Solution Found!
Determine the pOH of each solution.(a) [H3O+] = 8.3 × 10?10 M(b) [H3O+] = 1.6 × 10?7
Chapter 14, Problem 82P(choose chapter or problem)
Determine the pOH of each solution.
(a) \(\mathrm{[H_3O^+]=8.3 \times 10^{-10} ~M}\)
(b) \(\mathrm{[H_3O^+]=1.6 \times 10^{-7} ~M}\)
(c) \(\mathrm{[H_3O^+]=7.3 \times 10^{-2} ~M}\)
(d) \(\mathrm{[OH^-]=4.32 \times 10^{-4} ~M}\)
Equation Transcription:
Text Transcription:
[H_3O^+]=8.3 x 10^{-10} M
[H_3O^+]=1.6 x 10^{-7} M
[H_3O^+]=7.3 x 10^{-2} M
[OH^-]=4.32 x 10^{-4} M
Questions & Answers
QUESTION:
Determine the pOH of each solution.
(a) \(\mathrm{[H_3O^+]=8.3 \times 10^{-10} ~M}\)
(b) \(\mathrm{[H_3O^+]=1.6 \times 10^{-7} ~M}\)
(c) \(\mathrm{[H_3O^+]=7.3 \times 10^{-2} ~M}\)
(d) \(\mathrm{[OH^-]=4.32 \times 10^{-4} ~M}\)
Equation Transcription:
Text Transcription:
[H_3O^+]=8.3 x 10^{-10} M
[H_3O^+]=1.6 x 10^{-7} M
[H_3O^+]=7.3 x 10^{-2} M
[OH^-]=4.32 x 10^{-4} M
ANSWER:
Solution: Here, we are going to determine the pOH of the given solutions.
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.
moreover, pH + pOH =14
Where, pOH = - log[OH- ] and pH = - log[H+ ]
Step2:
- Given, [H3O+] = 8.3 x 10-10 M
From equation (1), [OH−] = Kw / [H3O+]