Be sure to answer all parts. The molecular scene at right depicts the relative concentrations of H3O+ (purple) and OH− (green) in an aqueous solution at 25° C. (Counter ions and solvent molecules are omitted for clarity.)
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Solution:
Part A:
Number of H3O+ ions, n1 = 2
Number of OH- ions, n2 = 20
Ratio of their concentrations,
We know that
Kw = [H3O+] × [OH-]
10-14 = [H3O+] × [OH-]
10-14 = [H3O+] × ([H3O+]/0.1)
([H3O+])2 = 1015
[H2O+] = 0.00000003162
pH = - log [H3O+]
pH = - log(0.00000003162)
pH = 7.50
Part B:
pH of the solution = 4.90
We know that
pH + pOH = 14
4.90 + pOH = 14
pOH = 9.1
or - log[OH-] = 9.1
[OH-] = 10-9.1
[OH-] = 794.328234724 x 10-12
Also,
Kw = [H3O+] × [OH-]
10-14 = [H3O+] × [OH-]
10-14 = [H3O+] × (7.94328 x 10-10)
[H3O+] = 0.00001258925
Therefore,
1.58489 x 104 ions would have to be draw forevery OH-for the given solution.
Be sure to answer all parts. The molecular scene at right depicts the relative concentrations of...
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