×
Log in to StudySoup
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 14 - Problem 118p
Join StudySoup for FREE
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 14 - Problem 118p

Already have an account? Login here
×
Reset your password

Lakes that have been acidified by acid rain (see Chemistry in the Environment box in

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 118P Chapter 14

Introductory Chemistry | 5th Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

4 5 1 344 Reviews
30
5
Problem 118P

Problem 118P

Lakes that have been acidified by acid rain (see Chemistry in the Environment box in Section 5.9) can be neutralized by liming, the addition of limestone (CaCO3). How much limestone in kilograms is required to completely neutralize a 3.8 × 109 L lake with a pH of 5.5?

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the mass of limestone required to completely neutralize the lake.

Step1:

pH is the negative logarithm to base 10 of hydrogen ion concentration.

pH = - log[H+ ]

Step2:

Given, pH of the Lake = 5.5

                pH =  - log[H+]

Therefore, [H+] = 10-pH 

                     = 10-5.5

                     = 3.16 x 10-6 M

Therefore, number of moles of hydrogen ion = Volume of the lake x [H+]

                                                = (3.8 x 109 L) x (3.16 x 10-6 mol/L)

                                                = 12.0 x 103 mol

Step3:

The neutralization reaction taking place in the lake is:

                        

CaCO3(s) + 2H+(aq) → Ca2+ + H2O + CO2

From the reaction, it is clear that for neutralization of 2 mol of hydrogen ion, 1 mole of CaCO3 is required.

Therefore, to neutralize 12 x 103 mol of hydrogen ion, number of moles of CaCO3 required = ½ x (1.2 x 103) = 6.0 x 103 mol.

Step4:

1 mol CaCO3 = molar mass of CaCO3 

Therefore, 6.0 x 103 mol CaCO3  = 6.0 x 103  x molar mass of CaCO3 

                                     = 6.0 x 103  x 100.0869 g/mol

                                     = 600.52 x 103  g

                                     = 600.52 kg        [1000 g = 1 kg]

Thus, 600.52 kg of limestone will be required.

                                                ---------------------

                                

Step 2 of 3

Chapter 14, Problem 118P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

The answer to “Lakes that have been acidified by acid rain (see Chemistry in the Environment box in Section 5.9) can be neutralized by liming, the addition of limestone (CaCO3). How much limestone in kilograms is required to completely neutralize a 3.8 × 109 L lake with a pH of 5.5?” is broken down into a number of easy to follow steps, and 48 words. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. Since the solution to 118P from 14 chapter was answered, more than 399 students have viewed the full step-by-step answer. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. This full solution covers the following key subjects: limestone, lake, addition, been, box. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. The full step-by-step solution to problem: 118P from chapter: 14 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM.

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Lakes that have been acidified by acid rain (see Chemistry in the Environment box in