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92 Hydrolysis of Salts

Learning Objectives

By the end of this section, you will be able to:

  • Predict whether a salt solution will be acidic, basic, or neutral
  • Calculate the concentrations of the various species in a salt solution
  • Describe the acid ionization of hydrated metal ions

 Neutralization Reactions

The reaction between an acid and a base is called an acid-base neutralization reaction. For example, if equivalent quantities of sodium hydroxide (NaOH, a strong base) and hydrogen chloride (HCl, a strong acid) react, the acid and the base are both neutralized:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Note that the major products of the neutralization reaction between an acid and a base will be water and a salt (an ionic compound composed of cations and anions).

Aqueous salt solutions may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt’s constituent ions. In the case when equimolar quantities of a strong acid and a strong base react, the resulting solution is neutral (i.e., sodium chloride is a “neutral salt”).  However, some salts are capable of undergoing further reaction with water to form acidic or basic aqueous solutions:  the cations, anions, or both can undergo acid-base reactions with water. When ions interact with water to form acidic or basic solutions, they are said to “hydrolyze” in water, or to undergo “hydrolysis”.

If the resulting solution is acidic, the salt is said to be an “acidic salt”.  If the resulting solution is basic, the salt is said to be a “basic salt”.  The ultimate pH of the solution can be determined based on the Ka and/or Kb values of the cations and anions in the salt.

Salts with Acidic Ions

A classic example of an acidic salt is ammonium chloride.  Ammonium chloride is a salt that could be formed by the reaction between the weak base ammonia (NH3) and the strong acid hydrochloric acid (HCl):

NH3(aq) + HCl(aq) → NH4Cl(aq)

Ammonium chloride full fully dissociate in water, as described by the equation

{\text{NH}}_{4}\text{Cl}\left(s\right)\rightarrow{\text{NH}}_{4}{}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)

The ammonium ion (NH4+) is the conjugate acid of the weak base ammonia (NH3) and is, itself, a weak acid.  (Recall that the conjugate acids of weak bases are weak acids.)  Thus the ammonium ion will, to a small extent, dissociate to transfer protons to water:  it is said to undergo acid ionization (or acid hydrolysis):

{\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{NH}}_{3}\left(aq\right)

Since ammonia is a weak base, Kb is measurable and the ammonium ion is a weak acid (with a Ka > 0).

The chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis) reaction is represented by

{\text{Cl}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons\text{HCl}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{b}}={K}_{\text{w}}\text{/}{K}_{\text{a}}

Since HCl is a strong acid, Ka is immeasurably large.  Thus, chloride is an extremely weak base, with a Kb ≈ 0. Chloride ions do not undergo significant reaction with water and do not affect the pH of the aqueous solution.  (The chloride ion is “pH neutral”).

Thus, dissolving ammonium chloride in water yields a solution of weak acid cations (NH4+) and inert anions (Cl−), resulting in an acidic solution.  Ammonium chloride is said to be an “acidic salt”.

Calculating the pH of an Acidic Salt Solution

Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, \left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]\text{Cl}, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of anilinium chloride

{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\left(aq\right)

 

Solution

The Ka for anilinium ion is derived from the Kb for its conjugate base, aniline (see Appendix H):

{K}_{\text{a}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{{K}_{\text{w}}}{{K}_{\text{b}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{4.3\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-10}}\phantom{\rule{0.2em}{0ex}}=2.3\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-5}

Using the provided information, an ICE table for this system is prepared:

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium ( M ). The second column has the header of “C subscript 6 H subscript 5 N H subscript 3 superscript positive sign plus sign H subscript 2 O equilibrium sign C subscript 6 H subscript 5 N H subscript 2 plus sign H subscript 3 O superscript positive sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.233, negative x, 0.233 minus x. The second column is blank for all three rows. The third column has the following: 0, positive x, x. The fourth column has the following: approximately 0, positive x, x.

Substituting these equilibrium concentration terms into the Ka expression gives

\begin{array}{l}{\text{K}}_{a}=\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]\text{/}\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{+}\right]\\ 2.3\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{\text{−}5}=\left(x\right)\left(x\right)\text{/}0.233-x\right)\end{array}

Assuming x << 0.233, the equation is simplified and solved for x:

\begin{array}{}\\ \\ 2.3\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{\text{−}5}={x}^{2}\text{/}0.233\\ x=0.0023\phantom{\rule{0.2em}{0ex}}M\end{array}

The ICE table defines x as the hydronium ion molarity, and so the pH is computed as

\text{pH}=-\text{log}\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\text{-}\text{log}\left(0.0023\right)=2.64

 

Check Your Learning

What is the hydronium ion concentration in a 0.100 M solution of ammonium nitrate, NH4NO3, a salt composed of the ions {\text{NH}}_{4}{}^{\text{+}} and {\text{NO}}_{3}{}^{\text{-}}. Which is the stronger acid {\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}} or {\text{NH}}_{4}{}^{\text{+}}?

Answer: [H3O+] = 7.5 x 10−6M; {\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}} is the stronger acid.

Salts with Basic Ions

As another example, consider dissolving sodium acetate in water:

NaCH3CO2(s) → Na+(aq) + CH3CO2–(aq)

The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion’s formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.

The acetate ion, CH3CO2– is the conjugate base of acetic acid, CH3CO2H, and so its base ionization (or base hydrolysis) reaction is represented by

{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+\text{OH}^{-}\left(aq\right)\phantom{\rule{1em}{0ex}}{K}_{\text{b}}={K}_{\text{w}}\text{/}{K}_{\text{a}}

Because acetic acid is a weak acid, its Ka is measurable and acetate ion is a weak base (Kb > 0).

Dissolving sodium acetate in water yields a solution of inert cations (Na+) and weak base anions (CH3CO2–) resulting in a basic solution.

Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base

Determine the acetic acid concentration in a solution with \left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]=0.050\phantom{\rule{0.2em}{0ex}}M and [OH−] = 2.5 x 10−6M at equilibrium. The reaction is:

{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)

 

Solution

The provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which

{K}_{\text{b}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{{K}_{\text{w}}}{{K}_{\text{a}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{1.8\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-5}}\phantom{\rule{0.2em}{0ex}}=5.6\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-10}

Substituting the available values into the Kb expression gives

{K}_{\text{b}}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{-}}\right]}\phantom{\rule{0.2em}{0ex}}=5.6\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-10}
=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left(2.5\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)}{\left(0.050\right)}\phantom{\rule{0.2em}{0ex}}=5.6\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-10}

Solving the above equation for the acetic acid molarity yields [CH3CO2H] = 1.1 x 10−5M.

Check Your Learning

What is the pH of a 0.083 M solution of NaCN?

Answer: 11.11

Salts with Acidic and Basic Ions

Some salts are composed of both acidic and basic ions, and so the pH of their solutions will depend on the relative strengths of these two species. Likewise, some salts contain a single ion that is amphiprotic, and so the relative strengths of this ion’s acid and base character will determine its effect on solution pH. For both types of salts, a comparison of the Ka and Kb values allows prediction of the solution’s acid-base status, as illustrated in the following example exercise.

Determining the Acidic or Basic Nature of Salts

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) KBr

(b) NaHCO3

(c) Na2HPO4

(d) NH4F

Solution

Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:

(a) The K+ cation is inert (a Group I cation) and will not affect pH — it is “pH neutral”. The bromide ion is the conjugate base of a strong acid, and so it is of negligible base strength (no appreciable base ionization). The solution is neutral.

(b) The Na+ cation is pH neutral (a Group I cation) and will not affect the pH of the solution; while the HCO3– anion is amphiprotic. The Ka of HCO3– is 4.7 x 10−11,and its Kb is \frac{1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{4.3\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-7}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}2.3\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-8}.

Since Kb >> Ka, the solution is basic.

(c) The Na+ cation is a Group I cation, and will not affect the pH of the solution. Thehe {\text{HPO}}_{4}{}^{\text{2−}} anion is amphiprotic. The Ka of {\text{HPO}}_{4}{}^{\text{2−}} is 4.2 x 10−13,

and its Kb is \frac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{6.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1.6\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-7}. Because Kb >> Ka, the solution is basic.

(d) The {\text{NH}}_{4}{}^{\text{+}} ion is acidic (see above discussion) and the F− ion is basic (conjugate base of the weak acid HF). Comparing the two ionization constants: Ka of {\text{NH}}_{4}{}^{\text{+}} is 5.6 x 10−10 and the Kb of F− is 1.4 x 10−11, so the solution is acidic, since Ka > Kb.

Check Your Learning

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) K2CO3

(b) CaCl2

(c) KH2PO4

(d) (NH4)2CO3

Answer: (a) basic; (b) neutral; (c) acidic; (d) basic

The Ionization of Hydrated Metal Ions

Unlike the group 1 and 2 metal ions of the preceding examples (Na+, Ca2+, etc.), some metal ions function as acids in aqueous solutions. These ions are not just loosely solvated by water molecules when dissolved, instead they are covalently bonded to a fixed number of water molecules to yield a complex ion (see chapter on coordination chemistry). As an example, the dissolution of aluminum nitrate in water is typically represented as

Al(NO3)(s) \rightleftharpoons Al3+(aq) + 3NO3–(aq)

However, the aluminum(III) ion actually reacts with six water molecules to form a stable complex ion, and so the more explicit representation of the dissolution process is

Al(NO3)(s)  + 6H2O(l) \rightleftharpoons Al(H2O)63+(aq) + 3NO3–(aq)

As shown in Figure 92-1, the Al(H2O)63+ ions involve bonds between a central Al atom and the O atoms of the six water molecules. Consequently, the bonded water molecules’ O–H bonds are more polar than in nonbonded water molecules, making the bonded molecules more prone to donation of a hydrogen ion:

\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left(\text{O}{\text{H}\right)}^{2+}\left(aq\right)\phantom{\rule{1em}{0ex}}
Ka = 1.4 x 10-5

The conjugate base produced by this process contains five other bonded water molecules capable of acting as acids, and so the sequential or step-wise transfer of protons is possible as depicted in few equations below:

\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{O}\text{H}\right)}^{2+}\left(aq\right)
\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{2+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{4}\left({\text{OH}\right)}_{2}{}^{\text{+}}\left(aq\right)
\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{4}\left({\text{OH}\right)}_{2}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{3}\left({\text{OH}\right)}_{3}\left(aq\right)
This is an example of a polyprotic acid, the topic of discussion in a later section of this chapter.

A reaction is shown using ball and stick models. On the left, inside brackets with a superscript of 3 plus outside to the right is structure labeled “[ A l ( H subscript 2 O ) subscript 6 ] superscript 3 plus.” Inside the brackets is s central grey atom to which 6 red atoms are bonded in an arrangement that distributes them evenly about the central grey atom. Each red atom has two smaller white atoms attached in a forked or bent arrangement. Outside the brackets to the right is a space-filling model that includes a red central sphere with two smaller white spheres attached in a bent arrangement. Beneath this structure is the label “H subscript 2 O.” A double sided arrow follows. Another set of brackets follows to the right of the arrows which have a superscript of two plus outside to the right. The structure inside the brackets is similar to that on the left, except a white atom is removed from the structure. The label below is also changed to “[ A l ( H subscript 2 O ) subscript 5 O H ] superscript 2 plus.” To the right of this structure and outside the brackets is a space filling model with a central red sphere to which 3 smaller white spheres are attached. This structure is labeled “H subscript 3 O superscript plus.”

FIGURE 92-1: When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.

 

Aside from the alkali metals (group 1) and some alkaline earth metals (group 2), most other metal ions will undergo acid ionization to some extent when dissolved in water. The acid strength of these complex ions typically increases with increasing charge and decreasing size of the metal ions. The first-step acid ionization equations for a few other acidic metal ions are shown below:

Fe(H2O)63+(aq) + H2O(l) \rightleftharpoons H3O+(aq) + Fe(H2O)5(OH)2+(aq)    pKa = 2.74
Cu(H2O)62+(aq) + H2O(l) \rightleftharpoons H3O+(aq) + Cu(H2O)5(OH)2+(aq)    pKa = 6.3
Zn(H2O)42+(aq) + H2O(l) \rightleftharpoons H3O+(aq) + Zn(H2O)3(OH)2+(aq)    pKa = 2.74

Hydrolysis of [Al(H2O)6]3+

Calculate the pH of a 0.10 M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion [Al(H2O)6]3+ in aqueous solution.

Solution

The equation for the reaction and Ka are:

Al(H2O)63+(aq) + H2O(l) \rightleftharpoons H3O+(aq) + Al(H2O)5(OH)2+(aq)    Ka = 1.4 x 10-5


 An ICE table with the provided information is

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of “A l ( H subscript 2 O ) subscript 6 superscript 3 positive sign plus H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus A l ( H subscript 2 O ) subscript 5 ( O H ) superscript 2 positive sign.” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus x. The second column has the following: approximately 0, positive x, x. The third column has the following: 0, positive x, x.


 Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:

{K}_{\text{a}}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{2+}\right]}{\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\right]}
=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.10-x}\phantom{\rule{0.2em}{0ex}}=1.4\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-5}


 Assuming x << 0.10 and solving the simplified equation gives:

x=1.2\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}M


 The ICE table defined x as equal to the hydronium ion concentration, and so the pH is calculated to be

\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=0+x=1.2\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}M
\text{pH}=\text{−log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\phantom{\rule{0.2em}{0ex}}=2.92\phantom{\rule{0.2em}{0ex}}\left(\text{an acidic solution}\right)

 

Check Your Learning

What is \left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{2+}\right] in a 0.15 M solution of Al(NO3)3 that contains enough of the strong acid HNO3 to bring [H3O+] to 0.10 M?

Answer: 2.1 x 10−5M

Key Concepts and Summary

The ions composing salts may possess acidic or basic character, ionizing when dissolved in water to yield acidic or basic solutions. Acidic cations are typically the conjugate partners of weak bases, and basic anions are the conjugate partners of weak acids. Many metal ions bond to water molecules when dissolved to yield complex ions that may function as acids.

 

Chemistry End of Chapter Exercises

Instructions

1. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) Al(NO3)3

(b) RbI

(c) KHCO2

(d) CH3NH3Br

 

2. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) FeCl3

(b) K2CO3

(c) NH4Br

(d) KClO4

Answer(s): (a) acidic; (b) basic; (c) acidic; (d) neutral

3. Determine whether the following 0.10 M aqueous solutions are acidic, basic or neutral. For each solution, indicate what the pH-determining reaction, the major species present, and the pH at 25 oC. Note that Ka and Kb  values for the weak acids and bases are available in the Appendices to this text.

(a) ammonium chloride, NH4Cl

(b) hydrogen chloride, HCl

(c) lithium nitrite, LiNO2

(d) sodium hydroxide, NaOH

(e) barium hydroxide, Ba(OH)2

4. Determine whether the following 0.10 M aqueous solutions are acidic, basic or neutral. For each solution, indicate what the pH-determining reaction, the major species present, and the pH at 25 oC. Note that Ka and Kb  values for the weak acids and bases are available in the Appendices to this text.

(a) sodium nitrate, NaNO3

(b) sodium benzoate, NaC6H5CO2)

(c) potassium fluoride, KF

(d) methylammonium chloride, CH3NH3Cl

(e) sodium cyanide, NaCN

5. Novocaine, C13H21O2N2Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7.0 x 10−6. Is a solution of novocaine acidic or basic? What are [H3O+], [OH−], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g mL-1.

 

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Hydrolysis of Salts Copyright © by D. Latimer; D. Vanderwel; and J. Hollett is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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