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?
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
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
Moles of SO42- = 2.83 x 10-3 mol/L x 75.0 L = 0.212 mol
Textbook: Introductory Chemistry
Author: Nivaldo J Tro
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 282 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.