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How much of a 1.50 M sodium sulfate solution in

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

Solution for problem 118P Chapter 13

Introductory Chemistry | 5th Edition

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

Introductory Chemistry | 5th Edition

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Problem 118P

How much of a 1.50 M sodium sulfate solution in milliliters is required to completely precipitate all of the barium in 150.0 mL of a 0.250 M barium nitrate solution?

Step-by-Step Solution:

Problem 118P :

Step 1:

Given :

Molarity of Sodium sulfate(Na2SO4) solution = 1.50 M

Molarity of barium nitrate(Ba(NO3)2) solution = 0.250 M

Volume of silver nitrate(AgNO3) solution = 150.0 mL

Volume of solution is in mL, so let’s convert to L :

                 1 L = 1000 mL

Therefore, 150.0 mL in L will be :

                = 150 mL

                = 0.150 L.

First, Let’s write the chemical reaction that’s happening :

Na2SO4 (aq) + Ba(NO3)2 (aq) →  2NaNO3 (aq) + BaSO4 (s)

Now, let’s find the number of moles of barium nitrate( Ba(NO3)2) using the molarity formula as barium nitrate is the solute :

Molarity =

Therefore, the number of moles(m) can be derived from the above formula as :

Moles of solute = Molarity Volume of Solution (in L)

 

n = 0.250 M0.150 L

n = 0.0375 moles

Step 2 of 3

Chapter 13, Problem 118P is Solved
Step 3 of 3

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

Since the solution to 118P from 13 chapter was answered, more than 557 students have viewed the full step-by-step answer. This full solution covers the following key subjects: barium, solution, precipitate, much, nitrate. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. The answer to “How much of a 1.50 M sodium sulfate solution in milliliters is required to completely precipitate all of the barium in 150.0 mL of a 0.250 M barium nitrate solution?” is broken down into a number of easy to follow steps, and 30 words. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. The full step-by-step solution to problem: 118P from chapter: 13 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5.

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How much of a 1.50 M sodium sulfate solution in