## The ionic product of water, K_{w}

Water is able to act as either a Brønsted–Lowry acid (by donating protons, H^{+}) or a Brønsted–Lowry base (by accepting protons), as shown by the equilibrium reaction:

H_{2}O(l) + H_{2}O(l) ⇌ H_{3}O^{+}(aq) + OH^{–}(aq)

which can be simplified by removing the spectator ions to form:

H_{2}O(l) ⇌ H^{+}(aq) + OH^{–}(aq)

The equilibrium expression for this reaction is:

$latex \displaystyle K_c = \frac{[H^+(aq)][OH^-(aq)]}{[H_2O(l)]}$

Because the extent of ionisation of water is very low, the concentration of H^{+}(aq) ions and OH^{–}(aq) ions in pure water is extremely small. Hence the value of K_{c} is also extremely small. By regarding the concentration of water as being constant we created a new equilibrium constant called the ionic product of water, K_{w}.

K_{w} = [H^{+}][OH^{–}]

The ionic product of water has a value at 1.00 × 10^{–14}mol^{2}dm^{–6} at 298K.

Using this equation we can find the hydrogen ion concentration in pure water, since we know that for each molecule of water that ionises, one H^{+} ion and one OH^{–} ion are produced.

Hence [H^{+}] = [OH^{–}]

Using the equilibrium expression K_{w} = [H^{+}][OH^{–}] and [H^{+}] = [OH^{–}] we get:

K_{w} = [H^{+}]^{2}

And therefore: [H^{+}] = $latex \sqrt{K_w}$

^{+}] = $latex \sqrt{1.00 \times 10^{-14}mol^2dm^{-6}}$

Therefore the hydrogen ion concentration in water is [H^{+}] = $latex 1.00 \times 10^{-7}moldm^{-3}$

## How to calculate pH

**The pH of a substance** is the negative common logarithm of the hydrogen ion concentration in that substance.

It can be calculated as follows:

pH = –log_{10}[H^{+}]

Since the acidity or alkalinity of a solution depends on the concentration of H^{+}(aq) ions, it is measured using the pH scale.

### Example – how to calculate pH given concentration

Calculate the pH of the following solutions:

- [H+] = 3.00 × 10
^{–4}moldm^{–3} - [H+] = 1.00 × 10
^{–2}moldm^{–3} - [H+] = 4.00 × 10
^{–8}moldm^{–3} - [H+] = 5.40 × 10
^{–12}moldm^{–3} - [H+] = 7.80 × 10
^{–10}moldm^{–3}

### Solution

Calculate the pH of the following solutions:

- pH = -log
_{10}(3.00 × 10^{–4}) - pH = -log
_{10}(1.00 × 10^{–2}) - pH = -log
_{10}(4.00 × 10^{–8}) - pH = -log
_{10}(5.40 × 10^{–12}) - pH = -log
_{10}(7.80 × 10^{–10})

= 3.5

= 2

= 7.4

= 11.3

= 9.1

In the example above:

- solutions a and b are acids are acids because they have pH values less than 7.
- solutions in parts c to e, are alkalis because they all have pH values greater than 7.

One more thing to note is that alkalis have a small concentration of H^{+} ions, and we use this concentration to calculate the pH of the alkali.

Another thing to note from the example is the simplicity of working with pH values than with the hydrogen ion concentration when expressing the acidity of a substance. Using the common logarithm of the hydrogen ion concentration allows us to express the acidity of a substance using natural looking numbers, such as saying "pH of the solution = 2" instead of awkward numbers like "[H^{+}] = 1.00 × 10^{–2} mol/dm^{3}".

The minus sign in the formula *pH = –log _{10}[H^{+}]* makes all resultant pH values positive and as a result, the pH scale has a range of 0 to 14, in which:

- the smaller the pH, the greater the concentration of H
^{+}(aq), the more acidic the solution is. - the greater the pH, the lower the concentration of H
^{+}(aq), the more alkaline the solution is. - a difference of one pH number means ten times difference in [H
^{+}] so that, for example, pH 1 has ten times the H^{+}concentration of pH 2.

## How to calculate the hydrogen ion concentration, [H+]

Hydrogen ion concentration can be calculated from pH using the equation pH = –log_{10}[H^{+}] and making [H^{+}] the subject of formula as follows:

- pH = –log
_{10}[H^{+}] - -pH = log
_{10}[H^{+}] - 10
^{-pH}= [H^{+}]*(using the theory of logarithms)*

### Example 2 – Calculating the hydrogen ion concentration, [H+] given the pH value.

Calculate the concentration of hydrogen ions in solutions having the following pH values:

- 2.90
- 3.70
- 11.2
- 5.40
- 12.9

### Solution

- [H
^{+}] = 10^{-2.90}= 0.00126 mol/dm^{3} - [H
^{+}] = 10^{-3.70}= 0.0002 mol/dm^{3} - [H
^{+}] = 10^{-11.2}= 6.31 × 10^{-12}mol/dm^{3} - [H
^{+}] = 10^{-5.40}= 0.000004 mol/dm^{3} - [H
^{+}] = 10^{-12.9}= 1260×10^{-16}mol/dm^{3}

## How to calculate pOH

The **pOH of a substance** is the negative common logarithm of the OH^{–} ion concentration in that substance.

It can be calculated as follows:

pOH = –log_{10}[OH^{–}]

Just like the pH of a substance, pOH is a measurement of the acidity or alkalinity of a substance.

### Relationship between pH and pOH

Since we know that at equilibrium, K_{w} = [H^{+}][OH^{–}],

- [H
^{+}][OH^{–}] = 1.00 × 10^{–14}mol^{2}dm^{–6}at 298K.

*taking the negative common logarithm of both sides, we get:*

- -log
_{10}([H^{+}][OH^{–}]) = -log_{10}(1.00 × 10^{–14})

*applying the laws of logarithms, we get:*

- -(log
_{10}[H^{+}]+ log_{10}[OH^{–}]) = 14log_{10}(1.00 × 10) - -log
_{10}[H^{+}] – log_{10}[OH^{–}] = 14

*since -log _{10}[H^{+}] = pH and -log_{10}[OH^{–}] = pOH*

- pH + pOH = 14

This means that if a solution has a pH of 2, its pOH = 14 -2 = 12.