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 - Page 161: 4.82a

Answer

Approximately 84.16667 ml of $HCl$.

Work Step by Step

$C(H^+)*V(H^+) = C(OH^-)*V(OH^-)$ Because each $Ba(OH)_2$ has 2 OH, the concentration of $OH^-$ will be the double of the $Ba(OH)_2$ concentration. $[OH^-] = 2*[Ba(OH)_2] = 2 * 0.101M = 0.202M$ $[H^+] = 1*[HCl] = 1*0.12M = 0.12M$ $0.12 * V(H^+) = 0.202*50$ $V(H^+) = \frac{10.1}{0.12}$ $V(H^+) \approx 84.1667ml$
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