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# 21.4: Acid-Base Neutralization

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Write balanced molecular and net ionic equations for acid-base neutralization reactions.
• Know the steps of an acid-base titration experiment.
• Calculate the concentration of an acid or a base from data obtained in a titration.
• Describe the appearance of a titration curve.
• Explain the chemistry of acid-base indicators.

## Lesson Vocabulary

• end point
• equivalence point
• indicator
• neutralization reaction
• salt
• standard solution
• titration
• titration curve

### Recalling Prior Knowledge

• What is a double-replacement reaction?
• How are net ionic equations written?

Your stomach contains hydrochloric acid, which aids in the digestion of food. Occasionally, one can suffer from an excess of stomach acid that leads to nausea or another condition called acid reflux. A simple way to combat excess stomach acid is to ingest an antacid. Antacids (Figure below) contain a base like calcium hydroxide that neutralizes the stomach acid. In this lesson, you will learn about acid-base neutralization reactions.

Over-the-counter antacids are bases that are used to neutralize excess acid in the stomach.

## Neutralization Reactions

While neutralization reactions are all alike in many ways, it is useful to look at those involving only strong acids and bases separately from those involving a weak acid, a weak base, or both.

### Strong Acid-Strong Base Reactions

When equal amounts of a strong acid such as hydrochloric acid are mixed with a strong base such as sodium hydroxide, the result is a neutral solution. In other words, the products of the reaction do not have the characteristics of either an acid or a base. Here is the balanced molecular equation.

HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l)\begin{align*}\mathrm{HCl}(aq) + \mathrm{NaOH}(aq) \rightarrow \mathrm{NaCl}(aq) + \mathrm{H_2O}(l)\end{align*}

A neutralization reaction is a reaction in which an acid and a base react in an aqueous solution to produce a salt and water. Note that the neutralization reaction above is an example of a double-replacement reaction. The aqueous sodium chloride that is produced in the reaction is called a salt. A salt is an ionic compound composed of a cation from a base and an anion from an acid. However, because essentially any anion or cation can be paired with hydrogen or hydroxide ions to make an acid or base, the term salt can be used to describe any ionic compound, and the two terms are often used interchangeably.

Recall that chemical reactions occurring in aqueous solution can often be written as a net ionic equation. The full ionic equation for the neutralization of hydrochloric acid with sodium hydroxide is written as follows:

H+(aq)+Cl(aq)+Na+(aq)+OH(aq)Na+(aq)+Cl(aq)+H2O(l)\begin{align*}\mathrm{H^+}(aq) + \mathrm{Cl^-}(aq) + \mathrm{Na^+}(aq) + \mathrm{OH^-}(aq) \rightarrow \mathrm{Na^+}(aq) + \mathrm{Cl^-}(aq) + \mathrm{H_2O}(l)\end{align*}

Since the acid and base are both strong, they are fully ionized, so they can be written as ions. Because the NaCl product is also water-soluble, it stays dissociated. The sodium and chloride ions are spectator ions, leaving the following net ionic reaction:

H+(aq)+OH(aq)H2O(l)\begin{align*}\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \rightarrow \mathrm{H_2O}(l)\end{align*}

All neutralization reactions between a strong acid and a strong base simplify to this net ionic equation.

What if the acid is a diprotic acid, such as sulfuric acid? The balanced molecular equation now involves a 1:2 ratio between acid and base.

H2SO4(aq)+2NaOH(aq)Na2SO4(aq)+2H2O(l)\begin{align*}\mathrm{H_2SO_4}(aq) + \mathrm{2NaOH}(aq) \rightarrow \mathrm{Na_2SO_4}(aq) + \mathrm{2H_2O}(l)\end{align*}

In order for the reaction to be a full neutralization, twice as many moles of NaOH must react with the H2SO4. The sodium sulfate salt is soluble, and so the net ionic reaction is again the same. Different mole ratios can occur when using polyprotic acids or bases that include multiple hydroxide ions per formula unit, such as Ca(OH)2.

### Reactions Involving a Weak Acid or Weak Base

Reactions where at least one of the components is weak do not generally result in a neutral solution. The reaction between weak nitrous acid and strong potassium hydroxide is shown below.

HNO2(aq)+KOH(aq)KNO2(aq)+H2O(l)\begin{align*}\mathrm{HNO_2}(aq) + \mathrm{KOH}(aq) \rightarrow \mathrm{KNO_2}(aq) + \mathrm{H_2O}(l)\end{align*}

In order to write the net ionic equation, the weak acid must be written as a molecule, since it does not ionize to a great extent in water. The base and the salt are fully dissociated.

HNO2(aq)+K+(aq)+OH(aq)K+(aq)+NO2(aq)+H2O(l)\begin{align*}\mathrm{HNO_2}(aq) + \mathrm{K^+}(aq) + \mathrm{OH^-}(aq) \rightarrow \mathrm{K^+}(aq) + \mathrm{NO^-_2}(aq) + \mathrm{H_2O}(l)\end{align*}

The only spectator ion is the potassium ion, resulting in the net ionic equation:

HNO2(aq)+OH(aq)NO2(aq)+H2O(l)\begin{align*}\mathrm{HNO_2}(aq) + \mathrm{OH^-}(aq) \rightarrow \mathrm{NO^-_2}(aq) + \mathrm{H_2O}(l)\end{align*}

The strong hydroxide ion essentially “forces” the weak nitrous acid to become ionized. The hydrogen ion from the acid combines with the hydroxide ion to form water, leaving the nitrite ion as the other product. The resulting solution is not neutral (pH = 7) but, instead, is slightly basic. We will investigate why this is the case in the next lesson.

Reactions can also involve a weak base and strong acid, resulting in a solution that is slightly acidic. The molecular and net ionic equations for the reaction of hydrochloric acid and ammonia are shown below.

HCl(aq)+NH3(aq)NH4Cl(aq)\begin{align*}\mathrm{HCl}(aq) + \mathrm{NH_3}(aq) \rightarrow \mathrm{NH_4Cl}(aq)\end{align*}
H+(aq)+NH3(aq)NH+4(aq)\begin{align*}\mathrm{H^+}(aq) + \mathrm{NH_3}(aq) \rightarrow \mathrm{NH^+_4}(aq)\end{align*} (Cl is spectator ion)

Reactions between acids and bases that are both weak may result in solutions that are neutral, acidic, or basic.

## Titrations

When hydrochloric acid is reacted with sodium hydroxide, an acid/base mole ratio of 1:1 is required for full neutralization.

HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l)\begin{align*}\mathrm{HCl}(aq) + \mathrm{NaOH}(aq) \rightarrow \mathrm{NaCl}(aq) + \mathrm{H_2O}(l)\end{align*}

If instead the hydrochloric acid were reacted with barium hydroxide, the mole ratio would be 2:1.

2HCl(aq)+Ba(OH)2(aq)BaCl2(aq)+2H2O(l)\begin{align*}\mathrm{2HCl}(aq) + \mathrm{Ba(OH)_2}(aq) \rightarrow \mathrm{BaCl_2}(aq) + \mathrm{2H_2O}(l)\end{align*}

Two moles of HCl are required to completely neutralize one mole of Ba(OH)2. In a balanced neutralization equation, the moles of H+ ions supplied by the acid will be equal to the moles of OH ions supplied by the base. The equivalence point is the point in a neutralization reaction where the number of moles of hydrogen ions is equal to the number of moles of hydroxide ions.

In the laboratory, you may need to determine the concentration of an acid or a base whose concentration is not known. This can be accomplished by performing a controlled neutralization reaction. A titration is an experiment in which a solution, whose concentration is known, is gradually added to a measured volume of another solution in order to determine its concentration. Many titrations are acid-base neutralization reactions, although other types of titrations can also be performed.

In order to perform an acid-base titration, the chemist must have a way to visually detect that the neutralization reaction has reached the equivalence point. An indicator is a substance that has a distinctly different color when in an acidic or basic solution. A commonly used indicator for strong acid-strong base titrations is phenolphthalein. Solutions in which a few drops of phenolphthalein have been added, turn from colorless to brilliant pink as the solution turns from acidic to basic (Figure below). The steps in a titration reaction are outlined below.

1. A measured volume of an acidic solution whose concentration is unknown is added to an Erlenmeyer flask.
2. Several drops of an indicator are added to the acid and mixed by swirling the flask.
3. A buret is filled with a basic solution of known molarity.
4. The stopcock of the buret is opened and base is slowly added to the acid while the flask is constantly swirled to ensure mixing. The stopcock is closed at the exact point at which the indicator just changes color. If the color change does not remain after swirling, the neutralization reaction has not yet reached completion.

In the titration of an acid by a base, the addition of the base causes the indicator phenolphthalein to change from colorless to pink.

The standard solution is the solution in a titration whose concentration is known. In the titration described above, the base solution is the standard solution. It is very important in a titration to add the solution from the buret slowly so that the point at which the indicator changes color can be found accurately. The end point of a titration is the point at which the indicator changes color. When phenolphthalein is the indicator, the end point will be signified by a faint pink color.

### Titration Calculations

At the equivalence point in a neutralization, the moles of acid are equal to the moles of base, assuming they react in a 1:1 ratio according to the balanced neutralization equation.

moles acid = moles base

Recall that the molarity (M) of a solution is defined as the moles of the solute divided by the liters of solution (L). The moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters.

moles solute = M × L

We can then set the moles of acid equal to the moles of base.

MA × VA = MB × VB

MA is the molarity of the acid, while MB is the molarity of the base. VA and VB are the volumes of the acid and base, respectively.

Suppose that a titration is performed and 20.70 mL of 0.500 M NaOH is required to reach the end point when titrated against 15.00 mL of HCl of unknown concentration. The above equation can be used to solve for the molarity of the acid.

MA=MB×VBVA=0.500M×20.70mL15.00mL=0.690M\begin{align*}\mathrm{M_A=\dfrac{M_B \times V_B}{V_A}=\dfrac{0.500\:M \times 20.70\:mL}{15.00\:mL}=0.690\:M}\end{align*}

The higher molarity of the acid compared to the base in this case means that a smaller volume of the acid is required to reach the equivalence point.

The above equation works only for neutralizations in which there is a 1:1 ratio between the acid and the base. Sample Problem 21.6 demonstrates a titration problem for which this is not the case.

Sample Problem 21.6: Titration

In a titration of sulfuric acid with sodium hydroxide, 32.20 mL of 0.250 M NaOH is required to neutralize 26.60 mL of the H2SO4 solution. Calculate the molarity of the sulfuric acid.

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

Known

• molarity of the NaOH solution = 0.250 M
• volume of the NaOH solution = 32.20 mL
• volume of the H2SO4 solution = 26.60 mL

Unknown

• molarity of the H2SO4 solution = ?
Equation: H2SO4(aq)+2NaOH(aq)Na2SO4(aq)+2H2O(l)\begin{align*}\mathrm{H_2SO_4}(aq) + \mathrm{2NaOH}(aq) \rightarrow \mathrm{Na_2SO_4}(aq) + \mathrm{2H_2O}(l)\end{align*}

First, determine the moles of NaOH that were consumed in the reaction. From the mole ratio, calculate the moles of H2SO4 that reacted. Finally, divide the moles of H2SO4 by the volume of the sample of acid to get the molarity.

Step 2: Solve.

molNaOH=M×L=0.250M×0.03220L=8.05×103molNaOH\begin{align*}\mathrm{mol\:NaOH=M \times L=0.250\:M \times 0.03220\:L=8.05 \times 10^{-3}\:mol\:NaOH}\end{align*}
8.05×103molNaOH×1molH2SO42molNaOH=4.03×103molH2SO4\begin{align*}\mathrm{8.05 \times 10^{-3}\:mol\:NaOH \times \dfrac{1\:mol\:H_2SO_4}{2\:mol\:NaOH}=4.03 \times 10^{-3}\:mol\:H_2SO_4}\end{align*}
4.03×103molH2SO40.02660L=0.151MH2SO4\begin{align*}\mathrm{\dfrac{4.03 \times 10^{-3}\:mol\:H_2SO_4}{0.02660\:L}=0.151\:M\:H_2SO_4}\end{align*}

The volume of H2SO4 required is smaller than the volume of NaOH because of the two hydrogen ions contributed by each molecule.

Practice Problem
1. What is the concentration of a Ba(OH)2 solution, if 17.25 mL is required to neutralize 19.10 mL of 0.520 M HBr?

### Titration Curves

As base is added to acid at the beginning of a titration, the pH rises very slowly. Nearer to the equivalence point, the pH begins to rapidly increase. If the titration is a strong acid with a strong base, the pH at the equivalence point is equal to 7. A bit past the equivalence point, the rate of change of the pH again slows down. A titration curve is a graphical representation of the pH of a solution during a titration. The figure below (Figure below) shows two different examples of a strong acid-strong base titration curve. On the left is a titration in which the base is added to the acid, so the pH progresses from low to high. On the right is a titration in which the acid is added to the base. In this case, the pH starts out high and decreases during the titration. In both cases, the equivalence point is reached when the moles of acid and base are equal, and the pH is 7. This also corresponds to the color change of the indicator.

A titration curve shows the pH changes that occur during the titration of an acid with a base. On the left, a strong base is being added to a strong acid. On the right, a strong acid is being added to a strong base. In both cases, the pH has a value of 7 at the equivalence point.

Titration curves can also be generated for weak acid-strong base or strong base-weak acid titrations. The general shapes are similar, but the graph has additional inflection points, and the pH at the equivalence point is not exactly 7. In a weak acid-strong base titration, the pH is greater than 7 at the equivalence point, and it is less than 7 at the equivalence point for a strong acid-weak base titration.

### Indicators

As noted earlier, an acid-base indicator is a substance that displays different colors when in the presence of an acid or a base. How does that work? An indicator is a weak acid that ionizes within a known pH range, usually about 2 pH units. We can represent the protonated form of the indicator molecule as HInd and the deprotonated form as Ind. The following equilibrium exists for the indicator.

\begin{align*}\underset{(\text{colorless})}{\text{HInd}(aq)} \mathrm{\rightleftharpoons H^+}(aq) \mathrm{+\underset{(pink)}{Ind^-}}(aq)\end{align*}

According to Le Châtelier’s principle, the addition of H+ ions (as in a low pH solution) drives the equilibrium to the left and the protonated HInd predominates. The addition of OH (as in a high pH solution) decreases the H+ concentration and drives the equilibrium to the right, so the deprotonated form Ind- predominates. To be useful as an indicator, the two forms must be different colors. In the case of phenolphthalein, the protonated form is colorless, while the deprotonated form is pink.

The figure below (Figure below) shows a variety of acid-base indicators that can be used in titration experiments.

Indicators are substances that change color at a certain pH. A wide variety of indicators is available.

Depending on the pH at the equivalence point, the appropriate indicator must be chosen. For example, bromophenol blue has a yellow color below a pH of about 3 and a blue-violet color above a pH of about 4. Bromophenol blue would not be a good choice as the indicator for a strong acid-strong base titration, because the pH is 7 at the equivalence point. However, it could be used for a strong acid-weak base titration, where the pH at the equivalence point is lower.

Most indicators, as shown above (Figure above), have two colored forms. A universal indicator makes use of a mixture of indicators so that an entire rainbow of colors is displayed when going from low pH to high pH (Figure below). A universal indicator is used to make pH paper, which can be used to quickly test solutions for their approximate pH.

A universal indicator solution gradually changes color from red in acidic conditions to violet in basic conditions.

Some naturally occurring substances act as acid-base indicators. Red cabbage juice, for example, can be used as a natural indicator (Figure below).

The juice of the red cabbage is red in acidic solution, purple in neutral solutions, and green to yellow in basic solutions.

## Lesson Summary

• Acids and bases neutralize each other, forming a salt and water. A strong acid-strong base neutralization results in a neutral solution with a pH of 7.
• A titration is an experiment in which a controlled acid-base neutralization reaction is used to determine the unknown concentration of an acid or a base. The equivalence point is reached when the number of hydrogen ions is equal to the number of hydroxide ions.
• A titration curve shows the pH of a solution during the course of a titration.
• Acid-base indicators are used in a titration in order to detect the end point of the titration.

## Lesson Review Questions

### Reviewing Concepts

1. What are the products of a reaction between an acid and a base?
2. How are weak acids and strong acids represented differently in the net ionic equation of a neutralization reaction?
3. What is the pH or approximate pH at the equivalence point for each titration listed below?
1. strong acid-weak base
2. weak acid-strong base
3. strong acid-strong base
4. A certain titration has its end point at pH = 5. Choose an appropriate indicator for this titration.
5. For the indicator phenol red, what is the color of HInd? What is the color of Ind?

### Problems

1. Finish and balance the following molecular equations.
1. \begin{align*}\mathrm{HBr}(aq) + \mathrm{Al(OH)_3}(aq) \rightarrow\end{align*}
2. \begin{align*}\mathrm{H_2SO_4}(aq) + \mathrm{LiOH}(aq) \rightarrow\end{align*}
3. \begin{align*}\mathrm{H_3PO_4}(aq) + \mathrm{Sr(OH)_2}(aq) \rightarrow\end{align*}
2. Write balanced net ionic equations for the neutralizations below. Assume aqueous solutions unless indicated otherwise.
1. Hydrochloric acid + potassium hydroxide
2. Hydrocyanic acid (weak acid) + potassium hydroxide
3. Nitric acid (strong acid) + solid sodium hydroxide
4. Carbonic acid (weak acid) + ammonia (weak base)
3. What is the molarity of a sodium hydroxide solution if 20.00 mL of it is fully neutralized by each of the following solutions?
1. 10.00 mL of 0.500 M HCl
2. 10.00 mL of 0.500 M H2SO4
3. 10.00 mL of 0.500 M H3PO4
4. What is the molarity of a solution of phosphoric acid (H3PO4) if 34.27 mL of it is neutralized in a titration by 28.85 mL of 0.350 M barium hydroxide (Ba(OH)2)?
5. 15.52 g of sodium hydroxide is dissolved into enough water to make 1.00 L of solution. Then, 25.00 mL of this solution is titrated against a solution of CH3COOH. If the endpoint is reached after adding 18.05 mL of the CH3COOH solution, what is the molarity of the acid?

## Points to Consider

Salt solution can be acidic, basic, or neutral.

• How can the acidity or basicity of a salt solution be determined?
• What is a hydrolysis reaction, and how does it affect the pH?

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Date Created:
Mar 29, 2013