What is the concentration (in M) of hydroxide ions in a solution at 25°C with pH = 4.282?
I would do it this way.
pH + pOH = pKw = 14
Solve for pOH.
Then pOH = -log(OH^-) and solve for OH^-
You could convert pH to H^+, then
(H^+)(OH^-) = 1 x 10^-14
and solve for OH^-
To find the concentration of hydroxide ions in a solution, we can use the equation for the relationship between the concentration of hydrogen ions (H+) and hydroxide ions (OH-) in water, which is described by the equation Kw = [H+][OH-].
At 25°C, the Kw constant is equal to 1.0 x 10^-14.
In order to determine the concentration of hydroxide ions (OH-), we can take advantage of the fact that the pH of a solution is defined as the negative logarithm of the hydrogen ion concentration (pH = -log[H+]). Rearranging this equation gives us [H+] = 10^(-pH).
So, in this case, [H+] = 10^(-4.282).
Since Kw = [H+][OH-], we can substitute [OH-] with Kw/[H+] to find [OH-].
Therefore, [OH-] = Kw / [H+].
Substituting the given values, we have [OH-] = (1.0 x 10^-14) / (10^(-4.282)).
Calculating this expression using scientific notation, we have:
[OH-] ≈ 2.34 x 10^(-11) M.
Therefore, the concentration of hydroxide ions in the solution at 25°C with a pH of 4.282 is approximately 2.34 x 10^(-11) M.