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20.3: The pH Concept

Created by: CK-12

Lesson Objectives

The student will:

  • calculate [\mathrm{H}^+] and [\mathrm{OH}^-] for a solution of acid or base.
  • define autoionization.
  • state the [\mathrm{H}^+], [\mathrm{OH}^-], and K_w values for the autoionization of water.
  • define pH and describe the pH scale.
  • write the formulas for pH and pOH and express their values in a neutral solution at 25^\circ\mathrm{C}.
  • explain the relationships among pH, pOH, and K_w.
  • calculate [\mathrm{H}^+], [\mathrm{OH}^-], pH, and pOH given the value of any one of the other values.
  • explain the relationship between the acidity or basicity of a solution and the hydronium ion concentration, [\mathrm{H}_3\mathrm{O}^+] , and the hydroxide ion concentration, [\mathrm{OH}^-], of the solution.
  • predict whether an aqueous solution is acidic, basic, or neutral from [\mathrm{H}_3\mathrm{O}^+], [\mathrm{OH}^-], or the pH.

Vocabulary

  • autoionization
  • hydronium ion
  • ion product constant for water (K_w)
  • pH
  • pOH

Introduction

We have learned many properties of water, such as pure water does not conduct electricity. The reason pure water does not conduct electricity is because the concentration of ions present when water ionizes is small. In this lesson, we will look a little closer at this property of water and how it relates to acids and bases.

The Hydronium Ion

Recall that ions in solution are hydrated. That is, water molecules are loosely bound to the ions by the attraction between the charge on the ion and the oppositely charged end of the polar water molecules, as illustrated in the figure below. When we write the formula for these ions in solution, we do not show the attached water molecules. It is simply recognized by chemists that ions in solution are always hydrated.

As with any other ion, a hydrogen ion dissolved in water will be closely associated with one or more water molecules. This fact is sometimes indicated explicitly by writing the hydronium ion, \mathrm{H}_3\mathrm{O}^+, in place of the hydrogen ion, \mathrm{H}^+. Many chemists still use \mathrm{H}^+_{(aq)} to represent this situation, but it is understood that this is just an abbreviation for what is really occurring in solution. You are likely to come across both, and it is important for you to understand that they are actually describing the same entity. When using the hydronium ion in a chemical equation, you may need to add a molecule of water to the other side so that the equation will be balanced. This is illustrated in the equations below. Note that you are not really adding anything to the reaction. The (aq) symbol indicates that the various reaction components are dissolved in water, so writing one of these water molecules out explicitly in the equation does not change the reaction conditions.

\mathrm{HCl}_{(aq)} \rightarrow \mathrm{H}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} (not showing hydronium)
\mathrm{HCl}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightarrow \mathrm{H}_3\mathrm{O}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} (showing hydronium)

Relationship Between [H+] and [OH-]

Even totally pure water will contain a small amount of \mathrm{H}^+ and \mathrm{OH}^-. This is because water undergoes a process known as autoionization. Autoionization occurs when the same reactant acts as both the acid and the base. Look at the reaction below.

\mathrm{H}_2\mathrm{O}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(aq)} \rightarrow \mathrm{H}_3\mathrm{O}^+_{(aq)} + \mathrm{OH}^-_{(aq)}

The ionization of water is frequently written as:

\mathrm{H}_2\mathrm{O}_{(l)} \rightarrow \mathrm{H}^+ + \mathrm{OH}^-.

The equilibrium constant expression for this dissociation would be K_w = [\mathrm{H}^+][\mathrm{OH}^-]. From experimentation, chemists have determined that in pure water, [\mathrm{H}^+] = 1 \times 10^{-7} \ \mathrm{mol/L} and [\mathrm{OH}^-] = 1 \times 10^{-7} \ \mathrm{mol/L}.

Because this is a particularly important equilibrium, the equilibrium constant is given a subscript to differentiate it from other reactions. K_w, also known as the ion product constant for water, always refers to the autoionization of water. We can the calculate K_w because we know the value of [\mathrm{H}^+] and [\mathrm{OH}^-] for pure water at 25^\circ\mathrm{C}.

K_w = [\mathrm{H}^+][\mathrm{OH}^-]
K_w = (1 \times 10^{-7})(1 \times 10^{-7})
K_w = 1 \times 10^{-14}

A further definition of acids and bases can now be made:

When [\mathrm{H}_3\mathrm{O}^+] = [\mathrm{OH}^-] (as in pure water), the solution is neutral.
When [\mathrm{H}_3\mathrm{O}^+] > [\mathrm{OH}^-], the solution is an acid.
When [\mathrm{H}_3\mathrm{O}^+] < [\mathrm{OH}^-], the solution is a base.

Stated another way, an acid has a [\mathrm{H}_3\mathrm{O}^+] that is greater than 1 \times 10^{-7} and a [\mathrm{OH}^-] that is less than 1 \times 10^{-7}. A base has a [\mathrm{OH}^-] that is greater than 1 \times 10^{-7} and a [\mathrm{H}_3\mathrm{O}^+] that is less than 1 \times 10^{-7}.

The equilibrium between \mathrm{H}^+, \mathrm{OH}^-, and \mathrm{H}_2\mathrm{O} will exist in all water solutions, regardless of anything else that may be present in the solution. Some substances that are placed in water may become involved with either the hydrogen or hydroxide ions and alter the equilibrium state. However, as long as the temperature is kept constant at 25^\circ\mathrm{C}, the equilibrium will shift to maintain the equilibrium constant, K_w, at exactly 1 \times 10^{-14}.

For example, a sample of pure water at 25^\circ\mathrm{C} has [\mathrm{H}^+] equal to 1 \times 10^{-7} \ \mathrm{M} and [\mathrm{OH}^-] = 1 \times 10^{-7} \ \mathrm{M}. The K_w for this solution, of course, will be 1 \times 10^{-14}. Suppose some \mathrm{HCl} gas is added to this solution so that the \mathrm{H}^+ concentration increases. This is a stress to the equilibrium system. Since the concentration of a product is increased, the reverse reaction rate will increase and the equilibrium will shift toward the reactants. The concentrations of both ions will be reduced until equilibrium is re-established. If the final [\mathrm{H}^+] = 1 \times 10^{-4} \ \mathrm{M}, we can calculate the [\mathrm{OH}^-] because we know that the product of [\mathrm{H}^+] and [\mathrm{OH}^-] at equilibrium is always 1 \times 10^{-14}.

K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1 \times 10^{-14}
[\mathrm{OH}^-] = \frac {1 \times 10^{-14}} {[\mathrm{H}^+]} = \frac {1 \times 10^{-14}} {1 \times 10^{-4}} = 1 \times 10^{-10} \ \mathrm{M}

Suppose, on the other hand, something is added to the solution that reduces the hydrogen ion concentration. As soon as the hydrogen ion concentration begins to decrease, the reverse rate decreases and the forward rate will shift the equilibrium toward the products. The concentrations of both ions will be increased until equilibrium is re-established. If the final hydrogen ion concentration is 1 \times 10^{-12} \ \mathrm{M}, we can calculate the final hydroxide ion concentration.

K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1 \times 10^{-14}
[\mathrm{OH}^-] = \frac {1 \times 10^{-14}} {[\mathrm{H}^+]} = \frac {1 \times 10^{-14}} {1 \times 10^{-12}} = 1 \times 10^{-2} \ \mathrm{M}

Using the K_w expression and our knowledge of the K_w value, as long as we know either the [\mathrm{H}^+] or the [\mathrm{OH}^-] in a water solution, we can always calculate the value for the other one.

Example:

What would be the [\mathrm{H}^+] for a grapefruit found to have a [\mathrm{OH}^-] of 1.26 \times 10^{-11} \ \mathrm{mol/L}? Is the solution acidic, basic, or neutral?

Solution:

K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1.00 \times 10^{-14}
[\mathrm{H}^+] = \frac {1 \times 10^{-14}} {[\mathrm{OH}^-]} = \frac {1 \times 10^{-14}} {1.26 \times 10^{-11}} = 7.94 \times 10^{-4} \ \mathrm{M}

Since the [\mathrm{H}^+] in this solution is greater than 1 \times 10^{-7} \ \mathrm{M}, the solution is acidic.

pH and pOH

There are a few very concentrated acid and base solutions used in industrial chemistry and laboratory situations. For the most part, however, acid and base solutions that occur in nature, used in cleaning, and used in biochemistry applications are relatively dilute. Most of the acids and bases dealt with in laboratory situations have hydrogen ion concentrations between 1.0 \ \mathrm{M} and 1.0 \times 10^{-14} \ \mathrm{M}. Expressing hydrogen ion concentrations in exponential numbers can become tedious, so a Danish chemist named Søren Sørensen developed a shorter method for expressing acid strength or hydrogen ion concentration with a non-exponential number. This value is referred to as pH and is defined by the following equation:

\mathrm{pH} = -\log [\mathrm{H}^+],

where \mathrm{p} = -\log and \mathrm{H} refers to the hydrogen ion concentration. The p from pH comes from the German word potenz, meaning power or the exponent of. Rearranging this equation to solve for [\mathrm{H}^+], we get [\mathrm{H}^+] = 10^{-\mathrm{pH}}. If the hydrogen ion concentration is between 1.0 \ \mathrm{M} and 1.0 \times 10^{-14} \ \mathrm{M}, the value of the pH will be between 0 and 14.

Example:

Calculate the pH of a solution where [\mathrm{H}^+] = 0.01 \ \mathrm{mol/L}.

Solution:

\mathrm{pH} = -\log (0.01)
\mathrm{pH} = -\log (1 \times 10^{-2})
\mathrm{pH} = 2

Example:

Calculate the [\mathrm{H}^+] if the pH is 4.

Solution:

[\mathrm{H}^+] = 10^{-\mathrm{pH}}
[\mathrm{H}^+] = 10^{-4}
[\mathrm{H}^+] = 1 \times 10^{-4} \ \mathrm{mol/L}

Example:

Calculate the pH of saliva, where [\mathrm{H}^+] = 1.58 \times 10^{-6} \ \mathrm{mol/L}.

Solution:

\mathrm{pH} = -\log [\mathrm{H}^+] = -\log (1.58 \times 10^{-6})
\mathrm{pH} = 5.8

Example:

Fill in the rest of Table below.

Hydrogen ion concentration and corresponding pH.
[\mathrm{H}^+] in mol/L -\log [\mathrm{H}^+] \mathrm{pH}
0.1 1.00 1.00
0.2 0.70 0.70
1 \times 10^{-5} ? ?
? ? 6.00
0.065 ? ?
? ? 9.00

Solution:

The completed table is shown below (Table below).

Hydrogen ion concentration and corresponding pH.
[\mathrm{H}^+] in mol/L -\log [\mathrm{H}^+] \mathrm{pH}
0.1 1.00 1.00
0.2 0.70 0.70
1.00 \times 10^{-5} 5 5
1.00 \times 10^{-6} 6.00 6.00
0.065 1.19 1.19
1.00 \times 10^{-9} 9.00 9.00

An acid with pH = 1, then, is stronger than an acid with pH = 2 by a factor of 10. Simply put, lower pH values correspond to higher \mathrm{H}^+ concentrations and more acidic solutions, while higher pH values correspond to higher \mathrm{OH}^- concentrations and more basic solutions. This is illustrated in the figure below. It should be pointed out that there are acids and bases that fall outside the pH range depicted. However, we will confine ourselves for now to those falling within the 0-14 range, which covers [\mathrm{H}^+] values from 1.0 \ \mathrm{M} all the way down to 1 \times 10^{-14} \ \mathrm{M}.

pH versus Acidity
pH level Solution
\mathrm{pH} < 7 Acid
\mathrm{pH} = 7 Neutral
\mathrm{pH} > 7 Basic

Have you ever cut an onion and had your eyes water up? This is because of a compound with the formula \mathrm{C}_3\mathrm{H}_6\mathrm{OS} that is found in onions. When you cut the onion, a variety of reactions occur that release a gas. This gas can diffuse into the air and eventfully mix with the water found in your eyes to produce a dilute solution of sulfuric acid. This is what irritates your eyes and causes them to water. There are many common examples of acids and bases in our everyday lives. Look at the pH scale below to see how these common examples relate in terms of their pH.

Even though both acidic and basic solutions can be expressed by pH, an equivalent set of expressions exists for the concentration of the hydroxide ion in water. This value, referred to as pOH, is defined as:

\mathrm{pOH} = -\log [\mathrm{OH}^-]

If the pOH is greater than 7, the solution is acidic. If the pOH is equal to 7, the solution is neutral. If the pOH is less than 7, the solution is basic.

If we take the negative log of the complete K_w expression, we obtain:

K_w = [\mathrm{H}^+][\mathrm{OH}^-]
-\log K_w = (-\log [\mathrm{H}^+]) + (-\log [\mathrm{OH}^-])
-\log (1 \times 10^{-14}) = (-\log [\mathrm{H}^+]) + (-\log [\mathrm{OH}^-])
14 = \mathrm{pH} + \mathrm{pOH}

Therefore, the sum of the pH and the pOH is always equal to 14 (at 25^\circ\mathrm{C}). Remember that the pH scale is written with values from 0 to 14 because many useful acid and base solutions fall within this range. Now let’s go through a few examples to see how this calculation works for problem-solving in solutions with an added acid or base.

Example:

What is the [\mathrm{H}^+] for a solution of \mathrm{NH}_3 whose [\mathrm{OH}^-] = 8.23 \times 10^{-6} \ \mathrm{mol/L}?

Solution:

[\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-] = 1.00 \times 10^{-14}
[\mathrm{H}_3\mathrm{O}^+] = \frac {1.00 \times 10^{-14}} {[\mathrm{OH}^-]} = \frac {1.00 \times 10^{-14}} {8.23 \times 10^{-6}} = 1.26 \times 10^{-9} \ \mathrm{M}

Example:

Black coffee has a [\mathrm{H}_3\mathrm{O}^+] = 1.26 \times 10^{-5} \ \mathrm{mol/L}. What is the pOH?

Solution:

\mathrm{pH} = -\log [\mathrm{H}^+] = -\log 1.26 \times 10^{-5} = 4.90
\mathrm{pH} + \mathrm{pOH} = 14
\mathrm{pOH} = 14 - \mathrm{pH} = 14 - 4.90 = 9.10

For a classroom demonstration of pH calculations (5d, 5f; 1e I&E Stand.), see http://www.youtube.com/watch?v=lca_puB1R8k (9:45).

Lesson Summary

  • Autoionization is the process where the same molecule acts as both an acid and a base.
  • Water ionizes to a very slight degree according to the equation \mathrm{H}_2\mathrm{O}_{(l)} \leftrightharpoons [\mathrm{H}^+] + [\mathrm{OH}^-].
  • In pure water at 25^\circ\mathrm{C}, [\mathrm{H}^+] = [\mathrm{OH}^-] = 1.00 \times 10^{-7} \ \mathrm{M}.
  • The equilibrium constant for the dissociation of water, K_w, is equal to 1.00 \times 10^{-14} at 25^\circ\mathrm{C}.
  • \mathrm{pH} = -\log [\mathrm{H}^+]
  • \mathrm{pOH} = -\log [\mathrm{OH}^-]
  • \mathrm{p}K_w = -\log K_w
  • \mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_w = 14.0

Further Reading / Supplemental Links

The websites below have more information about pH.

Review Questions

  1. What is the [\mathrm{H}^+] ion concentration in a solution of 0.350 \ \mathrm{mol/L} \ \mathrm{H}_2\mathrm{SO}_4?
    1. 0.175 \ \mathrm{mol/L}
    2. 0.350 \ \mathrm{mol/L}
    3. 0.700 \ \mathrm{mol/L}
    4. 1.42 \times 10^{-14} \ \mathrm{mol/L}
  2. A solution has a \mathrm{pH} of 6.54. What is the concentration of hydronium ions in the solution?
    1. 2.88 \times 10^{-7} \ \mathrm{mol/L}
    2. 3.46 \times 10^{-8} \ \mathrm{mol/L}
    3. 6.54 \ \mathrm{mol/L}
    4. 7.46 \ \mathrm{mol/L}
  3. A solution has a \mathrm{pH} of 3.34. What is the concentration of hydroxide ions in the solution?
    1. 4.57 \times 10^{-4} \ \mathrm{mol/L}
    2. 2.19 \times 10^{-11} \ \mathrm{mol/L}
    3. 3.34 \ \mathrm{mol/L}
    4. 10.66 \ \mathrm{mol/L}
  4. A solution contains 4.33 \times 10^{-8} \ \mathrm{M} hydroxide ions. What is the \mathrm{pH} of the solution?
    1. 4.33
    2. 6.64
    3. 7.36
    4. 9.67
  5. Fill in Table below and rank the solutions in terms of increasing acidity.
Table for Problem 5
Solutions [\mathrm{H}^+] \ \mathrm{(mol/L)} -\mathrm{log} \ [\mathrm{H}^+] \mathrm{pH}
A 0.25 0.60 0.60
B ? 2.90 ?
C 1.25 \times 10^{-8} ? ?
D 0.45 \times 10^{-3} ? ?
E ? 1.26 ?
  1. It has long been advocated that red wine is good for the heart. Wine is considered to be an acidic solution. Determine the concentration of hydronium ions in wine with \mathrm{pH} \ 3.81.
  2. What does the value of K_w tell you about the autoionization of water?
  3. If the \mathrm{pH} of an unknown solution is 4.25, what is the \mathrm{pOH}?
    1. 10^{-4.25}
    2. 10^{-9.75}
    3. 9.75
    4. 14.0 - 10^{-9.75}
  4. A solution contains a hydronium ion concentration of 3.36 \times 10^{-4} \ \mathrm{mol/L}. What is the \mathrm{pH} of the solution?
    1. 3.36
    2.  3.47
    3. 10.53
    4. none of the above
  5. A solution contains a hydroxide ion concentration of 6.43 \times 10^{-9} \ \mathrm{mol/L}. What is the \mathrm{pH} of the solution?
    1. 5.80
    2. 6.48
    3. 7.52
    4. 8.19
  6. An unknown solution was found in the lab. The \mathrm{pH} of the solution was tested and found to be 3.98. What is the concentration of hydroxide ions in this solution?
    1. 3.98 \ \mathrm{mol/L}
    2. 0.67 \ \mathrm{mol/L}
    3. 1.05 \times 10^{-4} \ \mathrm{mol/L}
    4. 9.55 \times 10^{-11} \ \mathrm{mol/L}

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