Please answer 18 and 19! and show work
18)
pH = 6.630
use:
pH = -log [H3O+]
6.630 = -log [H3O+]
[H3O+] = 2.344*10^-7 M
SINCE water is neutral,
[OH-] = [H3O+]
So,
[OH-] = 2.344*10^-7 M
Now use:
Kw = [H+] [OH-]
= (2.344*10^-7)*(2.344*10^-7)
= 5.50*10^-14
Answer: b
19)
use:
[H3O+] = (1.0*10^-14)/[OH-]
= (1.0*10^-14) / (5.5*10^-5)
= 1.8*10^-10 M
Since [OH-] > [H+], it is basic
Answer: b
Please answer 18 and 19! and show work What are the Bronsted-Lowry acids in the following...
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