Solution Found!
Calculate [OH ] given [H3O+] in each aqueous solution and classify the solution as
Chapter 14, Problem 64P(choose chapter or problem)
Calculate \(\mathrm{[OH^-]}\) given \(\mathrm{[H_3O^+]}\) in each aqueous solution and classify the solution as acidic or basic.
(a) \(\mathrm{[H_3O^+]=1.3 \times 10^{-3} ~M}\)
(b) \(\mathrm{[H_3O^+]=9.1 \times 10^{-12} ~M}\)
(c) \(\mathrm{[H_3O^+]=5.2 \times 10^{-4} ~M}\)
(d) \(\mathrm{[H_3O^+]=6.1 \times 10^{-9} ~M}\)
Equation Transcription:
Text Transcription:
[OH^-]
[H_3O^+]
[H_3O^+]=1.3 x 10^{-3} M
[H_3O^+]=9.1 x 10^{-12} M
[H_3O^+]=5.2 x 10^{-4} M
[H_3O^+]=6.1 x 10^{-9} M
Questions & Answers
QUESTION:
Calculate \(\mathrm{[OH^-]}\) given \(\mathrm{[H_3O^+]}\) in each aqueous solution and classify the solution as acidic or basic.
(a) \(\mathrm{[H_3O^+]=1.3 \times 10^{-3} ~M}\)
(b) \(\mathrm{[H_3O^+]=9.1 \times 10^{-12} ~M}\)
(c) \(\mathrm{[H_3O^+]=5.2 \times 10^{-4} ~M}\)
(d) \(\mathrm{[H_3O^+]=6.1 \times 10^{-9} ~M}\)
Equation Transcription:
Text Transcription:
[OH^-]
[H_3O^+]
[H_3O^+]=1.3 x 10^{-3} M
[H_3O^+]=9.1 x 10^{-12} M
[H_3O^+]=5.2 x 10^{-4} M
[H_3O^+]=6.1 x 10^{-9} M
ANSWER:
Solution: Here, we are going to calculate hydroxyl ion concentration in each of the solution.
Step1:
In pure water, one H2O molecule donates proton and acts as an acid and another water molecule accepts a proton and acts as a base at the same time. The following equilibrium exists:
H2O(l) + H2O(l) ------> H3O+(aq) + OH-(aq)
The equilibrium constant for the reaction is given by:
Kw= [H3O+][OH−] -----(1)
Where, Kw is called the ionic product of water and its value is 1 x 10-14 M2.
We can distinguish acidic, neutral and basic aqueous solutions by the relative values of the H3O+ and OH− concentrations:
Acidic: [H3O+] > [OH−]
Neutral: [H3O+] = [OH−]
Basic: [H3O+] < [OH−]