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21.5: Salt Solutions

Difficulty Level: At Grade Created by: CK-12

Lesson Objectives

• Predict whether a certain salt solution is acidic, basic, or neutral.
• Write balanced equations for hydrolysis reactions.
• Calculate the pH of a salt solution when the Ka or the Kb of the hydrolyzing ion is known.
• Use equations to show how buffers resist changes in pH by using up additional hydrogen or hydroxide ions.

Lesson Vocabulary

• buffer
• buffer capacity
• salt hydrolysis

Recalling Prior Knowledge

• How is the strength of an acid or a base related to its Ka or Kb?
• How is the pH of a weak acid or a weak base calculated?

Many salts are soluble and dissociate completely. The ions produced from the dissociation of a salt are sometimes capable of undergoing reactions with water that change the pH of the solution. In this lesson, you will learn about hydrolysis reactions and about an important class of solutions called buffers.

Hydrolysis of Salts

A salt is an ionic compound that is formed when an acid and a base neutralize each other. While it may seem that salt solutions would always be neutral, they can frequently be either acidic or basic.

Consider the salt formed when the weak acid hydrofluoric acid is neutralized by the strong base sodium hydroxide. The molecular and net ionic equations are shown below.

$\mathrm{HF}(aq) + \mathrm{NaOH}(aq) \rightarrow \mathrm{NaF}(aq)+ \mathrm{H_2O}(l)$
$\mathrm{HF}(aq) + \mathrm{OH^-}(aq) \rightarrow \mathrm{F^-}(aq) + \mathrm{H_2O}(l)$

Since sodium fluoride is soluble, the sodium ion is a spectator ion in the neutralization reaction. The fluoride ion is capable of reacting to a small extent with water, accepting a proton.

$\mathrm{F^-}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{HF}(aq) + \mathrm{OH^-}(aq)$

The fluoride ion is acting as a weak Brønsted-Lowry base. The hydroxide ion that is produced as a result of the above reaction makes the solution slightly basic. Salt hydrolysis is a reaction in which one of the ions from a salt reacts with water, forming either an acidic or basic solution.

Salts That Form Basic Solutions

When solid sodium fluoride is dissolved into water, it completely dissociates into sodium ions and fluoride ions. The sodium ions do not react with water, but a minor portion of the fluoride ions will deprotonate water, producing small amounts of hydrofluoric acid and hydroxide ions.

$\mathrm{F^-}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{HF}(aq) + \mathrm{OH^-}(aq)$

The pH of the resulting solution can be determined if the Kb of the fluoride ion is known.

Sample Problem 21.7: Salt Hydrolysis

20.0 g of sodium fluoride is dissolved in enough water to make 500.0 mL of solution. Calculate the pH of the solution. The Kb of the fluoride ion is 1.4 × 10−11.

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

Known

• mass of NaF = 20.0 g
• molar mass of NaF = 41.99 g/mol
• volume of solution = 0.5000 L
• Kb of F = 1.4 × 10−11

Unknown

• pH of solution = ?

The molarity of the F solution can be calculated from the mass, molar mass, and solution volume. Since NaF completely dissociates, the molarity of the NaF is equal to the molarity of the F ion. An ICE table can be used to calculate the concentration of OH produced, and the pH of the solution can be determined from that value.

Step 2: Solve.

$\mathrm{20.0\: \cancel{\mathrm{g\:NaF}} \times \dfrac{1\: \cancel{\mathrm{mol\:NaF}}}{41.99\: \cancel{\mathrm{g\:NaF}}} \times \dfrac{1\:mol\:F^-}{1\: \cancel{\mathrm{mol\:NaF}}}=0.476\:mol\:F^-}$
$\mathrm{\dfrac{0.476\:mol\:F^-}{0.5000\:L}=0.953\:M\:F^-}$
Hydrolysis equation: $\mathrm{F^-}(aq) + \mathrm{H_2O}(l) \mathrm{HF}(aq) + \mathrm{OH^-}(aq)$
Concentrations [F] [HF] [OH]
Initial 0.953 0 0
Change −x +x +x
Equilibrium 0.953 − x x x
$\mathrm{K_b=1.4 \times 10^{-11}=\dfrac{(x)(x)}{0.953-x}=\dfrac{x^2}{0.953-x} \approx \dfrac{x^2}{0.953}}$
$\mathrm{x=[OH^-]=\sqrt{1.4 \times 10^{-11}(0.953)}=3.65 \times 10^{-6}\:M}$
pOH = -log(3.65 × 10-6) = 5.44
pH = 14 - 5.44 = 8.56

The solution is slightly basic due to the hydrolysis of the fluoride ion.

Practice Problem
1. The Kb of the acetate ion (CH3COO) is 5.6 × 10−10. Calculate the pH of a 0.200 M solution of sodium acetate.

Salts that are derived from the neutralization of a weak acid (HF) by a strong base (NaOH) will always produce salt solutions that are basic. Now we will see what happens with a salt that is formed from the reaction of a strong acid with a weak base.

Salts That Form Acidic Solutions

Ammonium chloride (NH4Cl) is a salt that is formed when the strong acid HCl is neutralized by the weak base NH3. Ammonium chloride is soluble in water. The chloride ion cannot undergo an acid-base reaction with water because it is the conjugate base of the strong acid HCl. In other words, the Cl ion cannot accept a proton from water to form HCl and OH, as the fluoride ion did in the previous section. However, the ammonium ion is capable of reacting slightly with water by acting as a proton donor.

$\mathrm{NH^+_4}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{NH_3}(aq)$

The production of hydronium ions causes the resulting solution to be acidic. The pH of a solution of ammonium chloride can be found in a very similar way to the pH of the sodium fluoride solution in Sample Problem 21.7. However, since ammonium chloride is acting as an acid, it is necessary to know the Ka of NH4+, which is 5.6 × 10−10. As an example, we will find the pH of a 2.00 M solution of NH4Cl. Because the NH4Cl completely dissociates in water, the concentration of the ammonium ion is also 2.00 M.

$\mathrm{NH_4Cl}(s) \rightarrow \mathrm{NH^+_4}(aq) + \mathrm{Cl^-}(aq)$

Again, an ICE table is set up in order to solve for the concentration of the hydronium (or H+) ion produced (Table below).

Concentrations [NH4+] [H+] [NH3]
Initial 2.00 0 0
Change −x +x +x
Equilibrium 2.00 − x x x

Substituting into the Ka expression gives the following:

$\mathrm{K_a=5.6 \times 10^{-10}=\dfrac{x^2}{2.00-x} \approx \dfrac{x^2}{2.00}}$
$\mathrm{x=[H^+]=\sqrt{5.6 \times 10^{-10}(2.00)}=3.3 \times 10^{-5}\:M}$
pH = -log(3.3 × 10-5) = 4.48

A salt produced from a strong acid and a weak base yields a solution that is acidic.

Salts That Form Neutral Solutions

A salt that is derived from the reaction of a strong acid with a strong base forms a solution that has a pH of 7. An example is sodium chloride, formed from the neutralization of HCl by NaOH. A solution of NaCl in water has no acidic or basic properties, since neither ion is capable of reacting with water to produce any appreciable amounts of H3O+ or OH-. Other salts that form neutral solutions include potassium nitrate (KNO3) and lithium bromide (LiBr). The table below (Table below) summarizes how to determine the acidity or basicity of a salt solution.

Salt formed from: Salt Solution
Strong acid + Strong base Neutral
Strong acid + Weak base Acidic
Weak acid + Strong base Basic

Salts, formed from the reaction of a weak acid and a weak base, are more difficult to analyze because of competing hydrolysis reactions between the cation and the anion. These salts will not be considered in this text.

Buffers

If only 1.0 mL of 0.10 M hydrochloric acid is added to 1.0 L of pure water, the pH drops drastically from 7.0 to 4.0. This is a 1000-fold increase in the acidity of the solution. For many purposes, it is desirable to have a solution that is capable of resisting large changes in pH when relatively small amounts of acid or base are added to them. Such a solution is called a buffer. A buffer is a solution that contains significant amounts of both an acid and its conjugate base. Both components must be present for the system to resist changes in pH. Commercial buffer solutions are available with a wide variety of pH values.

One example of a buffer is a solution made of acetic acid (a weak acid) and sodium acetate (a salt containing its conjugate base). The pH of a buffer consisting of 0.50 M CH3COOH and 0.50 M CH3COONa is 4.74. If 10.0 mL of 1.0 M HCl is added to 1.0 L of the buffer, the pH only decreases to 4.73. This ability to “soak up” the additional hydrogen ions from the HCl that was added is due to the reaction below.

$\mathrm{CH_3COO^-}(aq) + \mathrm{H^+}(aq) \rightarrow \mathrm{CH_3COOH}(aq)$

Since both the acetate ion and the acetic acid were already present in the buffer, the only thing that changes is the ratio of one to the other. Small changes in that ratio have only very minor effects on the pH.

If 10.0 mL of 1.0 M NaOH were added to another 1.0 L of the same buffer, the pH would only increase to 4.76. In this case the buffer takes up the additional hydroxide ions.

$\mathrm{CH_3COOH}(aq) + \mathrm{OH^-}(aq) \rightarrow \mathrm{CH_3COO^-}(aq) + \mathrm{H_2O}(l)$

Again, the ratio of acetate ion to acetic acid changes slightly, this time causing a very small increase in the pH.

It is possible to add so much acid or base to a buffer that its ability to resist a significant change in pH is overwhelmed. The buffer capacity is the amount of acid or base that can be added to a buffered solution before a large change in pH occurs. The buffer capacity is exceeded when the number of moles of H+ or OH that are added to the buffer exceeds the number of moles of the buffer components.

Some common buffer systems are listed below (Table below). Buffers are especially critical in biological systems. For example, the pH of blood must be maintained within a fairly narrow range of about 7.35 to 7.45. There are several buffers at work in blood, but the most important one involves carbonic acid (H2CO3) and the hydrogen carbonate ion (HCO3). The ratio between the two is carefully maintained through metabolic processes so that the pH of blood is nearly constant.

Some Common Buffers
Buffer system Buffer components pH of buffer (equal molarities of both components)
Acetic acid / acetate ion CH3COOH / CH3COO 4.74
Carbonic acid / hydrogen carbonate ion H2CO3 / HCO3 6.38
Dihydrogen phosphate ion / hydrogen phosphate ion H2PO4 / HPO42− 7.21
Ammonia / ammonium ion NH3 / NH4+ 9.25

Lesson Summary

• Salt solutions can be acidic, basic, or neutral. Certain ions can undergo acid-base reactions with water, causing a change in the pH of the resulting solution.
• Salts containing anions that are conjugate bases of weak acids can accept hydrogen ions from water to form their conjugate acid and the hydroxide ion. These solutions are slightly basic, and the pH can be calculated if the Kb of the anion is known.
• Salts containing cations that are conjugate acids of weak bases can donate hydrogen ions to water to form their conjugate base and the hydronium ion. These solutions are slightly acidic, and the pH can be calculated if the Ka of the cation is known.
• Salts formed by reacting a strong acid with a strong base do not have sufficiently acidic or basic components, so a solution containing this type of salt will have a pH of 7.
• A buffer is a solution of a weak acid or base together with one of its salts. Buffers resist changes in pH when additional acid or base is added.

Lesson Review Questions

Reviewing Concepts

1. What type of salt produces an acidic solution? A basic solution?
2. Predict whether an aqueous solution of the following salts will be acidic, basic, or neutral.
1. K2CO3
2. KCl
3. NH4NO3
4. CsI
5. NH4Br
6. NaNO2
3. Explain why an anion derived from the neutralization of a weak acid produces a basic solution when dissolved in water.
4. Explain why a solution containing both HCN and NaCN is a buffer, but a solution containing both HBr and NaBr is not a buffer.

Problems

1. Write balanced net ionic equations showing the hydrolysis of the following salts.
1. NaNO2
2. NH4NO3
3. K2SO3
2. The Kb of the carbonate ion (CO32−) is 2.1 × 10−4. Calculate the pH of a 0.400 M solution of sodium carbonate (Na2CO3).
3. The Ka of the methyl ammonium ion (CH3NH3+) is 2.3 × 10−11. Calculate the pH of a 1.50 M solution of methyl ammonium chloride (CH3NH3Cl).
4. The pH of a 1.00 M solution of a certain salt is measured to be 9.81. Calculate the Kb of the anion of the salt.
5. Use the buffer between carbonic acid and the hydrogen carbonate ion (H2CO3 / HCO3) to show how a buffer system works. Show with two separate equations how this buffer would respond to small additions of acid or base.
6. Many household bleaches are a 6.0% by mass solution of sodium hypochlorite (NaClO). The Kb of ClO is 2.9 × 10−7. Assuming that the density of the bleach is 1.0 g/mL, calculate the pH of household bleach.

Points to Consider

Acid-base reactions involve the transfer of a hydrogen ion from one substance to another. An oxidation-reduction reaction is another common type of reaction that involves the transfer of one or more electrons from one substance to another.

• How can an oxidation-reduction reaction be recognized?
• What is corrosion, and how does it relate to a transfer of electrons?

Date Created:

Mar 29, 2013

Feb 16, 2015
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