The molarities of hydronium ions, H_{3}O^{+1}, and hydroxide ions, OH^{–1}, that are present in a solution are multiplicatively-related in the Ion-Product Constant Equation. Furthermore, the molarity of hydronium ions, H_{3}O^{+1}, that is present in a solution is related to the pH of that solution by the pH Equation. Finally, the molarity of hydroxide ions, OH^{–1}, that is present in a solution cannot be directly related to the pH of that solution. Therefore, the given pH must first be used to calculate the concentration of hydronium ions, H_{3}O^{+1}, that are present in the solution, and the resultant value can be used to determine the molarity of hydroxide ions, OH^{–1}, that are present in the solution.
In order to calculate the molar concentration of hydronium ions, H_{3}O^{+1}, from a pH, the given value, 13.54, is incorporated into the anti-log pH Equation, as shown below. As established in the previous sections of this chapter, the numerical solution, which corresponds to a concentration of hydronium ions, H_{3}O^{+1}, must be expressed in scientific notation and reported to three significant figures. Therefore, in order to calculate a properly-formatted concentration value, the quantity on the right side of the second equation that is shown below must be entered into a calculator. Finally, the resultant value is a molarity, and, therefore, is labeled with a unit of *"**M*,*"* *even though this unit is not established through the mathematical processes that are described **above*.

\(\rm{[H_3O^{+1}]}\) = \({10^{\rm{–pH}}}\)

\(\rm{[H_3O^{+1}]}\) = \({10^{–13.54}}\)

\(\rm{[H_3O^{+1}]}\) = \({2.88403... \times 10^{–14} M} ≈ {2.88 \times 10^{–14} M}\)

In order to calculate the concentration of hydroxide ions, OH^{–1}, that are present in this solution, the calculated molar concentration of hydronium ions, H_{3}O^{+1}, 2.88 x 10^{–14} *M*, is incorporated into the Ion-Product Constant Equation, as shown below. The numerical solution is calculated by dividing the value of K_{w}, 1.00 x 10^{–14}, by the calculated molarity, and, when using a calculator, any quantity that is expressed in scientific notation should be offset by parentheses. In order to align with the standards that were established in the previous sections of this chapter, the final answer, which is expressed in scientific notation, is reported to three significant figures. Finally, the numerical solution corresponds to a molarity, and, therefore, is labeled with a unit of *"**M*,*"* *even though this unit is not established through the mathematical process that is described above*.

\({1.00 \times 10^{–14}}\) = \(\rm{[H_3O^{+1}]}\)\(\rm{[OH^{–1}]}\)

\({1.00 \times 10^{–14}}\) = (\({2.88 \times 10^{–14} M}\))\(\rm{[OH^{–1}]}\)

\(\rm{[OH^{–1}]}\) = \({3.47222... \times 10^{–1} M} ≈ {3.47 \times 10^{–1} M}\)

Finally, in order to classify this solution as acidic, basic, or neutral, the hydronium, H_{3}O^{+1}, and hydroxide, OH^{–1}, ion concentrations can be compared to one another. Alternatively, the given pH can be analyzed, relative to the values on the pH Scale.
The magnitude of a number that is written in scientific notation *increases *as the value of a *negative power* decreases. Since –1 is a smaller negative exponent than –14, the calculated molarity of hydroxide ions, OH^{–1}, 3.47 x 10^{–1} *M*, which can also be written as 0.347 *M*, is greater than the calculated concentration of hydronium ions, H_{3}O^{+1}, 2.88 x 10^{–14} *M*, which can also be represented as 0.0000000000000288 *M*. Therefore, the solution is classified as basic.

The pH value of a neutral solution is *exactly* 7.00, any solution that has a pH that is *greater than* 7.00 can be classified as basic, and an acidic solution must have a pH value that is *less than than* 7.00. Therefore, because the given pH, 13.54, is *greater than* 7.00, this solution is classified as basic.