Chapter 16: Additional Aqueous Chemistry and the Oceans – Equilibria, Buffers, and Acid-Base Calculations

Global Warming and Ocean Acidification

Atmospheric CO2 and Carbonic Acid Formation

Atmospheric carbon dioxide (CO2) plays a significant role in ocean acidification. When CO2 dissolves in water, it reacts to form carbonic acid, which affects the pH of natural waters.

• Key Reaction: $\mathrm{CO_2(g) + H_2O(l) \rightleftharpoons H_2CO_3(aq)}$

• Application: This equilibrium is central to understanding how increased atmospheric CO2 leads to lower ocean pH.

• Example: The graph shows a steady increase in atmospheric CO2 from 1960 to 2000, with annual cycles.

The Common Ion Effect

Definition and Impact on Equilibrium

The common ion effect refers to the shift in the position of an equilibrium caused by the addition of an ion already present in the reaction mixture. This effect is important in controlling the ionization of weak acids and bases.

• Example Reaction: $\mathrm{H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)}$

• Equilibrium Constant: $K_{a1} = 4.3 \times 10^{-7}$

• Effect: Adding HCO3- shifts the equilibrium to the left, decreasing ionization of carbonic acid.

Equilibrium Calculations with the Common Ion Effect

When a solution contains both a weak acid and its conjugate base, equilibrium calculations must account for the common ion effect.

Condition	H2CO3(aq)	HCO3-(aq)	H3O+(aq)
Initial (R)	2.7 × 10-3 M	0	0
Change (C)	-x	+x	+x
Equilibrium (E)	(2.7 × 10-3) - x	x	x

The Henderson-Hasselbalch Equation

Buffer Equilibrium and pH Calculation

The Henderson-Hasselbalch equation is a useful tool for calculating the pH of buffer solutions, which contain a weak acid and its conjugate base.

• General Acid Dissociation: $\mathrm{HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)}$

• Equation: $\mathrm{pH = pK_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)}$

• Application: Used to estimate the pH of natural waters and laboratory buffer solutions.

Example: Calculating pH of River Water

Given: [HCO3-] = 1.0 × 10-4 M, [H2CO3] = 1.4 × 10-5 M, pKa = 6.37

• Acid-Base Pair: $\mathrm{H_2CO_3(aq) \rightleftharpoons HCO_3^-(aq) + H_3O^+(aq)}$

• Calculation: $\mathrm{pH = 6.37 + \log \left( \frac{1.0 \times 10^{-4}}{1.4 \times 10^{-5}} \right) = 7.22}$

• Interpretation: The calculated pH is near neutral, as expected for natural waters buffered by carbonic acid and bicarbonate.

Summary Table: Key Equations and Constants

Concept	Equation	Typical Value
Carbonic Acid Dissociation	$\mathrm{H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)}$	$K_{a1} = 4.3 \times 10^{-7}$
Henderson-Hasselbalch	$\mathrm{pH = pK_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)}$	pKa = 6.37 (carbonic acid)

Additional info:

• The notes cover topics from Chapter 16, which is relevant to General Chemistry, specifically acid-base equilibria, buffer solutions, and the chemistry of natural waters.

• Examples and calculations are provided for real-world systems, such as river water and atmospheric CO2 effects.