If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming that it is the only acid contributing to the pH? For sulfuric acid, Ka1 is very large and Ka2 is 0.012.

Question

05-04-2018
Chemistry

contestada

If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming that it is the only acid contributing to the pH? For sulfuric acid, Ka1 is very large and Ka2 is 0.012.

Respuesta :

Otras preguntas

One known myth about Business Ethics is, 'Our employees are ethical so we don't need pay attention to business ethics.'a) Trueb) False

On the grid, draw the region which satisfies all three of the Inequalities. x < 3 y > -2 y < x

Sine electrical values can be described in vectors and complex numbers. Please describe the following topics; * How can we describe Sinus values using Vector to

What are the notable features and lifespan of the alligator snapping turtle, the largest freshwater turtle in North America?

Lori’s family is on a road trip. They split their drive into the five legs listed in the table. Find the average velocity for each leg of the trip. Then arrange

The two shapes shown below aresimilar.Shape F has a perimeter of 14 cm.Calculate the perimeter of shape G.2 cm F20 cmGNot drawn accurately

Questions1. Describe the function of colon.and how it works

ListenRead the excerpt from "Natural Beauty at Risk."When most of us think of National Parks, we think of open spaces and healthy, abundant vegetation. We think

Apply the Concept of Natural Selectionto a Population of BirdsFlock XFlock YFlock ZThe birds shown above have beaks with threedifferent shapes and sizes. These

Discuss the broader factors driving the adoption of ERP systems in organizations across different industries and geographical regions.

zoexoe zoexoe · Answer 1 · 2018-04-14T21:23:41+02:00

There are 2.32 x 10^6 kg sulfuric acid in the rainfall.

Solution:
We can find the volume of the solution by the product of 1.00 in and 1800 miles2:
1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m
1.00 in * 1 m / 39.3701 in = 0.0254 m
Volume = 4.662 x 10^9 m^2 * 0.0254 m
= 1.184 x 10^8 m^3 * 1000 L / 1 m3
= 1.184 x 10^11 Liters

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70:
[H+] = 10^-pH = 10^-3.7 = 0.000200 M
[H2SO4] = 0.000100 M

By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid:
1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4

We can now calculate for the mass of sulfuric acid in the rainfall:
mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
= 2.32 x 10^9 g * 1 kg / 1000 g
= 2.32 x 10^6 kg H2SO4

Annainn Annainn · Answer 2 · 2019-02-13T06:45:32+01:00

Ans: Mass of H2SO4 present = 231.5*10⁴ kg

Height of rainfall = 1.00 inch = 0.0254 m

Area of rainfall = 1800 miles²

1 mile² = 2.59*10⁶ m²

Therefore area = 1800*2.59*10⁶ = 4.662*10⁹ m²

Volume (V) of rainfall = area (A) * height(h) = 0.0254 * 4.662*10⁹ = 1.184*10⁸ m³

1 m³ = 1000 L

V = 1.184*10¹¹ L

It is given that the pH = 3.70

pH = -log[H+]

[H+] = 10^-pH = 10⁻³.⁷⁰ = 1.995*10⁻⁴M

Now:

Molarity of H2SO4 = moles of H2SO4/volume

moles = 1.995*10⁻⁴moles.L-1 * 1.184*10¹¹ L = 2.362*10⁷ moles

Molar mass of H2SO4 = 98 g/mol

Mass of H2SO4 = 2.362*10⁷ moles * 98 g/mol = 231.48*10⁷ g = 231.5 *10⁴ kg