# 21.3: Acid and Base Strength

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Explain the difference between a strong acid or base and a weak acid or base.
• Write equilibrium expressions for the ionizations of weak acids and weak bases. Explain how the value of Ka or Kb relates to the strength of the acid or base.
• Calculate an unknown Ka or Kb from the solution concentration and the pH.
• Calculate the expected pH of a solution containing a given concentration of a weak acid or base that has a known Ka or Kb value.

## Lesson Vocabulary

• acid ionization constant
• base ionization constant
• strong acid
• strong base
• weak acid
• weak base

### Recalling Prior Knowledge

• What is meant by the ionization (or dissociation) of a polar or ionic compound that is dissolved in water?
• How is the pH of a solution containing a strong acid or strong base determined?

Some acids, such as sulfuric acid and nitric acid, are strong acids. Working with strong acids requires special care and protective clothing and eyewear. Other acids, such as acetic acid and citric acid, are encountered in many foods and beverages. These weak acids are considerably less dangerous. In this lesson, you will learn about the differences between strong and weak acids and bases.

## Strong and Weak Acids and Bases

Acids are classified as either strong or weak, based on the extent to which they ionize in water. A strong acid is an acid which is completely ionized in aqueous solution. As we saw earlier, hydrogen chloride (HCl) is a gas as a pure compound, but it ionizes into hydrogen ions and chloride ions upon being dissolved into water.

\begin{align*}\mathrm{HCl}(g) \rightarrow \mathrm{H^+}(aq) + \mathrm{Cl^-}(aq)\end{align*}

The directional arrow in the equation above indicates that the HCl reactant is converted completely into the product ions. In other words, there is 100% ionization of HCl.

A weak acid is an acid that ionizes only slightly in aqueous solution. Acetic acid, the acid found in vinegar, is a very common weak acid. Its ionization is shown below.

\begin{align*}\mathrm{CH_3COOH}(aq) \rightleftharpoons \mathrm{H^+}(aq) + \mathrm{CH_3COO^-}(aq)\end{align*}

The ionization of acetic acid is incomplete, and so the equation is shown with a double arrow, indicating equilibrium between the reactant and products. The extent of ionization varies for different weak acids, but it is generally less than 10%. A 0.10 M solution of acetic acid is only about 1.3% ionized, meaning that the equilibrium strongly favors the reactants.

Weak acids, like strong acids, ionize to yield the H+ ion and a conjugate base. In the examples so far, the conjugate bases are the chloride ion (Cl) and the acetate ion (CH3COO). Because HCl is a strong acid, its conjugate base (Cl) is extremely weak. The chloride ion is essentially incapable of taking the H+ ion off of H3O+ and becoming HCl again. In general, the stronger the acid, the weaker its conjugate base. Likewise, the weaker the acid, the stronger its conjugate base. The Table below shows a listing of some common acids and their conjugate bases. The acids are listed from strongest at the top to weakest at the bottom, while the order for the conjugate bases is reversed.

Relative Strengths of Acids and their Conjugate Bases
Acid Conjugate Base
Strong Acids
HCl (hydrochloric acid) Cl (chloride ion)
H2SO4 (sulfuric acid) HSO4 (hydrogen sulfate ion)
HNO3 (nitric acid) NO3 (nitrate ion)
Weak Acids
H3PO4 (phosphoric acid) H2PO4 (dihydrogen phosphate ion)
CH3COOH (acetic acid) CH3COO (acetate ion)
H2CO3 (carbonic acid) HCO3 (hydrogen carbonate ion)
HCN (hydrocyanic acid) CN (cyanide ion)

The top three acids in the table above (Table above) are strong acids and are thus all 100% ionized in solution. The others are weak acids and are only slightly ionized. Phosphoric acid is stronger than acetic acid, so it is ionized to a greater extent. Acetic acid is stronger than carbonic acid, and so on.

### The Acid Ionization Constant, Ka

The ionization for a generic weak acid, HA, can be written in one of two ways.

\begin{align*}\mathrm{HA}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{A^-}(aq)\end{align*}
\begin{align*}\mathrm{HA}(aq) \rightarrow \mathrm{H^+}(aq) + \mathrm{A^-}(aq)\end{align*}

The water molecule is omitted for simplicity in the second case. This will be the way that acid ionization equations will be written for the remainder of the chapter. Because the acid is weak, an equilibrium expression can be written. An acid ionization constant (Ka) is the equilibrium constant for the ionization of an acid.

\begin{align*}\mathrm{K_a=\dfrac{[H^+][A^-]}{[HA]}}\end{align*}

The acid ionization represents the fraction of the original acid that has been ionized in solution. Therefore, the numerical value of Ka is a reflection of the strength of the acid. Weak acids with relatively higher Ka values are stronger than acids with relatively lower Ka values. Because strong acids are essentially 100% ionized, the concentration of the acid in the denominator is nearly zero and the Ka value approaches infinity. For this reason, Ka values are generally reported only for weak acids.

The Table below is a listing of acid ionization constants for several acids. Note that polyprotic acids have a distinct ionization constant for each ionization step. Each successive ionization constant for a polyprotic acid is always smaller than the previous one.

Acid Ionization Constants at 25°C
Name of Acid Ionization Equation Ka
Sulfuric acid

\begin{align*}\mathrm{H_2SO_4 \rightleftharpoons H^+ + HSO^-_4}\end{align*}

\begin{align*}\mathrm{HSO^-_4 \rightleftharpoons H^+ + SO^{2-}_4}\end{align*}

very large

1.3 × 10−2

Oxalic acid

\begin{align*}\mathrm{H_2C_2O_4 \rightleftharpoons H^+ + HC_2O^-_4}\end{align*}

\begin{align*}\mathrm{HC_2O^-_4 \rightleftharpoons H^+ + C_2O^{2-}_4}\end{align*}

6.5 × 10−2

6.1 × 10−5

Phosphoric acid

\begin{align*}\mathrm{H_3PO_4 \rightleftharpoons H^+ + H_2PO^-_4}\end{align*}

\begin{align*}\mathrm{H_2PO^-_4 \rightleftharpoons H^+ + HPO^{2-}_4}\end{align*}

\begin{align*}\mathrm{HPO^{2-}_4 \rightleftharpoons H^+ + PO^{3-}_4}\end{align*}

7.5 × 10−3

6.2 × 10−8

4.8 × 10−13

Hydrofluoric acid \begin{align*}\mathrm{HF \rightleftharpoons H^+ + F^-}\end{align*} 7.1 × 10−4
Nitrous acid \begin{align*}\mathrm{HNO_2 \rightleftharpoons H^+ + NO_2^-}\end{align*} 4.5 × 10−4
Benzoic acid \begin{align*}\mathrm{C_6H_5COOH \rightleftharpoons H^+ + C_6H_5COO^-}\end{align*} 6.5 × 10−5
Acetic acid \begin{align*}\mathrm{CH_3COOH \rightleftharpoons H^+ + CH_3COO^-}\end{align*} 1.8 × 10−5
Carbonic acid

\begin{align*}\mathrm{H_2CO_3 \rightleftharpoons H^+ + HCO^-_3}\end{align*}

\begin{align*}\mathrm{HCO^-_3 \rightleftharpoons H^+ + CO^{2-}_3}\end{align*}

4.2 × 10−7

4.8 × 10−11

Hydrocyanic acid \begin{align*}\mathrm{HCN \rightleftharpoons H^+ + CN^-}\end{align*} 4.9 × 10−10

### The Base Ionization Constant, Kb

As with acids, bases can either be strong or weak, depending on their extent of ionization. A strong base is a base which ionizes completely in aqueous solution. The most common strong bases are soluble metal hydroxide compounds, such as potassium hydroxide. Some metal hydroxides are not as strong because they are not as soluble. For example, calcium hydroxide is only slightly soluble in water. However, the portion that does dissolve completely dissociates into ions.

A weak base is a base that ionizes only slightly in aqueous solution. Recall that a base can be defined as a substance that accepts a hydrogen ion from another substance. When a weak base such as ammonia is dissolved in water, it accepts an H+ ion from water, forming the hydroxide ion and the conjugate acid of the base, the ammonium ion.

\begin{align*}\mathrm{NH_3}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{NH^+_4}(aq) + \mathrm{OH^-}(aq)\end{align*}

The equilibrium greatly favors the reactants, and the extent of ionization of the ammonia molecule is very small.

An equilibrium expression can be written for the reactions of weak bases with water. Again, because the concentration of liquid water is essentially constant, the water is not included in the expression. A base ionization constant (Kb) is the equilibrium constant for the ionization of a base. For ammonia, the expression is:

\begin{align*}\mathrm{K_b=\dfrac{[NH^+_4][OH^-]}{[NH_3]}}\end{align*}

The numerical value of Kb is a reflection of the strength of the base. Weak bases with relatively high Kb values are stronger than bases with relatively low Kb values. The Table below lists base ionization constants for several weak bases.

Base Ionization Constants at 25°C
Name of Base Ionization Equation Kb
Methylamine \begin{align*}\mathrm{CH_3NH_2 + H_2O \rightleftharpoons CH_3NH^+_3 + OH^-}\end{align*} 5.6 × 10−4
Ammonia \begin{align*}\mathrm{NH_3 + H_2O \rightleftharpoons NH^+_4 + OH^-}\end{align*} 1.8 × 10−5
Pyridine \begin{align*}\mathrm{C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^-}\end{align*} 1.7 × 10−9
Acetate ion \begin{align*}\mathrm{CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-}\end{align*} 5.6 × 10−10
Fluoride ion \begin{align*}\mathrm{F^- + H_2O \rightleftharpoons HF + OH^-}\end{align*} 1.4 × 10−11
Urea \begin{align*}\mathrm{H_2NCONH_2 + H_2O \rightleftharpoons H_2CONH^+_3 + OH^-}\end{align*} 1.5 × 10−14

Notice that the conjugate base of a weak acid is also a weak base. For example, the acetate ion has a small tendency to accept a hydrogen ion from water to form acetic acid and the hydroxide ion.

## Calculations with Ka and Kb

The numerical value of Ka or Kb can be determined by experiment. A solution of known concentration is prepared, and its pH is measured with an instrument called a pH meter (Figure below).

A pH meter is a laboratory device that provides quick, accurate measurements of the pH of solutions.

Sample Problem 21.4 shows the steps involved to determine the Ka of formic acid (HCOOH).

Sample Problem 21.4: Calculation of an Acid Ionization Constant

A 0.500 M solution of formic acid is prepared, and its pH is measured to be 2.04. Determine the Ka for formic acid.

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

Known

• initial [HCOOH] = 0.500 M
• pH = 2.04

Unknown

• Ka = ?

First, the pH is used to calculate the value of [H+] at equilibrium. An ICE table is set up in order to determine the concentrations of HCOOH and HCOO at equilibrium. All concentrations are then substituted into the Ka expression, and the Ka value is calculated.

Step 2: Solve.

[H+] = 10-pH = 10-2.04 = 9.12 × 10−3

Since each formic acid molecule that ionizes yields one H+ ion and one formate ion (HCOO), the concentrations of H+ and HCOO are equal at equilibrium. We assume that the initial concentrations of each ion are zero, resulting in the following ICE table (Table below). Although the concentration of H+ would technically be 1.0 × 10−7 before any dissociation occurs, we can ignore this very small amount because, in this case, it does not make a noticeable contribution to the final value of [H+].

Concentrations [HCOOH] [H+] [HCOO]
Initial 0.500 0 0
Change −9.12 × 10−3 +9.12 × 10−3 +9.12 × 10−3
Equilibrium 0.491 9.12 × 10−3 9.12 × 10−3

Substituting the equilibrium values into the Ka expression gives the following:

\begin{align*}\mathrm{K_a=\dfrac{[H^+][HCOO^-]}{[HCOOH]}=\dfrac{(9.12 \times 10^{-3})(9.12 \times 10^{-3})}{0.491}=1.7 \times 10^{-4}}\end{align*}

The value of Ka is consistent with that of a weak acid. Two significant figures are appropriate for the answer, since there are two digits after the decimal point in the reported pH.

Practice Problems
1. Hypochlorous acid (HClO) is a weak acid. What is its Ka if a 0.250 M solution of hypochlorous acid has a pH of 4.07?

Similar steps can be taken to determine the Kb of a base. For example, a 0.750 M solution of the weak base ethylamine (C2H5NH2) has a pH of 12.31.

\begin{align*}\mathrm{C_2H_5NH_2 + H_2O \rightleftharpoons C_2H_5NH^+_3 + OH^-}\end{align*}

Since one of the products of the ionization reaction is the hydroxide ion, we need to first find the value of [OH] at equilibrium. The pOH is 14 – 12.31 = 1.69. [OH] is then calculated to be 10−1.69 = 2.04 × 10−2 M. The ICE table is then set up as shown below (Table below).

Concentrations [C2H5NH2] [C2H5NH3+] [OH]
Initial 0.750 0 0
Change −2.04 × 10−2 +2.04 × 10−2 +2.04 × 10−2
Equilibrium 0.730 2.04 × 10−2 2.04 × 10−2

Substituting into the Kb expression yields the Kb for ethylamine.

\begin{align*}\mathrm{K_b=\dfrac{[C_2H_5NH^+_3][OH^-]}{[C_2H_5NH_2]}=\dfrac{(2.04 \times 10^{-2})(2.04 \times 10^{-2})}{0.730}=5.7 \times 10^{-4}}\end{align*}

### Calculating the pH of a Weak Acid or Weak Base

The Ka and Kb values have been determined for a great many acids and bases, as shown in the Acid and Base Ionization Constants Tables (Table above and Table above). These can be used to calculate the pH of any solution of a weak acid or base whose ionization constant is known.

Sample Problem 21.5: Calculating the pH of a Weak Acid

Calculate the pH of a 2.00 M solution of nitrous acid (HNO2). The Ka for nitrous acid can be found in the table above (Table above).

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

Known

• initial [HNO2] = 2.00 M
• Ka = 4.5 × 10−4

Unknown

• pH = ?

First, an ICE table is set up with the variable x used to signify the change in concentration of the substance due to ionization of the acid. Then, the Ka expression is used to solve for x and calculate the pH.

Step 2: Solve.

Concentrations [HNO2] [H+] [NO2]
Initial 2.00 0 0
Change −x +x +x
Equilibrium 2.00 − x x x

The Ka expression and value is used to set up an equation to solve for x.

\begin{align*}\mathrm{K_a=4.5 \times 10^{-4}=\dfrac{(x)(x)}{2.00-x}=\dfrac{x^2}{2.00-x}}\end{align*}

The quadratic equation is required to exactly solve this equation for x. However, a simplification can be made because of the fact that relatively little of the weak acid will be ionized at equilibrium. Since the reactants are heavily favored, the value of x will be much less than 2.00, so we can say that 2.00 - x is approximately equal to 2.00.

\begin{align*}\mathrm{4.5 \times 10^{-4}=\dfrac{x^2}{2.00-x} \approx \dfrac{x^2}{2.00}}\end{align*}
\begin{align*}\mathrm{x=\sqrt{4.5 \times 10^{-4}(2.00)}=2.9 \times 10^{-2}\:M=[H^+]}\end{align*}

Since the variable x represents the hydrogen-ion concentration, the pH of the solution can now be calculated.

pH = -log[H+] = -log[2.9 × 10-2] = 1.54

The pH of a 2.00 M solution of a strong acid would be equal to –log(2.00) = −0.30. The higher pH of the 2.00 M nitrous acid is consistent with it being a weaker acid.

Practice Problem
1. Calculate the pH of a 0.20 M solution of hydrocyanic acid.

The procedure for calculating the pH of a solution of a weak base is similar to that of the weak acid in Sample Problem 21.5. However, the variable x will represent the concentration of the hydroxide ion. The pH is found by taking the negative logarithm to get the pOH, followed by subtracting from 14 to get the pH.

## Lesson Summary

• Strong acids and bases are those that ionize completely in solution. Weak acids and bases ionize only to a very small extent in solution. A stronger acid has a correspondingly weaker conjugate base, and vice-versa.
• The acid ionization constant (Ka) and base ionization constant (Kb) are numerical measures of the strength of acids and bases. The larger the ionization constant, the stronger the acid or base.
• Solutions of known concentration and known pH can be used to find an unknown Ka or Kb.
• For acids or bases with known Ka or Kb, the pH of the solution can be determined as long as the concentration of the solution is also known.

## Lesson Review Questions

### Reviewing Concepts

1. Explain why a strong acid is not the same thing as a concentrated acid.
2. Acid HX is a strong acid, while HY is a weak acid.
1. Compare the relative amounts of HX, H+, and X that are present in solution at equilibrium.
2. Compare the relative amounts of HY, H+, and Y that are present in solution at equilibrium.
3. Write equations for the ionizations of the following acid and base in water.
1. bromous acid, HBrO2
2. bromite ion, BrO2
4. Use the table above (Table above) to list the bases F, CN, and HCO3 in order from weakest to strongest.
5. Write the Ka expressions for the following weak acids.
1. sulfurous acid, H2SO3
2. lactic acid, HC3H5O3
6. Write Kb expressions for the following weak bases.
1. carbonate ion, CO32-
2. aniline, C6H5NH2

### Problems

1. A 1.25 M solution of a certain unknown acid has a pH of 4.73. Calculate the Ka of the acid.
2. A 0.350 M solution of a certain unknown base has a pH of 11.22. Calculate the Kb of the base.
3. Determine the pH of the following solutions.
1. 2.40 M benzoic acid
2. 0.745 M pyridine
4. What concentration of a solution of hydrofluoric acid should be prepared in order to have a pH of 2.00?

## Points to Consider

Acids and bases react with each other in a reaction called a neutralization.

• What are the products of a neutralization reaction?
• How can a neutralization reaction be used to determine the unknown concentration of an acid or base in an aqueous solution?

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Date Created:
Mar 29, 2013