Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 4 - Reactions in Aqueous Solution - Worked Example - Page 130: 22


The molarity of the acetic acid solution is equal to 0.758 mol/L.

Work Step by Step

1. Find the number of moles of $NaOH$: $\frac{0.200mol (NaOH)}{1L} \times 94.7mL\times \frac{1L}{1000mL} = 0.01894$ $mol (NaOH)$ 2. Use the stoichiometry of the reaction to find the number of moles of acetic acid: The coefficients indicate a ratio of 1 to 1. $0.01894mol(NaOH) \times \frac{1mol(CH_3CO_2H)}{1mol(NaOH)} = 0.01894mol(CH_3CO_2H)$ 3. Find the concentration, using the value for the volume of vinegar used. $\frac{0.01894mol(CH_3CO_2H)}{25mL} \times \frac{1000mL}{1L} = 0.758 mol/L (CH_3CO_2H)$
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