[H^+] = 2.7 x 10^-3
Calculate for [OH^-]
[OH^-] = ? M
I do these by calculating pOH = -log(OH^-), then subtracting from 14.
pH + pOH = pKw = 14
The other way is [H^+]*[OH^-] = 10^-14
To calculate for [OH^-], we can use the relationship between [H^+] and [OH^-] in water, which is described by the equation:
[H^+] x [OH^-] = Kw
Where Kw is the ion product of water, which is equal to 1.0 x 10^(-14) at 25 degrees Celsius.
Given that [H^+] is 2.7 x 10^(-3) M, we can substitute this value into the equation to find [OH^-].
[H^+] x [OH^-] = Kw
(2.7 x 10^(-3)) x [OH^-] = 1.0 x 10^(-14)
To solve for [OH^-], we rearrange the equation:
[OH^-] = (1.0 x 10^(-14)) / (2.7 x 10^(-3))
Calculating this result, we get:
[OH^-] ≈ 3.7 x 10^(-12) M
Therefore, [OH^-] is equal to approximately 3.7 x 10^(-12) M.