10.5: 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.

Kw = [H⁺][OH^–] = 1 × 10^-14

Example 10.5.1

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.

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

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.

Exercise \(\PageIndex{1}\)

A solution is prepared by dissolving 15.0 grams of NaOH in enough water to make 500.0 mL of solution. Calculate the pH of the solution. 

Answer

M = mol NaOH/L soln = 0.375mol/0.500L = 0.750M 

[NaOH] = [OH^–] = 0.750 M

[H⁺][OH^–] = 1 × 10^-14 hence [H⁺] = 1 × 10^-14/[OH^–]

[H⁺] = 1 × 10^-14/0.750 = 1.33 × 10^-14

pH = −log[H⁺] = −log (1.33 × 10^-14) 
pH = 13.876

 

Summary

• Calculations of pH for acidic and basic solutions are described.

EXERCISES

1. A solution is prepared by dissolving 22.0 grams of HCl in enough water to make 300.0 mL of solution. Calculate the pH of the solution.

2. What is the pH of each of the following solutions? Each is either a strong acid or base.

a. 0.01 M HBr solution

b. 0.10 M LiOH solution

c. 0.001 M HNO₃ solution

d. 0.001 M KOH solution

3. The hydronium ion concentration in a sample of rainwater is found to be 1.7 × 10⁻⁶ M at 25 °C. What is the concentration of hydroxide ions in the rainwater?

4. The hydronium ion concentration in a sample of rainwater is found to be 1.7 × 10⁻⁶ M at 25 °C. What is the pH of the rainwater?

5. The hydroxide ion concentration in household ammonia is 3.2 × 10⁻³ M at 25 °C. What is the concentration of hydronium ions in the solution? And, what is its pH?

6.  State whether each solution is acidic, neutral, or basic.

a.   [H⁺]=8.6×10⁻³M

b.   [H⁺]=3.0×10⁻⁹M

c.    [H⁺]=2.1×10⁻⁷ M

d.   [H⁺]=1.4×10⁻⁶M

ANSWERS

1. HCl is a strong acid and completely dissociates (see reaction). Since it completely dissociates, the concentration of HCl equals the concentration of H⁺. We need to calculate the concentration of HCl. Molar mass of HCl is 36.46 g/mol

HCl(aq)  →  H⁺(aq)  +  Cl^–(aq)

22.6g HCl x (1mol/36.46g)=0.603 mol HCl

M = mol solute/Lsoln

M = 0.603mol/0.300L=2.01 M HCl

[HCl] = [H⁺] = 2.01 M

pH = −log[H⁺] = −log[2.01] = 0.303  

2.

a. strong acid: pH = 2

b. strong base: pH = 13

c. strong acid: pH = 3

d. strong base pH = 11

3. [OH-] =5.9 × 10⁻⁹ M

4. pH = 5.77

5.  [H⁺] = 3.1 × 10⁻¹² M and pH = 11.51

6. 

a. The solution is acidic. pH = 2.07

b. The solution is basic. pH = 8.52

c. The solution is acidic. pH = 6.68

d. The solution is acidic. pH= 5.85

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