Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 15 - Acids and Bases - Questions & Problems: 15.24

Answer

- This solution has $1.98 \times 10^{-3}$ moles of $KOH$. - Its pOH is 0.44.

Work Step by Step

- ** $5.5 ml = 5.5 \times 10^{-3} L$ $n (moles) = Concentration (M) * Volume (L)$ $n (moles) = 0.360 * 5.50 \times 10^{-3} = 1.98 \times 10^{-3}$ - $pOH = -log[OH^-]$ Since $KOH$ is a strong base: $[OH^-] = [KOH] = 0.360M$ $pOH = -log(0.360) = 0.44$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.