Problem 93 - Chemistry The Central Science

Question

If the \(\mathrm{pH}\) 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 \(\mathrm{pH}\) ?

Step-by-Step Solution

Answer

The mass of H2SO4 in the rainfall is calculated as follows: Moles of H2SO4 = \(\frac{10^{-3.5}}{2}\) * \(9.889 \times 10^8 \, L\) Mass of H2SO4 = Moles of H2SO4 * 98 g/mol Mass of H2SO4 (kg) = Mass of H2SO4 (g) * \(10^{-3} \, kg/g\) After calculating these values, we can determine the mass of H2SO4 in kilograms present in the rainfall.

Step 1: Find the concentration of H+ ions from the given pH value

We're given that the pH of the rainfall is 3.5. The equation to find the concentration of H+ ions is: \(pH = -\log[H^+]\) Now, let's calculate the concentration of H+ ions: \([H^+] = 10^{-pH}\) \[ [H^+] = 10^{-3.5} \]

Step 2: Find moles of H2SO4 in the solution assuming it's the only acid

Since H2SO4 is the only acid contributing to the pH, we can assume the molarity of H2SO4 is equal to the concentration of H+ ions we calculated in Step 1. The stoichiometry between H2SO4 and H+ ions is a 1:2 ratio, therefore the molarity of H2SO4 will be half the concentration of H+ ions. Moles of H2SO4: \[ [H_2SO_4] = \frac{[H^+]}{2}\] \[ [H_2SO_4] = \frac{10^{-3.5}}{2} \]

Step 3: Convert the rainfall volume to liters

We're given a 1.0-inch rainfall over 1500 sq. miles. To find the total volume of rainfall in liters, we need to convert the given values: 1 inch = 2.54 cm 1 mile = 1.609 km First, convert the area from square miles to square centimeters: \[1500 \, mi^2 * (1.609 \times 10^5 \, cm/mi)^2 = 3.89 \times 10^{11} \, cm^2\] Next, find the volume in cubic centimeters: \(V = 1.0 \, in * 3.89 \times 10^{11} \, cm^2 * (2.54 \, cm/in) = 9.889 \times 10^{11} \, cm^3\) Lastly, convert the volume to liters: \(V = 9.889 \times 10^{11} \, cm^3 * (0.001 \, L/cm^3) = 9.889 \times 10^{8} \, L\)

Step 4: Calculate the mass of H2SO4 in the solution

Now that we have the molarity of H2SO4 and the volume of the solution, we can calculate the mass of H2SO4: Moles of H2SO4 = Molarity * Volume Moles of H2SO4 = \(\frac{10^{-3.5}}{2}\) * \(9.889 \times 10^8 \, L\) Molecular weight of H2SO4 = 98 g/mol Mass of H2SO4 = Moles of H2SO4 * Molecular Weight Mass of H2SO4 = Moles of H2SO4 * 98 g/mol Convert the mass to kilograms: Mass of H2SO4 (kg) = Mass of H2SO4 (g) * \(10^{-3} \, kg/g\)

Step 5: Final Answer

Calculate the mass of H2SO4 in kilograms using the above formulas to find the amount of H2SO4 in the rainfall.

Key Concepts

Concentration of Hydrogen Ions

Understanding the concentration of hydrogen ions ([H^+]) is essential in determining the acidity of a solution. The [H^+] concentration is directly linked to the (\text{pH}) of the solution, with a lower (\text{pH}) indicating a higher concentration of hydrogen ions.

The relationship between [H^+] concentration and (\text{pH}) is logarithmic, expressed in the formula:

• \[ pH = -\log[H^+] \]

To find [H^+] concentration, we reverse this relationship to:

• \[ [H^+] = 10^{-\text{pH}} \]

By applying this formula to a solution with (\text{pH} = 3.5), you get:

\( [H^+] = 10^{-3.5} \) mol/L.

This demonstrates the sensitivity of the (\text{pH}) scale due to its logarithmic nature, illustrating how a small change in (\text{pH}) causes a large change in [H^+] concentration.

Stoichiometry of Acid Reactions

Stoichiometry helps us understand the relationships between reactants and products in chemical reactions. For acids, it's about determining how the reactants like (\text{H}_2\text{SO}_4) dissociate in water.

Sulfuric acid (\text{H}_2\text{SO}_4) is a diprotic acid, which means it can release two hydrogen ions ([H^+]) per molecule.

This results in a 1:2 stoichiometric ratio between \text{H}_2\text{SO}_4 and [H^+] ions. Therefore, the concentration of \text{H}_2\text{SO}_4 is half that of the [H^+] ions:

• \[ [H_2\text{SO}_4] = \frac{[H^+]}{2} \]

In our example,
\( [H]^+ \approx 10^{-3.5} \) mol/L,
Thus,
\( [H_2\text{SO}_4] \approx \frac{10^{-3.5}}{2} \) mol/L.

This understanding allows us to calculate exactly how much \text{H}_2\text{SO}_4 is present in the solution, ensuring accurate chemical calculations and predictions.

Conversions and Unit Analysis

Conversions and unit analysis are crucial in chemical calculations to ensure that measurements and calculations are consistent across different units.

In our task, we converted the rainfall volume from inches to liters. First, we needed to understand the area conversion:

• Converting 1500 square miles:\[1500 \, mi^2 * (1.609 \times 10^5 \, cm/mi)^2 = 3.89 \times 10^{11} \, cm^2\]

Next, converting the rainfall depth and calculating the volume:

• Volume:\[V = 1.0 \, in * 3.89 \times 10^{11} \, cm^2 * (2.54 \, cm/in) = 9.889 \times 10^{11} \, cm^3\]

Then, convert the volume to liters using:

• \[ V = 9.889 \times 10^{11} \, cm^3 * (0.001 \, L/cm^3) = 9.889 \times 10^{8} \, L \]

The application of these conversions ensures our calculations for chemical quantities in liters are precise and make the calculation of mass for \text{H}_2\text{SO}_4 practical and accurate.

Understanding these conversions is essential for executing accurate and systematic unit analysis in chemical calculations.