21.10: Calculating pH of Acids and Bases

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It's not uncommon to see a tropical fish tank in homes or businesses. These brightly-colored creatures are relaxing to watch, but do require a certain amount of maintenance in order for them to survive. Tap water is usually too alkaline when it comes out of the faucet, so some adjustments need to be made. The pH of the water will change over time while it is in the tank, which means it needs to be tested every so often. Any fish tank caretaker has a chance to be a chemist for their fish!

Calculating pH of Acids and Bases

Calculation of pH is simple when there is a \(1 \times 10^\text{power}\) problem. However, in real life, this is rarely the situation. If the coefficient is not equal to 1, a calculator must be used to find the pH. For example, the pH of a solution with \(\left[ \ce{H^+} \right] = 2.3 \times 10^{-5} \: \text{M}\) can be found as shown below.

\[\text{pH} = -\text{log} \left[ 2.3 \times 10^{-5} \right] = 4.64\nonumber \]

When the pH of a solution is known, the concentration of the hydrogen ion can be calculated. The inverse of the logarithm (or antilog) is the \(10^x\) key on a calculator.

\[\left[ \ce{H^+} \right] = 10^{-\text{pH}}\nonumber \]

For example, suppose that you have a solution with a pH of 9.14. To find the \(\left[ \ce{H^+} \right]\) use the \(10^x\) key.

\[\left[ \ce{H^+} \right] = 10^{-\text{pH}} = 10^{-9.14} = 7.24 \times 10^{-10} \: \text{M}\nonumber \]

Hydroxide Ion Concentration and pH

As we saw earlier, the hydroxide ion concentration of any aqueous solution is related to the hydrogen ion concentration through the value of \(K_\text{w}\). We can use that relationship to calculate the pH of a solution of a base.

Example \(\PageIndex{1}\)

Sodium hydroxide is a strong base. Find the pH of a solution prepared by dissolving \(1.0 \: \text{g}\) of \(\ce{NaOH}\) into enough water to make \(1.0 \: \text{L}\) of solution.

Solution

Step 1: List the known values and plan the problem.

Known

Mass \(\ce{NaOH} = 1.0 \: \text{g}\)
Molar mass \(\ce{NaOH} = 40.00 \: \text{g/mol}\)
Volume solution \(= 1.0 \: \text{L}\)
\(K_\text{w} = 1.0 \times 10^{-14}\)

Unknown

First, convert the mass of \(\ce{NaOH}\) to moles. Second, calculate the molarity of the \(\ce{NaOH}\) solution. Because \(\ce{NaOH}\) is a strong base and is soluble, the \(\left[ \ce{OH^-} \right]\) will be equal to the concentration of the \(\ce{NaOH}\). Third, use \(K_\text{w}\) to calculate the \(\left[ \ce{H^+} \right]\) in the solution. Lastly, calculate the pH.

Step 2: Solve.

\[\begin{align*} &1.00 \: \cancel{\text{g} \: \ce{NaOH}} \times \frac{1 \: \text{mol} \: \ce{NaOH}}{40.00 \: \cancel{\text{g} \: \ce{NaOH}}} = 0.025 \: \text{mol} \: \ce{NaOH} \\ &\text{Molarity} = \frac{0.025 \: \text{mol} \: \ce{NaOH}}{1.00 \: \text{L}} = 0.025 \: \text{M} \: \ce{NaOH} = 0.025 \: \text{M} \: \ce{OH^-} \\ &\left[ \ce{H^+} \right] = \frac{K_\text{w}}{\left[ \ce{OH^-} \right]} = \frac{1.0 \times 10^{-14}}{0.025 \: \text{M}} = 4.0 \times 10^{-13} \: \text{M} \\ &\text{pH} = -\text{log} \left[ \ce{H^+} \right] = -\text{log} \left( 4.0 \times 10^{-13} \right) = 12.40 \end{align*}\]

Step 3: Think about your result.

The solution is basic and so its pH is greater than 7. The reported pH is rounded to two decimal places because the original mass and volume has two significant figures.