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21.2: The pH Concept

Created by: CK-12

Lesson Objectives

  • Describe the equilibrium process of water ionizing into hydrogen and hydroxide ions, and explain the ion-product constant.
  • Calculate the concentration of either the hydrogen ion or the hydroxide ion in water when the other value is known.
  • Calculate the pH of an aqueous solution of an acid or a base.
  • Calculate the hydrogen-ion concentration of a solution with a known pH.
  • Calculate the pOH of acids and bases and know the relationships between pH, pOH, hydrogen-ion concentration, and hydroxide-ion concentration.

Lesson Vocabulary

  • acidic solution
  • basic solution
  • ion-product of water
  • pH
  • pOH
  • self-ionization

Check Your Understanding

Recalling Prior Knowledge

  • What ion is present in an acidic solution according to the Arrhenius theory?
  • What ion is present in a basic solution according to the Arrhenius theory?

The acidity or basicity of a solution is a very important property. Natural bodies of water like lakes and streams have many dissolved components. The acidity of the water is very important to the fish and other organisms that live in the water. The concept of pH is used by chemists to keep track of how acidic or basic aqueous solutions are, and adjusting the pH is important in many practical applications. In this lesson, you will learn the meaning of pH and how to calculate it.

Self-Ionization of Water

Water is a molecular compound, so you may not necessarily expect it to break apart into ions. However, sensitive experiments show that water is actually a very weak electrolyte. When two molecules of water collide, there can be a transfer of a hydrogen ion from one molecule to the other. The products are a positively charged hydronium ion and a negatively charged hydroxide ion.

\mathrm{H_2O}(l) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq)

The self-ionization of water is the process in which water ionizes to hydronium ions and hydroxide ions. As with other aqueous acid-base reactions, the process is often simplified to show the ionization of just one water molecule into a hydrogen ion and a hydroxide ion.

\mathrm{H_2O}(l) \rightleftharpoons \mathrm{H^+}(aq) + \mathrm{OH^-}(aq)

Either equation is adequate, though the first is more accurate, since hydrogen ions in aqueous solution will always be attached to water molecules. Further discussion of acids and acid ionizations in this book will primarily show hydrogen ions in aqueous solution as H+, but keep in mind that this is just a commonly used abbreviation for the more accurate hydronium structure.

The Ion-Product of Water

The self-ionization of water occurs to a very limited extent. In other words, the equilibrium position strongly favors the reactant water molecule. As with other equilibrium reactions, we can write an equilibrium expression for the self-ionization process. The equilibrium constant is referred to as the ion-product for water and is given the symbol Kw.

Kw = [H+][OH-]

The ion-product of water (Kw) is the mathematical product of the concentrations of hydrogen ions and hydroxide ions. Note that H2O is not included in the ion-product expression because it is a pure liquid.

The value of Kw is very small, which is consistent with a reaction that heavily favors the reactants. At 25°C, the experimentally determined value of Kw in pure water is 1.0 × 10-14.

Kw = [H+][OH-] = 1.0 × 10-14

This equation is true for all dilute aqueous solutions at 25°C. As we will see, the relative concentrations of H+ and OH are different in acidic and basic solutions. However, the product of the two concentrations must remain equal to 1.0 × 10−14.

In pure water, the concentrations of hydrogen and hydroxide ions are equal to one another. Any aqueous solution in which [H+]=[OH-] is said to be neutral. To find the molarity of each ion in a neutral solution, simply take the square root of Kw.

[H+] = [OH-] = 1.0 × 10-7 M

For any neutral solution at 25°C, each of these ions has a concentration of 1.0 × 10-7 M.

An acidic solution is a solution in which the concentration of hydrogen ions is greater than the concentration of hydroxide ions. For example, hydrogen chloride ionizes to produce H+ and Cl ions upon dissolving in water.

HCl(g) → H+(aq) + Cl-(aq)

This increases the concentration of H+ ions in the solution. According to Le Châtelier’s principle, the equilibrium represented by \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H^+}(aq) + \mathrm{OH^-}(aq) is forced to the left, toward the reactant. As a result, the concentration of the hydroxide ion decreases.

A basic solution is a solution in which the concentration of hydroxide ions is greater than the concentration of hydrogen ions. Solid potassium hydroxide dissociates in water to yield potassium ions and hydroxide ions.

KOH(s) → K+(aq) + OH-(aq)

Now, the increased concentration of the OH ions requires a decrease in the concentration of the H+ ions in order to keep the value of the ion-product [H+][OH] at its equilibrium value. Sample problem 21.1 illustrates how Kw can be used to find the concentration of either H+ or OH in an aqueous solution when the other concentration is known.

Sample Problem 21.1: Using Kw in an Aqueous Solution

Hydrochloric acid (HCl) is a strong acid, meaning that it is essentially 100% ionized in solution. What are the values of [H+] and [OH] in a 2.0 × 10−3 M solution of HCl?

Step 1: List the known values and plan the problem.

Known

  • [HCl] = 2.0 × 10−3 M
  • Kw = 1.0 × 10−14

Unknown

  • [H+] = ? M
  • [OH] = ? M

Because HCl is 100% ionized, the concentration of H+ ions in solution will be equal to the original concentration of HCl. Each HCl molecule that was originally present ionizes into one H+ ion and one Cl ion. The concentration of OH can then be determined from [H+] and Kw.

Step 2: Solve.

[H+] = 2.0 × 10−3 M
Kw = [H+][OH-] = 1.0 × 10−14
\mathrm{[OH^-]=\dfrac{K_w}{[H^+]}=\dfrac{1.0 \times 10^{-14}}{2.0 \times 10^{-3}}=5.0 \times 10^{-12}\:M}

Step 3: Think about your result.

[H+] is much higher than [OH] because the solution is acidic. As with other equilibrium constants, the units for Kw are customarily omitted.

Practice Problems
  1. State whether the following solutions are acidic or basic.
    1. [H+] = 4.0 × 10−8 M
    2. [OH] = 3.0 × 10−5 M
    3. [H+] = 7.1 × 10−6 M
  2. A certain aqueous solution has a hydroxide ion concentration of 5.6 × 10−5 M. Calculate [H+] in this solution.

The pH Scale

Expressing the acidity of a solution by using the molarity of the hydrogen ion is cumbersome because the quantities are generally very small. Danish scientist, Søren Sørenson (1868-1939), proposed an easier system for indicating the concentration of H+ called the pH scale. The letters pH stand for the power of the hydrogen ion. The pH of a solution is the negative logarithm of the hydrogen-ion concentration.

pH = -log[H+]

In pure water or a neutral solution [H+] = 1.0 × 10−7 M. Substituting this value into the pH expression:

pH = -log[1.0 × 10−7] = -(-7.00) = 7.00

The pH of pure water or any neutral solution is thus 7.00. Due to the somewhat less intuitive rules for dealing with significant figures in the context of logarithms, only the numbers to the right of the decimal point in the pH value are the significant figures. Since 1.0 × 10−7 has two significant figures, the pH can be reported as 7.00.

A logarithmic scale condenses the range of acidity to numbers that are easy to use. For example, a solution in which [H+] = 1.0 × 10−4 M has a hydrogen-ion concentration that is 1000 times higher than in pure water. The pH of such a solution is 4.00, a difference of just 3 pH units. Notice that when [H+] is written in scientific notation and the coefficient is 1, the pH is simply the exponent with the sign changed. The pH of a solution in which [H+] = 1 × 10−2 M is 2.0 and the pH of a solution in which [H+] = 1 × 10−10 M is 10.0. If the coefficient is not equal to 1, a calculator must be used to find the pH. For example, the pH of a solution in which [H+] = 2.3 × 10−5 M can be found as shown below.

pH = -log[2.3 × 10-5] = 4.64

As we saw earlier, a solution in which [H+] is higher than 1 × 10−7 M is acidic, while a solution in which [H+] is lower than 1 × 10−7 M is basic. Consequently, solutions with pH values of less than 7 are acidic, while solutions with pH values higher than 7 are basic. The figure below (Figure below) illustrates this relationship, along with some examples of the pH values for various solutions.

The pH scale is a logarithmic scale based on the concentration of hydrogen ions. The higher the H+ ion concentration is, the lower the pH of the solution.

The pH scale is generally presented as running from 0 to 14, though it is possible to have a pH of less than 0 or greater than 14. For example, a highly concentrated 3.0 M solution of HCl has a negative pH.

pH = -log(3.0) = -0.48

When the pH of a solution is known, the concentration of the hydrogen ion can be calculated. The inverse of the logarithm (or antilog) is the 10x key on a calculator.

[H+] = 10-pH

For example, suppose that you have a solution with a pH of 9.14. [H+] can be found as follows:

[H+] = 10-pH = 10-9.14 = 7.24 × 10-10 M

Hydroxide Ion Concentration and pH

As we saw earlier, the hydroxide ion concentration of any aqueous solution is related to the hydrogen ion concentration through the value of Kw. We can use that relationship to calculate the pH of a solution of a base.

Sample Problem 21.2: The pH of a Base

Sodium hydroxide is a strong base. Find the pH of a solution prepared by dissolving 1.0 g of NaOH into enough water to make 1.0 L of solution.

Step 1: List the known values and plan the problem.

Known

  • mass of NaOH = 1.0 g
  • molar mass of NaOH = 40.00 g/mol
  • volume of solution = 1.0 L
  • Kw = 1.0 × 10−14

Unknown

  • pH of solution = ?

First, convert the mass of NaOH to moles. Second, calculate the molarity of the NaOH solution. Because NaOH is a strong base and is soluble in water, all of the dissolved NaOH will be dissociated, so [OH] will be equal to the calculated concentration of the NaOH. Third, use Kw to calculate the [H+] in the solution. Lastly, calculate the pH.

Step 2: Solve.

1.0 \ \cancel{\text{g NaOH}} \times \dfrac{1 \ \text{mol NaOH}}{40.00 \ \cancel{\text{g NaOH}}}=0.025 \ \text{mol NaOH}
\text{Molarity}=\dfrac{0.025 \ \text{mol NaOH}}{1.0 \ \text{L}}=0.025 \ \text{M NaOH}=0.025 \ \text{M OH}^-
[\text{H}^+]=\dfrac{\text{K}_\text{w}}{[\text{OH}^-]}=\dfrac{1.0 \times 10^{-14}}{0.025 \ \text{M}}=4.0 \times 10^{-13} \ \text{M}
\text{pH}=-\log[\text{H}^+]=-\log(4.0 \times 10^{-13})=12.40

Step 3: Think about your result.

The solution is basic, so its pH is greater than 7. The reported pH is rounded to two decimal places because the original mass and volume each have two significant figures.

Practice Problems
  1. Calculate the pH values for solutions with the following concentrations.
    1. [H+] = 4.8 × 10-4 M
    2. [OH] = 2.5 × 10−6 M
  2. A certain solution has a pH of 1.93. Calculate its [H+] and [OH].

The pOH Concept

As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration.

pOH = -log[OH-]

The pH of a solution can be related to the pOH. Consider a solution with pH = 4.0. The [H+] of the solution would be 1.0 × 10-4 M. Dividing Kw by this value yields an [OH] value of 1.0 × 10−10 M. The pOH of the solution would therefore be -log(1.0 × 10-10) = 10. This example illustrates the following relationship.

pH + pOH = 14

The pOH scale is similar to the pH scale, in that a pOH of 7 is indicative of a neutral solution. A basic solution has a pOH of less than 7, while an acidic solution has a pOH of greater than 7. The pOH is convenient to use when finding the hydroxide ion concentration from a solution with a known pH.

Sample Problem 21.3: Using pOH

Find the hydroxide concentration of a solution with a pH of 4.42.

Step 1: List the known values and plan the problem.

Known

  • pH = 4.42
  • pH + pOH = 14

Unknown

  • [OH] = ? M

First, the pOH is calculated, followed by the [OH].

Step 2: Solve.

pOH = 14 - pH = 14 - 4.42 = 9.58
[OH-] = 10-pOH = 10-9.58 = 2.6 × 10-10 M

Step 3: Think about your result.

The pH is that of an acidic solution, and the resulting hydroxide-ion concentration is less than 1 × 10−7 M. The answer has two significant figures because the given pH has two decimal places.

Practice Problems
  1. Calculate the pOH of a 0.075 M solution of HCl.
  2. What is the hydroxide-ion concentration of a solution with a pH of 12.87?

The diagram below (Figure below) shows all of the interrelationships between [H+], [OH-], pH, and pOH.

The flowchart shows the way to convert between hydrogen ion concentration, hydroxide ion concentration, pH, and pOH.

Lesson Summary

  • Water self-ionizes into hydrogen ions and hydroxide ions. The equilibrium constant for this process is referred to as Kw and is equal to 1.0 × 10-14 at 25°C.
  • The concentrations of H+ and OH are both equal to 1.0 × 10−7 M in pure water. If the H+ concentration is higher than the OH concentration, the solution is acidic. If the OH concentration is higher than the H+ concentration, the solution is basic.
  • The pH of a solution is the negative logarithm of the H+ ion concentration. Pure water or a neutral solution has a pH of 7.0. Acidic solutions have pH values that are lower than 7, while basic solutions have pH values that are greater than 7.
  • The H+ ion concentration can be calculated when the pH is known by using the antilog function.
  • The pOH scale is based on the concentration of the OH ion.

Lesson Review Questions

Reviewing Concepts

  1. Describe the process by which water self-ionizes, and explain why pure water is considered to be neutral.
  2. Explain why the concentrations of the hydrogen ion and the hydroxide ion cannot both be greater than 1 × 10−7 M in an aqueous solution.
  3. Indicate whether solutions with the following pH values are acidic, basic, or neutral.
    1. pH = 9.4
    2. pH = 7.0
    3. pH = 5.0
  4. How can the pOH of a solution be determined if its pH is known?

Problems

  1. Calculate the OH ion concentration in solutions with the following H+ concentrations. Assume a temperature of 25°C. State whether the solution is acidic or basic.
    1. [H+] = 2.3 × 10−4 M
    2. [H+] = 8.7 × 10−10 M
  2. Calculate the H+ ion concentration in solutions with the following OH concentrations. Assume a temperature of 25°C. State whether the solution is acidic or basic.
    1. [OH] = 1.9 × 10−9 M
    2. [OH] = 0.60 M
  3. Calculate the pH of a solution with each of the following ion concentrations and indicate if the solution is acidic or basic.
    1. [H+] = 1.0 × 10−5 M
    2. [H+] = 2.8 × 10−11 M
    3. [OH] = 1.0 × 10−2 M
    4. [OH] = 4.4 × 10−9 M
  4. Calculate the pOH of the solutions listed in question 7.
  5. Determine [H+] and [OH] in aqueous solutions with the following pH or pOH values.
    1. pH = 1.87
    2. pH = 11.15
    3. pH = −0.95
    4. pOH = 6.21
    5. pOH = 14.42
    6. pOH = 7.03
  6. A solution is prepared by dissolving 15.0 grams of NaOH into enough water to make 500.0 mL of solution. Calculate the pH of the solution.
  7. You have prepared 1.00 L of a solution with a pH of 5.00. What is the pH of the solution if 0.100 L of additional water is added to it? (Hint: Calculate the moles of H+ ions present in the solution.)
  8. How much water would need to be added to the original solution in question 11 in order to bring the pH to 6.00?

Further Reading / Supplemental Links

Points to Consider

Acids and bases can be characterized as either strong or weak, based on the extent to which they ionize in water.

  • What are acid and base ionization constants?
  • How is the pH of a solution containing a weak acid or a weak base calculated?

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