Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179f: 72


[NaOH] = $0.167M$

Work Step by Step

This reaction takes place at the equivalence point; this means that that moles of KHC$_{8}$H$_{4}$O$_{8}$ = moles of NaOH Therefore to get [NaOH] we need the moles of NaOH used $\frac{0.902g KHC_{8}H_{4}O_{8}}{1}\times\frac{1 molKHC_{8}H_{4}O_{8} }{204.1gKHC_{8}H_{4}O_{8}}\times\frac{1 mol NaOH}{1 mol KHC_{8}H_{4}O_{8}} = 0.00442 mol NaOH$ To get [NaOH] divide moles by the concentration [NaOH] = $\frac{0.00442 mol NaOH}{0.02645 L} = 0.167M$
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