A 75.0-L solution is 0.0251 M in Ca2+. If Na2SO4 is added

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

Problem 106P Chapter 15

Introductory Chemistry | 5th Edition

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

Introductory Chemistry | 5th Edition

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Problem 106P

A 75.0-L solution is 0.0251 M in Ca2+. If Na2SO4 is added to the solution in order to precipitate the calcium, what minimum mass of Na2SO4 is required to get a precipitate?

Step-by-Step Solution:

Solution 106P

Step 1:

Precipitation occurs when the ionic product exceeds the solubility product constant.

CaSO4  → Ca2+ + SO42-

Solubility product for CaSO4   = 7.10 x 10-5

Concentration of 75.0 L Ca2+ is 0.0251 M

We know,

Ksp = [Ca2+][SO42-]

[SO42-] = Ksp/ Ca2+]

           = 7.10 x 10-5/ 0.0251 = 2.83  x 10-3 M

Thus, minimum concentration of SO42-  2.83  x 10-3 M is required to get a precipitate of  CaSO4

Step 2:  

Moles of SO42-  = 2.83  x 10-3  mol/L x 75.0 L = 0.212  mol

Step 3 of 3

Chapter 15, Problem 106P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo Tro
ISBN: 9780321910295

This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. The answer to “A 75.0-L solution is 0.0251 M in Ca2+. If Na2SO4 is added to the solution in order to precipitate the calcium, what minimum mass of Na2SO4 is required to get a precipitate?” is broken down into a number of easy to follow steps, and 32 words. This full solution covers the following key subjects: solution, precipitate, order, minimum, mass. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. Since the solution to 106P from 15 chapter was answered, more than 250 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 106P from chapter: 15 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM.

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