Determine the [OH-], pH, and pOH of a solution with a [H*] of 3.5 x 10-1" M at 25 °C.[OH] =MpH =РОН 3Determine the [H+], pH, and pOH of a solution with an [OH-] of 0.0024 M at 25 °C.[H*] =MpH =РОН 3Determine the [H*], [OH-], and pOH of a solution with a pH of 5.95 at 25 °C.

Answered: Determine the [OH-], pH, and pOH of a…

Introductory Chemistry: A Foundation

8th Edition

ISBN:9781285199030

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter16: Acids And Bases

Section: Chapter Questions

Problem 68AP: Which of the following conditions indicate a basic solution? pOH = 11.21 pH = 9.42 (OH’] > IH+| |OH]...

See similar textbooks

Related questions

Concept explainers

Question

100%

Determine the [OH-], pH, and pOH of a solution with a [H*] of 3.5 x 10-1" M at 25 °C.
[OH] =
M
pH =
РОН 3
Determine the [H+], pH, and pOH of a solution with an [OH-] of 0.0024 M at 25 °C.
[H*] =
M
pH =
РОН 3
Determine the [H*], [OH-], and pOH of a solution with a pH of 5.95 at 25 °C.

Expert Solution

Step 1

$F o r p a r t 1 : g i v e n [H^{+}] = 3.5 x 10^{- 11} M w e k n o w t h a t, [H^{+}] [O H^{-}] = 10^{- 14} [O H^{-}] = \frac{10^{- 14}}{[H^{+}]} [O H^{-}] = \frac{10^{- 14}}{3.5 x 10^{- 11}} [O H^{-}] = 2.857 x 10^{- 4} M$

$f o r p H p H = - \log [H^{+}] p H = - \log (3.5 \times 10^{- 11}) p H = - \log 3.5 - \log 10^{- 11} p H = - \log 3.5 + 11 \log 10 p H = - 0.544 + 11 p H = 10.456 f o r p O H p H + p O H = - \log [O H^{-}] 10.456 + p O H = 14 p O H = 14 - 10.456 p O H = 3.544$

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions