Solution Found!
Determine the pOH of each solution and classify it as acidic, basic, or neutral.(a)
Chapter 14, Problem 80P(choose chapter or problem)
Determine the pOH of each solution and classify it as acidic, basic, or neutral.
(a) \(\mathrm{[OH^-]=4.5 \times 10^{-2}~ M}\)
(b) \(\mathrm{[OH^-]=3.1 \times 10^{-12}~ M}\)
(c) \(\mathrm{[OH^-]=5.4 \times 10^{-5}~ M}\)
(d) \(\mathrm{[OH^-]=1.2 \times 10^{-2}~ M}\)
Equation Transcription:
Text Transcription:
[OH^-]=4.5 x 10^{-2} M
[OH^-]=3.1 x 10^{-12} M
[OH^-]=5.4 x 10^{-5} M
[OH^-]=1.2 x 10^{-2} M
Questions & Answers
QUESTION:
Determine the pOH of each solution and classify it as acidic, basic, or neutral.
(a) \(\mathrm{[OH^-]=4.5 \times 10^{-2}~ M}\)
(b) \(\mathrm{[OH^-]=3.1 \times 10^{-12}~ M}\)
(c) \(\mathrm{[OH^-]=5.4 \times 10^{-5}~ M}\)
(d) \(\mathrm{[OH^-]=1.2 \times 10^{-2}~ M}\)
Equation Transcription:
Text Transcription:
[OH^-]=4.5 x 10^{-2} M
[OH^-]=3.1 x 10^{-12} M
[OH^-]=5.4 x 10^{-5} M
[OH^-]=1.2 x 10^{-2} M
ANSWER:
Solution 80P:
Here, we are going to calculate the pOH of each solution and identify whether the reaction is acidic, basic and neutral.
We know, the ion product constant Kw for water is
Kw =[H3O+][OH-] = 1x10-14
pH + pOH =14
And pOH = - log[OH- ] and pH = - log[H+ ]
(a) [OH−] = 4.5 × 10−2 M
Step 1: We know,
pOH = - log[OH- ]
= -log(4.5 × 10−2 )
= 2 -log4.5
= 2 -0.653
=1.35
The pOH of the solution is 1.35