×
Log in to StudySoup
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 73e
Join StudySoup for FREE
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 73e

Already have an account? Login here
×
Reset your password

Determine whether each compound is soluble or insoluble.

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 73E Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

4 5 1 286 Reviews
20
5
Problem 73E

Determine whether each compound is soluble or insoluble. If the compound is soluble, list the ions present in solution.

a. AgNO3                 b. Pb(C2H3O2)2                 c. KNO3                 d. (NH4)2S

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the volume of stock solution used to prepare 1000 mL of 0.250 M ammonia. Also, we are going to find the concentration of the final solution if 10 mL of the stock solution is diluted to 0.500 L. Step1: In order to dilute a stock solution, we have to use the following dilution equation, M 1 1M V 2 2 -----------(1) Here, M an1V are t1 molarity and volume of the concentrated stock solution whereas M and 2 V are the molarity and volume of the diluted solution. 2 Step2: a) In the above problem, let us consider the stock solution of NH to be concentrated 3 solution and the solution of NH to be prep3ed as dilute solution. Therefore, M (con1ntrated) = 14.8 M V (concentrated) = 1 M 2ilute) = 0.250 M V 2ilute) = 1000.0 mL Putting these values in equation (1), we get, 14.8 M X V = 0.10 M X 1000 mL 14.8 M X V = 250 (M x mL) 1 V1 250 / 14.8 [(M x mL) / M] = 16.89 mL Thus, 16.89 mL of stock solution is required to make 1000.0 ml of 0.250 M ammonia. Step3: b) Again, for the second problem, let us consider the stock solution of NH to be 3 concentrated solution and the solution of NH to be prepared 3 dilute solution. Therefore, M (concentrated) = 14.8 M 1 V 1oncentrated) = 10.0 mL M 2ilute) = V (dilute) = 0.500 L 2 = 0.500 x 1000 mL = 500 mL Putting these values in equation (1), we get, 14.8 M X 10.0 mL = M X 500 mL 2 148 (M x mL) = M X 500 mL2 M 2148 / 500 [(M x mL) / mL] = 0.296 M Thus, the concentration of the final solution will be 0.296 M. ----------------------

Step 2 of 3

Chapter 4, Problem 73E is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

The answer to “?Determine whether each compound is soluble or insoluble. If the compound is soluble, list the ions present in solution. a. AgNO3 b. Pb(C2H3O2)2 c. KNO3 d. (NH4)2S” is broken down into a number of easy to follow steps, and 27 words. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. The full step-by-step solution to problem: 73E from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. Since the solution to 73E from 4 chapter was answered, more than 298 students have viewed the full step-by-step answer. This full solution covers the following key subjects: solution, dilute, stock, mnh, portion. This expansive textbook survival guide covers 82 chapters, and 9454 solutions.

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Determine whether each compound is soluble or insoluble.