/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 119 Atmospheric \(\mathrm{CO}_{2}\) ... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 119

Atmospheric \(\mathrm{CO}_{2}\) levels have risen by nearly \(20 \%\) over the past 40 years from \(315 \mathrm{ppm}\) to \(380 \mathrm{ppm}\). (a) Given that the average \(\mathrm{pH}\) of clean, unpolluted rain today is \(5.4\), determine the \(\mathrm{pH}\) 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

Expert verified
(a) The pH of unpolluted rain 40 years ago is approximately -log(\(304 \times \frac{10^{(-5.4)}}{380}\)). (b) The volume of CO鈧� in a 20.0-L bucket of today's rainwater is \(\frac{(3.8 \times 10^{-4} mol/L \times 20L)(0.0821 L atm/mol K)(298 K)}{1 atm}\) L.

Step by step solution

01

(a) Find the pH of unpolluted rain 40 years ago

1. Determine the CO2 concentration 40 years ago: Since CO2 levels have risen by 20%, CO2 concentration 40 years ago = Current concentration - 0.20 脳 Current concentration Current concentration = \(380 ppm\) \(CO2_{40years\_ago} = 380 - 0.20 \times 380 = 304 ppm\) 2. Determine the H鈧侰O鈧� concentration: The reaction between CO2 and H2O is in equilibrium. As the CO2 concentration increases, the H鈧侰O鈧� concentration increases as well. We assume that the ratio of CO2 concentration to H鈧侰O鈧� concentration is constant. \(\frac{H_{2}CO_{3_{today}}}{CO_{2_{today}}} = \frac{H_{2}CO_{3_{40years\_ago}}}{CO_{2_{40years\_ago}}}\) Given the pH of today's rainwater, we can calculate the concentration of H鈧侰O鈧�: pH = -log(\(H_{3}O^{+}\)), Thus, \(H_{3}O^{+}\) = \(10^{(-5.4)}\) The dissociation of carbonic acid H鈧侰O鈧� forms H鈧僌鈦� and HCO鈧冣伝 ions. Therefore, \(H_{2}CO_{3_{today}} = H_{3}O^{+}_{today}\). Now we can find the concentration of H鈧侰O鈧� in rainwater 40 years ago. \(H_{2}CO_{3_{40years\_ago}} = CO_{2_{40years\_ago}} \times \frac{H_{2}CO_{3_{today}}}{CO_{2_{today}}}\) \(H_{2}CO_{3_{40years\_ago}} = 304 \times \frac{10^{(-5.4)}}{380}\) 3. Calculate the pH of rain 40 years ago: The concentration of H鈧僌鈦� ions in the rainwater 40 years ago will be equal to the concentration of H鈧侰O鈧� as it dissociates into H鈧僌鈦� ions. Therefore, \(H_{3}O^{+}_{40years\_ago} = H_{2}CO_{3_{40years\_ago}}\) pH = -log(\(H_{3}O^{+}_{40years\_ago}\)) pH = -log(\(304 \times \frac{10^{(-5.4)}}{380}\))
02

(b) Calculate the volume of CO鈧� in a 20.0-L bucket of today's rainwater

1. Calculate the moles of CO鈧�: First, convert the CO2 concentration from ppm to moles: \(CO_{2}\) concentration (mol/L) = \(CO_{2}\) concentration (ppm) 脳 \(\frac{1mol}{10^6 L}\) \(CO_{2}\) concentration (mol/L) = \(380 ppm \times \frac{1 mol}{10^6 L} = 3.8 \times 10^{-4} mol/L\) Now calculate the moles of CO鈧� in 20.0-L bucket of rainwater: Moles of CO鈧� = \(CO_{2}\) concentration (mol/L) 脳 Volume of rainwater (L) Moles of CO鈧� = \(3.8 \times 10^{-4} mol/L \times 20L\) 2. Use the Ideal Gas Law to find the volume of CO鈧�: The Ideal Gas Law is: PV = nRT Where P = pressure (1 atm), V = volume (L), n = moles of CO鈧�, R = gas constant (0.0821 L atm/mol K), and T = temperature (298 K) Rearrange the formula and solve for volume V: V = \(\frac{nRT}{P}\) V = \(\frac{(3.8 \times 10^{-4} mol/L \times 20L)(0.0821 L atm/mol K)(298 K)}{1 atm}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atmospheric CO2 levels
Over the past 40 years, the levels of carbon dioxide (CO鈧�) in the atmosphere have increased significantly. This rise can have various implications on environmental and biological systems. Increased CO鈧� levels, climbing from 315 parts per million (ppm) to 380 ppm, means there's more CO鈧� available in the air to dissolve into water and form carbonic acid (H鈧侰O鈧�).
\[\text{CO}_2(g) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{CO}_3(aq)\]
Why is this important? CO鈧� plays a central role in the Earth's carbon cycle and has direct effects on climate change by trapping heat in the atmosphere. Higher CO鈧� can also lead to acidification of oceans and freshwater systems due to increased formation of carbonic acid, which can lower pH levels. Understanding CO鈧� levels is vital to assessing its impact on both natural systems and human health.
Rainwater pH
The pH of rainwater is an indication of its acidity or alkalinity. It's primarily influenced by the presence of CO鈧� in the atmosphere. When CO鈧� dissolves in rainwater, it forms carbonic acid, contributing to a lower pH level.
In unpolluted conditions, rainwater typically has a pH around 5.6 due to naturally occurring carbonic acid. Today's average rainwater pH of 5.4 suggests mild acidity, but 40 years ago, with lower CO鈧� levels at 304 ppm, the rain would have been less acidic, meaning a higher pH.
To calculate the pH change over the years, we assume:
  • Carbonic acid formation is the sole factor influencing pH.
  • The proportion between CO鈧� and H鈧侰O鈧� remains constant.
This can be calculated using: \[\text{pH} = -\log([\text{H}_3\text{O}^+])\] Where \[\text{H}_3\text{O}^+\] represents the concentration of hydronium ions, directly linked to H鈧侰O鈧� concentration. Greater atmospheric CO鈧� leads to more carbonic acid, shifting the equilibrium towards increased hydronium ions and thereby lowering the pH.
Ideal Gas Law
The Ideal Gas Law is a critical equation in chemistry, which relates the state of a gas to its pressure, volume, temperature, and amount. It's expressed as:\[PV = nRT\]Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant (0.0821 L atm/mol K), and \(T\) is the temperature in Kelvin.

This equation helps to determine the volume of CO鈧� dissolved in rainwater at specific conditions. Suppose we have rainwater containing 3.8 x 10鈦烩伌 mol/L of CO鈧� in a 20.0-L bucket at conditions of 25掳C and 1 atm pressure.
  • First, we calculate moles of CO鈧� by multiplying the concentration by the volume of the rainwater.
  • Next, rearrange the Ideal Gas Law to solve for volume:
  • \[V = \frac{nRT}{P}\]
The calculated volume is a measure of how much space this dissolved gas would occupy if it were in a purely gaseous state, offering insight into how gases behave under various conditions.

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Most popular questions from this chapter

The hypochlorite ion, \(\mathrm{ClO}^{-}\), acts as a weak base. (a) Is \(\mathrm{ClO}^{-}\)a stronger or weaker base than hydroxylamine? (b) When \(\mathrm{ClO}^{-}\) acts as a base, which atom, \(\mathrm{Cl}\) or \(\mathrm{O}\), acts as the proton acceptor? (c) Can you use formal charges to rationalize your answer to part (b)?

For solutions of a weak acid, a graph of \(\mathrm{pH}\) versus the logarithm of the initial acid concentration should be a straight line. What is the magnitude of the slope of that line?

(a) The hydrogen sulfite ion \(\left(\mathrm{HSO}_{3}^{-}\right)\)is amphiprotic. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base toward water. (b) What is the conjugate acid of \(\mathrm{HSO}_{3}^{-}\)? What is its conjugate base?

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), (b) \(\mathrm{HCO}_{3}^{-}\)

Calculate the molar concentration of \(\mathrm{OH}^{-}\)in a \(0.724 \mathrm{M}\) solution of hypobromite ion \(\left(\mathrm{BrO}^{-} ; K_{b}=4.0 \times 10^{-6}\right)\). What is the \(\mathrm{pH}\) of this solution?
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