Answered: How many mL of a 0.218 M Na3PO4…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

How many mL of a 0.218 M Na3PO4 solution are required to precipitate all the calcium ions in a 395 mL of a 0.254 M CaCl2 solution. Go to 3 sig figs and do not use scientific notation

Expert Solution

Step 1

Precipitation reaction

The reaction in which aqueous cationic species and anionic species combine to form insoluble precipitate is called precipitation reaction.

The amount of precipitate formed depends upon the mole ratio of the combining ions.

Eg:

${Ag}^{+} (aq) + {Cl}^{-} (aq) \to AgCl (s)$

Step 2

The reaction between Na₃PO₄ and CaCl₂ is;

$3 {CaCl}_{2} (aq) + 2 {Na}_{3} {PO}_{4} (aq) \to {Ca}_{3} {({PO}_{4})}_{2} (s) + 6 NaCl (aq)$

The precipitation reaction between Ca²⁺ and PO₄^3- ions is;

$3 {Ca}^{2 +} (aq) + 2 {PO}_{4}^{3 -} (aq) \to {Ca}_{3} {({PO}_{4})}_{2} (s)$

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