# Water and the pH Scale

Learning Goals

In this section you will learn about another method of quantifying the difference in strength of acids and bases. You will learn about pH, pOH, and how to calculate them.

Synopsis

For a Bronsted-Lowry acid or base, we know that the larger the value of the concentration of H3O+ or OH-, the stronger the acid or base, respectively. So if we can then devise a measure of stating the concentration of either then we will have a convenient quantitative method to state the strength of an acid or base. Since we will be working in water solutions, let's examine the water equilibrium in some detail by first writing the water-water reaction:

H2O(l) + H2O(l) --> H3O+(aq) + OH-(aq)

We can write the equilibrium constant for this reaction as

K = [H3O+][OH-]/[H2O]2 or K[H2O]2 = [H3O+[OH-]

But remember that we usually don't include the concentration of water in the equilibrium expression. Its concentration does not change appreciably and is very large for dilute solutions. So we will combine its constant concentration with K to give a new equilibrium constant, Kw. Thus,

Kw = [H3O+][OH-]

Because there are no other sources of protons in pure water, we know that [H3O]+ must equal [OH-].

Electrical conductivity measurements tell us that, at 25 oC, they are both equal to 1.0 x 10-7 so Kw = 1.0 x 10-14. [Remember that equilibrium constants are a function of temperature so Kw changes with temperature.]

We can now say that

• For a neutral solution, [H3O+] = [OH-] = 1.0 x 10-7
• For an acid solution, [H3O+] > 1.0 x 10-7 and [OH-] less than 1.0 x 10-7
• For a basic solution, [H3O+] less than 1.0 x 10-7 and [OH-] > 1.0 x 10-7

When we add an acid or base to this water equilibrium, we will upset the equilibrium condition. However remember LeChatelier's principle tells us that the reaction will shift to oppose the perturbation and will reestablish equilibrium.

Let's consider that we have a 0.0010 solution of NaOH and we want to know the concentration of the hydronium ion and the hydroxide ion. Consider that the ionization of water does not occur until after the addition of the NaOH so that we have initially [H3O+] = 0, and [OH-] = 0.0010. There will be shift in the reaction to reestablish equilibrium. Since the stoichiometric coefficients of the products are equal, the shift in the amounts of [H3O+] and [OH-] will be equal; let's call that shift x. After equilibrium is established we will have x M concentration of H3O+ and (0.0010+x) M concentration of OH-. Now the equilibrium constant expression will be

Kw = 1.0 x10-14 = x (0.0010+ x) = 0.0010 x + x2

We could use the quadratic equation to easily solve for x, but we can also note that x will be very small. This means 0.0010 + x is approximately equal to 0.0010. So we now have that 1.0 x 10-14 = 0.0010x or x = 1.0 x 10-11 M for the concentration of H3O+, and the concentration of OH- will be 0.0010 + 1.0 x 10-11 which is 0.001 or 1.0 x 10-3. Note that the product of concentrations is still Kw as it must be.

We know that the strength of the conjugate base of a strong acid is weak, so we know that Kb for this base will be less than Ka for the acid. It is easy to show that the product of the Ka of an acid and the Kb of the corresponding base is equal to Kw (page 806 in your text). So we have the very useful expression

Kw = KaKb

If we know either Ka or Kb we can easily use this equation to calculate the unknown.

It would be cumbersome to continually write all H+ concentrations in exponential notation, so we try to simplify our method. We know that we can remove the base ten exponentials by taking the log base ten so that is what we do. Also since the exponentials are all negative, we will take the negative of the log base ten of the concentrations of H3O+ and OH- and use the symbol "p" to indicate that operation. Thus for the above problem:

pH = - log [H3O+] = - log(1.0 x 10-11) = 11 and pOH = -log [OH-] = -log (1.0 x 10-3) = 3

Note that pH + pOH = 14 as it must at 25 oC.

We can experimentally determine the pH by two methods. One method uses indicators, which change color over a particular pH range as shown in Figure 17.11 on page 810. The use of an indicator will give an approximate value for the pH. An electronic pH meter will give an accurate measure of the pH.

Review Questions

1. Given that the pH of vinegar is 2.80, what is the hydroxide ion concentration of the vinegar?
2. Work Exercise 17.8 on page 806.
3. If the pOH of an HCl solution is 10.62 what is the hydronium ion concentration?
On-Line Activity

Here is a site which offers more information on pH:

Web Author: Dr. Leon L. Combs
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