ 0.0095 and 0.0095-y ≈ 0.0095).

   $ K_{a2} = \frac{(0.0095)(y)}{0.0095} = y = 1.0 \times 10^{-7} $

   $ [A^{2-}] = 1.0