/*! 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 81 If the pH of a \(1.0\) -in. rain... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 81

If the pH of a \(1.0\) -in. rainfall over \(1500 \mathrm{mi}^{2}\) is \(3.5\), how many kilograms of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are present, assuming that it is the only acid contributing to the \(\mathrm{pH}\) ?

Short Answer

Expert verified
The mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) in the rainfall, in kilograms, is given by: \[Mass\,\,of\,\,\mathrm{H}_{2}\mathrm{SO}_{4}\,\,(in\,\,kg) = \frac{98.079 \times \left(\frac{1}{2} \times 10^{-3.5} \times 0.0254 \times 3.880\times10^9\right)}{1000}\]Calculating the value of this expression will provide the mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) in the rainfall, in kilograms.

Step by step solution

01

Find the concentration of hydrogen ions

To find the concentration of hydrogen ions, we can use the formula that relates pH and hydrogen ion concentration: \[pH = - \log_{10} {[H^{+}]}\] where \({[H^{+}]}\) is the hydrogen ion concentration. Given pH is \(3.5\), we can calculate the \({[H^{+}]}\) as follows: \[{[H^{+}]} = 10^{-pH} = 10^{-3.5}\]
02

Determine the concentration of \(\mathrm{H}_{2}\mathrm{SO}_{4}\)

Since \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is assumed to be the only acid contributing to the pH, the concentration of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) can be equal to half of the concentration of hydrogen ions (as 1 molecule of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissociates into 2 hydrogen ions): \[[\mathrm{H}_{2}\mathrm{SO}_{4}] = \frac{1}{2}[H^{+}] = \frac{1}{2} \times 10^{-3.5}\]
03

Calculate the volume of rainfall

We are given the depth of rainfall (1.0 inch) and the area (1500 mi^2). Let's first convert the units to meters. Depth of 1 inch = 0.0254 meters Area of 1500 mi^2 = \(3.880\times10^9\,\) m^2 Now, we can calculate the volume of rainfall by multiplying the depth and area: \[Volume = Depth \times Area = 0.0254 \times 3.880\times10^9\,\]
04

Find the mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\)

First, we determine the total moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) in the volume of rainfall by multiplying the concentration by the volume: \[Moles\,\,of\,\,\mathrm{H}_{2}\mathrm{SO}_{4} = [\mathrm{H}_{2}\mathrm{SO}_{4}] \times Volume = \frac{1}{2} \times 10^{-3.5} \times 0.0254 \times 3.880\times10^9\,\] Now, we can find the mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) using the molar mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) which is \(98.079\,\text{g/mol}\): \[Mass\,\,of\,\,\mathrm{H}_{2}\mathrm{SO}_{4} = Moles\,\,of\,\,\mathrm{H}_{2}\mathrm{SO}_{4} \times Molar\,\,Mass = 98.079 \times \left(\frac{1}{2} \times 10^{-3.5} \times 0.0254 \times 3.880\times10^9\right)\] Finally, we can convert the mass in grams to kilograms by dividing it by 1000: \[Mass\,\,of\,\,\mathrm{H}_{2}\mathrm{SO}_{4}\,\,(in\,\,kg) = \frac{Mass\,\,of\,\,\mathrm{H}_{2}\mathrm{SO}_{4}\,\,(in\,\,g)}{1000}\] Calculate the value of the expression, and you will find the mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) in the rainfall, in kilograms.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

pH Measurement
Understanding pH is essential when discussing acidity. It's a measure of how acidic or basic a solution is. The pH scale ranges from 0 to 14. Here's a quick breakdown:
  • A pH less than 7 means the solution is acidic.
  • A pH of 7 indicates a neutral solution.
  • A pH greater than 7 indicates a basic or alkaline solution.
For acid rain with a pH of 3.5, it's clearly acidic. The formula connecting pH to hydrogen ion concentration is:\[pH = - \log_{10} {[H^{+}]}\]This equation allows us to find out how many hydrogen ions are present in any given solution. So, a lower pH implies a higher concentration of hydrogen ions.
In this exercise, the given pH of 3.5 can be plugged into the formula to find the concentration of hydrogen ions.
Hydrogen Ion Concentration
Hydrogen ion concentration is key to understanding a solution's acidity. The concentration \([H^{+}]\) tells us how many hydrogen ions are in the solution.
Using the pH value, we can find this concentration through:\[{[H^{+}]} = 10^{-pH}\]Plugging in our example:\[{[H^{+}]} = 10^{-3.5}\]This calculation gives us the concentration of hydrogen ions necessary to determine the level of acidity in the rain. Since sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) breaks down into two hydrogen ions, the concentration of sulfuric acid is half of that of the hydrogen ions. This is crucial because it helps in further steps like finding the mass of acid present in the rainfall.
Unit Conversion
Unit conversion is often required in chemistry problems to ensure consistency in calculations. Here, the dimensions of the rainfall need to be consistent.**Converting Rainfall Depth:**
- 1 inch to meters equals 0.0254 meters.
- This allows us to work with metric units, making calculations smoother.**Converting Area:**
  • 1500 square miles to square meters is \(3.880\times10^9\, \text{m}^2\).
Calculating the volume of rainfall requires multiplication of depth by area, both now in compatible metric units:
\[Volume = 0.0254 \times 3.880\times10^9\]
With this calculated volume, further steps, like calculating moles or mass, become much more straightforward.
Stoichiometry
Stoichiometry involves measuring the quantitative relationships in a chemical reaction. For this exercise, it's about counting the amount of sulfuric acid based on the hydrogen ion concentration. First, calculate moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) based on the hydrogen ion concentration and the volume of rain. The link between \(\mathrm{H}_{2}\mathrm{SO}_{4}\) and \([H^{+}]\) is given by dividing the concentration of hydrogen ions by 2, since each molecule of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) produces 2 hydrogen ions. Once moles are calculated:
  • Use the molar mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), which is 98.079 g/mol, to find the mass of acid.
  • Convert grams to kilograms by dividing by 1000.
This allows precise calculation of the total mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) in the rainfall, providing the final answer in a practical and usable form.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A reaction that contributes to the depletion of ozone in the stratosphere is the direct reaction of oxygen atoms with ozone: $$ \mathrm{O}(g)+\mathrm{O}_{3}(g) \longrightarrow 2 \mathrm{O}_{2}(g) $$ At \(298 \mathrm{~K}\) the rate constant for this reaction is \(4.8 \times 10^{5} M^{-1} \mathrm{~s}^{-1} .\) (a) Based on the units of the rate constant, write the likely rate law for this reaction. (b) Would you expect this reaction to occur via a single elementary process? Explain why or why not (c) From the magnitude of the rate constant, would you expect the activation energy of this reaction to be large or small? Explain. (d) Use \(\Delta H^{\circ}\), values from Appendix \(C\) to estimate the enthalpy change for this reaction. Would this reaction raise or lower the temperature of the stratosphere?

(a) Why is the fluorine present in chlorofluorocarbons not a major contributor to depletion of the ozone layer? (b) What are the chemical forms in which chlorine exists in the stratosphere following cleavage of the carbonchlorine bond?

(a) Distinguish between photodissociation and photoionization. (b) Use the energy requirements of these two processes to explain why photodissociation of oxygen is more important than photoionization of oxygen at altitudes below about \(90 \mathrm{~km}\).

Air pollution in the Mexico City metropolitan area is among the worst in the world. The concentration of ozone in Mexico City has been measured at 441 ppb \((0.441 \mathrm{ppm})\). Mexico City sits at an altitude of 7400 feet, which means its atmospheric pressure is only \(0.67\) atm. Calculate the partial pressure of ozone at 441 ppb if the atmospheric pressure is 067 atm.

An important reaction in the formation of photochemical smog is the photodissociation of \(\mathrm{NO}_{2}\) : $$ \mathrm{NO}_{2}+h \nu \longrightarrow \mathrm{NO}(g)+\mathrm{O}(g) $$ The maximum wavelength of light that can cause this reaction is \(420 \mathrm{~nm} .\) (a) In what part of the electromagnetic spectrum is light with this wavelength found? (b) What is the maximum strength of a bond, in \(\mathrm{kJ} / \mathrm{mol}\), that can be broken by absorption of a photon of \(420-\mathrm{nm}\) light? (c) Write out the photodissociation reaction showing Lewis-dot structures.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.