21.10: Calculating pH of Acids and Bases

Many people enjoy having tropical fish in their 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 you need to test it every so often. Then you get to be a chemist for your 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 that 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.

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

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.

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

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.

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

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 21.10.1

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

• pH of solution $$= ?$$

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}