pH, pOH, [H+], and [OH-] are all interrelated. For simplicity, we assume that we're working with aqueous solutions at 25 °C.
The following mathematical relations are relevant here:
- pH = -log[H⁺]
- pOH = -log[OH⁻]
- [H⁺] = 10^-pH
- [OH⁻] = 10^-pOH
- pH + pOH = 14
- [H⁺] × [OH⁻] = 10^-14
A pH of 7 is neutral; a pH less than 7 is acidic; and a pH greater than 7 is basic.
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We are given four solutions of varying [H⁺], and we want to find the pH and the [OH⁻]. The pH will tell us whether the solution is neutral, an acid, or a base.
Finding the [OH⁻]
Using the relationship [H⁺] × [OH⁻] = 10^-14, we can calculate the [OH⁻] of each solution by dividing 10^-14 by the given [H⁺]:
[OH⁻] = 10^-14/(3.3 × 10^-5) = 3.0 × 10^-10
[OH⁻] = 10^-14/(4.0 × 10^-8) = 2.5 × 10^-7
[OH⁻] = 10^-14/(7.2 × 10^-11) = 1.4 × 10^-4
[OH⁻] = 10^-14/(8.2 × 10^-4) = 1.2 × 10^-11
Note: While not indicated here, the units for [H⁺] and [OH⁻] are moles/liter (M).
Finding the pH
To find the pH of each solution, we take the negative logarithm of the [H⁺] with base 10, i.e., pH = -log₁₀[H⁺]:
pH = -log(3.3 × 10^-5) = 4.48
pH = -log(4.0 × 10^-8) = 7.40
pH = -log(7.2 × 10^-11) = 10.14
pH = -log(8.2 × 10^-4) = 3.09
Note that, for significant figures involving logarithms, the number of decimal places should equal the number of significant figures in the argument of our logarithm (the argument is what we're taking the -log of). Since we're taking the -log of quantities given to two significant figures, our corresponding pH values are given to two decimal places.
Acid, Base, or Neutral?
As mentioned above, a pH = 7 is neutral, a pH < 7 is acidic, and a pH > 7 is basic.
4.48 < 7, so the first solution is an acid.
7.40 > 7, so the second solution is a base.
10.14 > 7, so the third solution is a base.
3.09 < 7, so the fourth solution is an acid.