Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 15 - Sections 15.1-15.12 - Exercises - Problems by Topic - Page 746: 48c

Answer

$1.4\times10^{-3}$, Acidic

Work Step by Step

For this we need to use the equations $K_{w}=[OH^{-}][H_{3}O^{+}]$ and $K_{w}=1\times10^{-14}$. To calculate the concentration of $H_{3}O^{+}$ set the equations equal to each other and plug in the concentration of $OH^{-}$. $1\times10^{-14}=(6.9\times10^{-12})[H_{3}O^{+}]$ Rearranging, we get: $\frac{1\times10^{-14}}{6.9\times10^{-12}}=[H_{3}O^{+}]$ so $[H_{3}O^{+}]=1.4\times10^{-3}$ Since this value for $[H_{3}O^{+}]$ is greater than the value for $[OH^{-}]$, the solution is acidic.
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.