# Calculating pH and pOH

*Using Brønsted-Lowry Theory*.

### HSC Chemistry Syllabus

**Calculate pH, pOH, hydrogen ion concentration ([H**^{+}]) and hydroxide ion concentration ([OH^{–}]) for a range of solutions (ACSCH102)

### Calculations Involving pH and pOH in HSC Chemistry

**pH and Acidity**

- Acids by various definitions are hydrogen ion (H
^{+}) producers (Arrhenius Theory) or proton donors (Brønsted-Lowry Theory)

- Not all hydrogen containing molecules are acids. Molecules are only considered to be acids if their hydrogen ions can be ionised.

- The acidity of a solution is measured by pH which stands for the potential of hydrogen. It is calculated using the formula

$$pH = -log_{10}[H^+]$$

*Or the hydrogen ion concentration can be calculated using the formula*

* *

$$[H^+] = 10^{-pH}$$

- pH is measured on a logarithmic scale which means that a change in value of 1 corresponds to a ten-fold change

pH = 1 corresponds to [H^{+}] of 0.1 mol L^{–1}

pH = 2 corresponds to [H^{+}] of 0.01 mol L^{–1}

- Notice that a smaller pH corresponds to an increasing concentration of hydrogen ions. This means a smaller pH will mean a more acidic solution

### pOH and Basicity

- The basicity of a solution can be measured with pOH. Similar to pH, a change in value of 1 corresponds to a tenfold change in [OH
^{–}].

$$pOH = -log_{10}[OH^-]$$

*Or the hydroxide concentration can be calculated using the* formula

$$[OH^-] = 10^{-pOH}$$

### Significant Figures in Calculations of pH and pOH

It is important to recognise that significant figures for hydrogen ion and hydroxide ion concentration, are expressed as decimal places when calculating pH and pOH

### Example 1

Calculate the pH for a solution with hydrogen ion concentration of 0.0010 mol L^{–1}

*Solution:*

Looking at our example the hydrogen ion concentration 0.0010 is given as 2 significant figures. This means that we will have to express our pH to 2 decimal places

To calculate pH, use the formula

$$pH = -log_{10}[H^+]$$

$$pH = -log_{10}[0.0010] = 3$$

However remember that we must express our answer to 2 decimal places

$$\therefore pH = 3.00$$

### Example 2

Calculate the concentration of hydrogen ions in a solution of pH 4.35.

*Solution:*

This time we are going from pH to a hydrogen ion concentration. The pH has 2 decimal places so we need to give our answer in 2 significant figures.

To calculate the hydrogen ion concentration, we use the formula

$$[H^+] = 10^{-pH}$$

$$[H^+] = 10^{-4.35} = 4.466735922 \times 10^{-5} \hspace{0.1cm} mol \hspace{0.1cm} L^{-1}$$

However remember that we must express our answer to 2 significant figures

$$\therefore [H^+] = 4.5 \times 10^{-5} \hspace{0.1cm} mol \hspace{0.1cm} L^{-1} $$

### Relationship between pH and pOH

To understand the quantitative relationship between pH and pOH, we must introduce the self-ionisation of water.

Self-ionisation of water is represented by the reaction:

$$2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-$$

The equilibrium constant of this reaction `K_w = 1.0 \times 10^{-14}` at 25 ºC. This is extremely small which indicates that the equilibrium lies on the far-left side of the reaction, resulting in very small amounts of H

_{3}O

^{+}and OH

^{–}equilibrium.

$$K_w = [H_3O^+(aq)][OH^-(aq)] = 1.0 \times 10^{-14}$$

The relationship between pH and pOH can be derived by applying logarithmic function to both sides of the above equation:

$$\log{([H_3O^+][OH^-])} = \log{(10^{-14})}$$

$$\log{[H_3O^+]} + \log{[OH^-]} = -14$$

$$-\log{[H_3O^+]} - \log{[OH^-]} = 14$$

$$pH + pOH = 14$$

Note that pH + pOH = 14 is only true at 25ºC because the value of `K_w` changes with temperature.