Question

Atmospheric \(\mathrm{CO}_{2}\) levels have risen by nearly \(20 \%\) over the past 40 years from 315 ppm to 380 ppm. (a) Given that the average pH of clean, unpolluted rain today is 5.4, determine the pH of unpolluted rain 40 years ago. Assume that carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) formed by the reaction of \(\mathrm{CO}_{2}\) and water is the only factor influencing \(\mathrm{pH}\). $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$ (b) What volume of \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\) is dissolved in a 20.0-L bucket of today's rainwater?

Short answer

The pH of unpolluted rain 40 years ago was approximately 5.6. The volume of CO2 dissolved in a 20-L bucket of today's rainwater at 25°C and 1 atm is 4.22 mL.

Step-by-step solution

Calculate the CO2 concentrations 40 years ago

First, let's calculate the CO2 level 20% less than today's level (380 ppm). Old CO2 level = 380 ppm × (1 - 0.2) = 380 ppm × 0.8 = 304 ppm

Find the H2CO3 concentrations

According to the reaction CO2(g) + H2O(l) ⇌ H2CO3(aq), we can assume that the concentrations of CO2 and H2CO3 are approximately equal (since the reaction mostly goes to completion). So, we have: H2CO3 concentration today = 380 ppm H2CO3 concentration 40 years ago = 304 ppm

Calculate the pH for 40 years ago

Given the pH of today's rainwater (5.4), we can write: pH_Today = -log[H+]_Today We also know that the H+ concentration is directly proportional to the H2CO3 concentration. Thus, we can use the ratio of the H2CO3 concentrations: \(\frac{[H^+]_{40\ years\ ago}}{[H^+]_{Today}} = \frac{[H_2CO_3]_{40\ years\ ago}}{[H_2CO_3]_{Today}}\) \([H^+]_{40\ years\ ago} = \frac{[H_2CO_3]_{40\ years\ ago}}{[H_2CO_3]_{Today}} \times [H^+]_{Today}\) \[= \frac{304 \ ppm}{380 \ ppm} \times [H^+]_{Today}\] Now, we can find the pH of the rain 40 years ago: \(pH_{40\ years\ ago} = -log([H^+]_{40\ years\ ago})\)

Calculate the volume of CO2 at 25°C and 1 atm

Now, we need to find the volume of CO2 dissolved in a 20 L bucket of today's rain. The CO2 concentration is given as 380 ppm, which means 380 parts per million. First, we need to calculate the mass of CO2 in 20 L of rainwater: Mass of CO2 in 20 L = (20 L × 380 ppm) = 7.6 mg To find the volume, we will use the ideal gas law: PV = nRT where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature. First, we convert the mass of CO2 to moles, using the molar mass of CO2 (44.01 g/mol): Moles of CO2 = \(\frac{7.6 \times 10^{-3}\ g}{44.01\ g/mol} = 1.73 \times 10^{-4}\ mol\) Now, we can find the volume using the given conditions – P = 1 atm and T = 25°C (298 K). The value of the gas constant (R) is 0.0821 L.atm/mol.K: \(V = \frac{nRT}{P} = \frac{(1.73 \times 10^{-4}\ mol)(0.0821\ L\cdot atm/mol\cdot K)(298\ K)}{1\ atm} = 4.22 \times 10^{-3}\ L\) So, the volume of CO2 dissolved in a 20-L bucket of today's rainwater at 25°C and 1 atm is 4.22 mL.

Key concepts

Carbonic Acid

Carbonic acid, represented by the formula \(\text{H}_2\text{CO}_3\), is formed when carbon dioxide \(\text{CO}_2\) dissolves in water. This reaction is crucial for understanding the chemistry of rainwater. When \(\text{CO}_2\) from the atmosphere encounters water droplets in clouds, it gets converted into carbonic acid. This weak acid slightly decreases the pH of rainwater. The equation for this process is: \[ \text{CO}_2 (g) + \text{H}_2\text{O} (l) \rightleftharpoons \text{H}_2\text{CO}_3 (aq) \] Carbonic acid is a key player in maintaining the equilibrium between \(\text{CO}_2\) and water, making it an essential component in understanding environmental and biological systems. This weak acid influences both weathering processes in rocks and physiological mechanisms in organisms.

CO2 Levels

Over the past four decades, \(\text{CO}_2\) levels have climbed by about 20%, impacting the acidity of rain. Increased atmospheric \(\text{CO}_2\) translates to higher \(\text{H}_2\text{CO}_3\) concentrations in rain, altering the pH balance.

• 40 years ago: 304 ppm
• Today: 380 ppm

These shifts matter because they show how human activity and natural processes affect atmospheric chemistry. Increased \(\text{CO}_2\) levels not only modify the pH of rain but also play a role in climate change, ocean acidification, and plant growth dynamics. Understanding these levels is vital for predicting future environmental changes.

pH of Rainwater

The pH scale measures the acidity or basicity of a solution. A pH of 7 is neutral, below 7 is acidic, and above 7 is basic. Clean rainwater is slightly acidic, usually around a pH of 5.4 due to carbonic acid. To understand historical pH changes, consider the relationship: as \([\text{H}_2\text{CO}_3]\) increases, more \([\text{H}^+]\) ions are produced, lowering the pH. For example, if today's rainwater has a pH of 5.4, calculating the pH 40 years ago involves examining the ratio of \(\text{H}_2\text{CO}_3\) concentrations:\[ \frac{[\text{H}^+]_{40\ years\ ago}}{[\text{H}^+]_{today}} = \frac{[\text{H}_2\text{CO}_3]_{40\ years\ ago}}{[\text{H}_2\text{CO}_3]_{today}} \] This equation helps us understand how the pH of rainwater has evolved over time due to changes in atmospheric \(\text{CO}_2\) levels.

Ideal Gas Law

The ideal gas law helps us understand the relationship between pressure, volume, temperature, and the number of moles of a gas. It's expressed as: \[ PV = nRT \] Where:

• \(P\) is pressure in atm
• \(V\) is volume in L
• \(n\) is moles of gas
• \(R\) is the gas constant (0.0821 L.atm/mol.K)
• \(T\) is temperature in Kelvin

To find the volume of \(\text{CO}_2\) dissolved in rainwater, convert the mass to moles and apply this equation. For instance, at 25°C and 1 atm, 1.73 x \(10^{-4}\) moles of \(\text{CO}_2\) translates to a volume of 4.22 mL in 20 L of rain. This calculation illustrates how the ideal gas law allows us to estimate gas behaviors in various conditions, an essential tool for chemists and environmental scientists alike.