Introductory Chemistry (5th Edition)

Published by Pearson

Chapter 13 - Solutions - Exercises - Problems: 79

Answer

(a) 0.12 M of $Na_{2}SO_{4}$ dissociates to produce 0.24 M of Na+ ions and 0.12 M of $SO_{4}$-2 ions. (b) 0.25 M of $K_{2}CO_{3}$ dissociates to produce 0.5 M K+ ions and 0.25 M of $CO_{3}$-2 ions. (c) 0.11 M of RbBr dissociates to produce 0.11 M of Rb+ ions and 0.11 M of Br- ions.

Work Step by Step

(a) 0.12 M $Na_{2}SO_{4}$ $Na_{2}SO_{4}$ → 2 Na+ + $SO_{4}$-2 1 mole of $Na_{2}SO_{4}$ dissociates to form 2 moles of Na+2 ions and one mole of $SO_{4}$-2 ions. 0.12 M of $Na_{2}SO_{4}$ dissociates to produce 0.24 M of Na+ ions and 0.12 M of $SO_{4}$-2 ions. (b) 0.25 M $K_{2}CO_{3}$ $K_{2}CO_{3}$ → 2 K+ + $CO_{3}$-2 1 mole of $K_{2}CO_{3}$ dissociates to produce 2 moles of K+ and 1 mole of $CO_{3}$-2 ions. So,0.25 M of $K_{2}CO_{3}$ dissociates to produce 0.5 M K+ ions and 0.25 M of $CO_{3}$-2 ions. (c) 0.11 M RbBr RbBr → Rb+ + Br- 1 mole of RbBr dissociates to produce 1 mole of Rb+ ions and 1 mole of Br- ions. So, 0.11 M of RbBr dissociates to produce 0.11 M of Rb+ ions and 0.11 M of Br- ions.

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