Which of the following is the [OH-] in pure water at room at room temperature?
1.0 ´ 10-14 M
1.0 ´ 10-7 M
1.0 ´ 107 M
1.0 ´ 1014 M
Answer:
1.0 * 10^-7 M
Since Kw at room temp. (25ºC) is 10^-14,
Kw = [H+][OH-]
and the fact that pure water is neutral
[H+] = [OH-]
So,
10^-14 = [OH-]^2
[OH-] = sqrt(10^-14) = 10^-7 M
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1.0 ´ 10-7 M
1.0 ´ 107 M
1.0 ´ 1014 M
Answer:
1.0 * 10^-7 M
Since Kw at room temp. (25ºC) is 10^-14,
Kw = [H+][OH-]
and the fact that pure water is neutral
[H+] = [OH-]
So,
10^-14 = [OH-]^2
[OH-] = sqrt(10^-14) = 10^-7 M
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