# pH Calculations

## pH calculations

#### Tip:

The pH scale is a log scale. Remember from mathematics that a difference of one on a base $$\text{10}$$ log scale (the one on your calculator) is equivalent to a multiplication by 10. That is:

1 = log(10)

2 = log(100)

3 = log($$\text{1 000}$$)

4 = log($$\text{10 000}$$)

So a change from a pH of 2 to a pH of 6 represents a very large change in the $$\text{H}^{+}$$ concentration.

pH can be calculated using the following equation:

pH = -log[$$\text{H}^{+}$$]

[$$\text{H}^{+}$$] and [$$\text{H}_{3}\text{O}^{+}$$] can be substituted for one another:

pH = -log[$$\text{H}_{3}\text{O}^{+}$$]

The brackets in the above equation are used to show concentration in $$\text{mol.dm^{-3}}$$.

## Example 1: pH Calculations

### Question

Calculate the pH of a solution where the concentration of hydrogen ions is

$$\text{1} \times \text{10}^{-\text{7}}$$ $$\text{mol.dm^{-3}}$$.

### Step 1: Determine the concentration of hydrogen ions

In this example, the concentration has been given: $$\text{1} \times \text{10}^{-\text{7}}$$ $$\text{mol.dm^{-3}}$$

### Step 2: Substitute this value into the pH equation and calculate the pH value

pH = -log[$$\text{H}^{+}$$]

= -log($$\text{1} \times \text{10}^{-\text{7}}$$)

= 7

#### Tip:

Important: It may be useful to know that for calculations involving the pH scale, the following equations can also be used:

[$$\text{H}^{+}$$][$$\text{OH}^{-}$$] = $$\text{1} \times \text{10}^{-\text{14}}$$

[$$\text{H}_{3}\text{O}^{+}$$][$$\text{OH}^{-}$$] = $$\text{1} \times \text{10}^{-\text{14}}$$

pH = 14 – p[$$\text{OH}^{-}$$]

pH = 14 – (-log[$$\text{OH}^{-}$$])

## Example 2: pH Calculations

### Question

In a $$\text{162}$$ $$\text{cm^{3}}$$ solution of ethanoic acid, the following equilibrium is established:

$$\text{CH}_{3}\text{COOH}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})$$ $$\leftrightharpoons$$ $$\text{CH}_{3}\text{COO}^{-}(\text{aq}) + \text{H}_{3}\text{O}^{+}(\text{aq})$$

The number of moles of $$\text{CH}_{3}\text{COO}^{-}$$ is found to be $$\text{0.001}$$ $$\text{mol}$$. Calculate the pH of the solution.

### Step 1: Determine the number of moles of hydronium ions in the solution

According to the balanced equation for this reaction, the mole ratio of $$\text{CH}_{3}\text{COO}^{-}$$ ions to $$\text{H}_{3}\text{O}^{+}$$ ions is $$\text{1}$$:$$\text{1}$$, therefore the number of moles of these two ions in the solution will be the same.

So, n($$\text{H}_{3}\text{O}^{+}$$) = $$\text{0.001}$$ $$\text{mol}$$.

### Step 2: Determine the concentration of hydronium ions in the solution

$$\text{C (mol.dm}^{-3}{\text{)}} = \dfrac{\text{n (mol)}}{\text{V (dm}^{3}{\text{)}}}$$

V = $$\text{162}$$ $$\text{cm^{3}}$$ $$\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}$$ = $$\text{0.162}$$ $$\text{dm^{3}}$$

[$$\text{H}_{3}\text{O}^{+}$$] $$= \dfrac{\text{0.001} {\text{ mol}}}{\text{0.162} {\text{ dm}}^{3}} =$$ $$\text{0.0062}$$ $$\text{mol.dm^{-3}}$$

### Step 3: Substitute this value into the pH equation and calculate the pH value

pH = -log[$$\text{H}^{+}$$] = -log[$$\text{H}_{3}\text{O}^{+}$$]

= -log($$\text{0.0062}$$)

= 2.21

Understanding pH is very important. In living organisms, it is necessary to maintain a constant pH in the optimal range for that organism, so that chemical reactions can occur.

 pH $$\text{1}$$ $$\text{6}$$ $$\text{7}$$ $$\text{8}$$ $$\text{13}$$ $$[\text{H}^{+}]$$ $$\text{1} \times \text{10}^{-\text{1}}$$ $$\text{1} \times \text{10}^{-\text{6}}$$ $$\text{1} \times \text{10}^{-\text{7}}$$ $$\text{1} \times \text{10}^{-\text{8}}$$ $$\text{1} \times \text{10}^{-\text{13}}$$ $$[\text{OH}^{-}]$$ $$\text{1} \times \text{10}^{-\text{13}}$$ $$\text{1} \times \text{10}^{-\text{8}}$$ $$\text{1} \times \text{10}^{-\text{7}}$$ $$\text{1} \times \text{10}^{-\text{6}}$$ $$\text{1} \times \text{10}^{-\text{1}}$$ Solution $$\color{red}{\textbf{strongly acidic}}$$ $$\color{red}{\text{weakly acidic}}$$ neutral $$\color{blue}{\text{weakly basic}}$$ $$\color{blue}{\textbf{strongly basic}}$$

Table: The concentration of $$[\text{H}^{+}]$$ and $$[\text{OH}^{-}]$$ ions in solutions with different pH.

#### Fact:

A buildup of acid in the human body can be very dangerous. Lactic acidosis is a condition caused by the buildup of lactic acid in the body. It leads to acidification of the blood (acidosis) and can make a person very ill. Some of the symptoms of lactic acidosis are deep and rapid breathing, vomiting, and abdominal pain. In the fight against HIV, lactic acidosis is a problem. One of the antiretrovirals (ARV’s) that is used in anti-HIV treatment is Stavudine (also known as Zerit or d4T). One of the side effects of Stavudine is lactic acidosis, particularly in overweight women. If it is not treated quickly, it can result in death.

#### Fact:

Litmus paper can be used as a pH indicator. It is sold in strips. Purple litmus paper will become red in acidic conditions and blue in basic conditions. Blue litmus paper is used to detect acidic conditions, while red litmus paper is used to detect basic conditions.

In agriculture, it is important for farmers to know the pH of their soils so that they are able to plant the right kinds of crops. The pH of soils can vary depending on a number of factors, such as rainwater, the kinds of rocks and materials from which the soil was formed and also human influences such as pollution and fertilisers. The pH of rain water can also vary, and this too has an effect on agriculture, buildings, water courses, animals and plants. Rainwater is naturally acidic because carbon dioxide in the atmosphere combines with water to form carbonic acid. Unpolluted rainwater has a pH of approximately $$\text{5.6}$$. However, human activities can alter the acidity of rain and this can cause serious problems such as acid rain.