What is the [oh−] of a solution if its ph = 8.78? answer in units of m?

1 answer:

7 0

PH = -log([H+])
[H+] = 10^(-pH)

[H+] = 10^(-8.78) = 1.65*10^-9

[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.

[OH-] = Kw/[H+] = (1.0*10^-14)/(1.65*10^-9) = 6.06*10^-6

The concentration of OH- ions is 6.1*10^-6 M.

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