Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 4 - Reactions in Aqueous Solution - Exercises: 4.69b


30mL of 0.15M K2CrO4 has more K+ ions

Work Step by Step

The first solution is 30mL of 0.15M K2CrO4. First off, we find how many moles of K2CrO4 we have as follows: 30mL * ( 1L/1000mL) = 0.030 L 0.15M = 0.15mol/L * (0.030L) = 0.0045 mol K2CrO4 Then we see that in K2CrO4 there are 2 moles of K+ for each mole of K2CrO4 so 0.0045 mol K2CrO4 * (2 mol K+/1 mol K2CrO4) = 0.009 mole K+ The second solution is 25mL of 0.080 M K3PO4 Here again 25mL --> 0.025L 0.080 mol/L * 0.025L = 0.002 mole K3PO4 Last, there are 3 moles of K+ for each mole of K3PO4 so we have 0.002 mole K3PO4 * (3 moles K+/1mole K3PO4) = 0.006 moles K+ Since 0.009 > 0.006 then solution 1 has more
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