ABC - Autoionization of Water and the pH Scale (Lesson)

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Autoionization of Water and the pH Scale

Autoionization of Water

In the previous lesson, the concept of Bronsted-Lowry acids and bases was explored. In summary, those substances that have the ability to donate a proton are considered to be Bronsted-Lowry acids while those that can accept protons are considered to be Bronsted-Lowry bases. On some occasions, substances are encountered that have the ability to do both and are referred to as amphoteric compounds.

A specific example of an amphoteric substance that is especially pertinent to this topic is water (H₂O). Water has the ability to both accept and donate protons as shown in the equilibrium below:

2 H₂O (l) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ H₃O⁺ (aq) + OH^- (aq)

This process is referred to as the autoionization of water, and it results in the production of hydronium ions (H₃O⁺) as well as hydroxide ions (OH^-) in a sample of pure water. This special equilibrium will be relied upon heavily for the remainder of this module (although it is really just yet another specific example of a dynamic chemical equilibrium). Because of the importance of this equilibrium, it receives the special designation of K_w as further illustrated below:

K_w = [H₃O⁺][OH^-] = 1.0 x 10^-14 @ 25 $LaTeX: ^\circ$ $^\circ$ C

Notice that the equilibrium constant is extremely small indicating that in a sample of pure water, the vast majority of molecules remain unionized and only approximately one molecule in 100 trillion actually exists as either hydronium or hydroxide.

Calculating Hydrogen and Hydrogen Ion Concentrations

In circumstances where the concentration of either hydroxide or hydronium ions is known, the concentration of the other ion can be determined using the equilibrium expression for K_w. Take for example a scenario where the hydronium concentration is 0.40 M. Substituting this value in for the [H₃O⁺] allows us to calculate the concentration of hydroxide in this solution:

K_w = 1.0 x 10^-14 = [H₃O⁺][OH^-]

1.0 x 10^-14 = (0.40)[OH^-]

[OH^-] = 2.5 x 10^-14 M

You Try It!

In the self-assessment activity below, calculate the hydroxide and hydronium ion concentrations. Click on the plus sign to check your answer!

pH and the pH Scale

In a sample of pure water, the concentrations of both hydronium and hydroxide are the same. Using the Kw expression, this concentration can be determined:

K_w = 1.0 x 10^-14 = [H₃O⁺][OH^-]

Assume that the [H₃O⁺] = [OH^-] = x

K_w = 1.0 x 10^-14 = (x)(x) = x²

[H₃O⁺] = [OH^-] = 1.0 x 10^-7 M

When describing either the acidity or basicity of a solution, a convenient method for doing so is necessary. To avoid having to express the concentrations of ions that are so small, it is helpful to use a scale referred to as the pH scale.

The pH scale in water ranges from 0-14 where values of 0-7 are considered acidic and 7-14 are considered basic. A value of 7 indicates that the concentrations of hydronium and hydroxide are in perfect balance and the solution would be considered neutral.

The pH scale is a logarithmic scale such that a difference by one log unit is a factor of 10 different (e.g. pH = 3 is ten times more acidic than pH = 4). The equations to calculate pH are shown below:

pH = -log[H₃O⁺] or [H₃O⁺] = 10^-pH

In the same way that acidity can be described using the pH scale, basicity can be described using a pOH scale. The equations related to pOH are analogous to those of pH are as follows:

pOH = -log[OH^-] or [OH^-] = 10^-pOH

It then follows that since the concentrations of hydronium and hydroxide are related to each other through the autoionization of water equilibrium there must be a relationship between pH and pOH. This proves to be true, and the two are related through the relationships below:

[H₃O⁺][OH^-] = 1 x 10^-14 and pH + pOH = 14.00

Calculations Involving pH and pOH

An important concept to be mindful of is the fact that pH always refers to the concentration of hydronium in an aqueous solution. Technically speaking, it is not the concentration of an acid that is used to calculate pH; however, the presence of an acid will alter the concentration of hydronium present in that solution. Take for example an aqueous solution of HBr (as an aside, only strong acids will be considered in this lesson; HBr is one of the seven strong acids although weak acids and their effect on pH will be discussed in future lessons).

Assume that a solution of HBr has a concentration of 0.15 M. When HBr behaves as a Bronsted-Lowry acid, it donates a proton to water as shown below:

HBr (l) + H₂O (l) $LaTeX: \rightarrow$ $\rightarrow$ H₃O⁺ (aq) + Br^- (aq)

It is important to notice a few things from this process:

First of all, it should be noted that this is not an equilibrium process as evidenced by the one-directional arrow. By definition, strong acids dissociate 100% when placed into water.
Secondly, the addition of HBr to water results in the production of hydronium. It is this hydronium, produced by the addition of HBr, that results in an increase in the acidity of the solution.

So, how can the pH of this solution be calculated? Because there is a 1:1 mole ratio between HBr (l) and H₃O⁺ (aq) and because HBr is a strong acid, the concentration of hydronium produced in water is the same as the concentration of HBr. Substituting into the pH equation below allows for the calculation of the pH of this solution:

pH = -log[H₃O⁺] = -log(0.15)

pH = 0.82

Similar calculations can be performed for pOH as well. For example, an aqueous solution of potassium hydroxide has a concentration of 0.25 M:

KOH (s) $LaTeX: \rightarrow$ $\rightarrow$ K⁺ (aq) + OH^- (aq)

In this case, the strong base KOH dissolves in water to increase the amount of hydroxide in the solution and therefore its concentration. As before, the mole ratio between KOH and OH^- is also 1:1. As a result, the concentration of hydroxide in the solution would also be 0.25 M allowing for the calculation of pOH as shown below:

pOH = -log[OH^-] = -log(0.25)

pOH = 0.60

You Try It!

In the self-assessment activity below, calculate the pH. Click on the plus sign to check your answer!

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