Chapter 17:  Acids and Bases

 

Section 17-1: Arrhenius Theory, pH, and pOH

Section 17-2: Monoprotic Strong Acids and Strong Bases

Section 17-3: Monoprotic Weak Acids

Section 17-4: Polyprotic Weak Acids and Sulfuric Acid

Section 17-5: Bronsted Theory

Section 17-6: Weak Bases

Section 17-7: Lewis Structures and Acid Strength

Section 17-8: Hydrolysis Aqueous Ions as Acids and Bases

Section 17-9: Reactions of Nonmetal Oxides with Water to Create Oxoacids

Section 17-10: Experiment Determining the Molar Mass of an Unknown Monoprotic Weak Acid by Titration

Chapter 17 Practice Exercises and Review Quizzes

 

 

 

 

Section 17-1:  Arrhenius Theory, pH, and pOH


Arrhenius theory essentially states that an acid ionizes in water to produce H+, known as a hydrogen ion or a proton, while a base ionizes in water to produce OH-, hydroxide ion.  One way to express the level of acidity or basicity in an aqueous solution is to indicate the molarity of H+, [H+], or the molarity of OH-, [OH-].  Alternatively, pH and pOH (both expressed without a unit) can be used:

 

Equation 1:  pH = - log [H+]

 

Equation 2:  pOH = - log [OH-]

 

[Note:  When rounding the result y in a calculation y = - log x, the number of decimal places in y should equal the number of significant figures in x.]

 

In any aqueous solution, the equilibrium reaction known as the autoionization of water occurs:

 

H2O (l) H+ (aq) + OH- (aq)


The equilibrium constant Kc for the reaction above is typically replaced by Kw in the equilibrium expression:

 

Equation 3:  Kw = [H+][OH-]

 

At 25C, Kw = 1.0 x 10-14.  Therefore, we can derive an equation relating pH and pOH at 25C as follows:

 

- log (Kw) = -log ([H+][OH-])

- log (1.0 x 10-14) = (- log [H+]) + (-log [OH-])

Equation 4:  14.00 = pH + pOH

 

On the pH scale at 25C:

 

pH < 7 = acidic

pH 7 = neutral

pH > 7 = basic

 

In this chapter, unless otherwise indicated, assume all solutions are aqueous and at 25C.

 

 

 

 

Section 17-2:  Monoprotic Strong Acids and Strong Bases


If an acid molecule is only capable of releasing one H+ (has only one ionizable proton), the acid is said to be monoprotic.  We can assume that all molecules of the monoprotic strong acids HCl, HBr, HI, HNO3, and HClO4 will ionize completely in solution to produce one proton and an anion having a 1- charge, with the reverse reaction being negligible (and, thus, we will use a single forward arrow rather that a double equilibrium arrow in the equation).  Therefore, the initial molarity (Mi) of the strong acid will become the molarity of H+ present in the solution at equilibrium, as demonstrated below on the RICE chart for a nitric acid solution:

 

R

    HNO3 (aq)           H+ (aq)         +       NO3- (aq)     

I

Mi

0

0

C

-Mi

+Mi

+Mi

E

0

Mi

Mi

 

Once we know the molarity of H+ at equilibrium, we can also calculate the molarity of hydroxide ion (which, even in an acidic solution, is produced due to the autoionization of water), pH, and pOH as follows:

 

Sample Exercise 17A:

 

A 44 mL sample of HCl gas, measured at 45°C and 711 torr, was dissolved in water to yield 18.5 mL of solution.  Calculate the molarity of hydrogen ion, the molarity of hydroxide ion, pH, and pOH in this solution.

 

Solution:

 

First, use the Ideal Gas Law to calculate the moles of HCl dissolved, then divide by the volume of solution to determine the initial molarity of HCl:

 

 

 

Since HCl is a strong acid, we now complete a RICE chart showing complete ionization similar to that for HNO3 above:

 

R

      HCl (aq)            H+ (aq)        +         Cl- (aq)     

I

0.085

0

0

C

-0.085

+0.085

+0.085

E

0

0.085

0.085

 

We can now complete the calculations using the equations above:

 

 

(Note that we could have also calculated pOH using Equation 2.)

 

Soluble metal hydroxides are strong bases that will ionize completely in water.  If there is only one hydroxide ion per formula unit, the initial molarity of the metal hydroxide will become the molarity of hydroxide ion present in the solution at equilibrium, as demonstrated below on the RICE chart for a sodium hydroxide solution:

 

R

     NaOH (aq)            Na+ (aq)      +         OH- (aq)     

I

Mi

0

0

C

-Mi

+Mi

+Mi

E

0

Mi

Mi

 

However, if there are two hydroxide ions per formula unit, the molarity of hydroxide ion present in the solution at equilibrium will be twice the initial molarity of the metal hydroxide:

 

Sample Exercise 17B:


A 0.0070 gram sample of calcium hydroxide was dissolved in water to create 64 mL of solution.  Calculate the molarity of hydroxide ion, the molarity of hydrogen ion, pOH, and pH in this solution.

 

Solution:

 

First, convert the mass of calcium hydroxide to moles, then divide by the volume of solution to determine the initial molarity of calcium hydroxide:

 

 

Since calcium hydroxide is a strong base, we now complete a RICE chart showing complete ionization similar to that for NaOH above:

 

R

  Ca(OH)2 (aq      Ca2+ (aq)       +     2 OH- (aq)     

I

0.0015

0

0

C

-0.0015

+0.0015

+2(0.0015)

E

0

0.0015

0.0030

 

 

 

 

 

Section 17-3:  Monoprotic Weak Acids

 

For the remainder of this chapter, assume that any acid that is not included on the strong acid list of HCl, HBr, HI, HNO3, or HClO4 is a weak acid, meaning that only a fraction of the acid molecules ionize in water.  For the ionization of a monoprotic weak acid, the RICE chart and pH calculation will differ from that of a monoprotic strong acid in the following ways:

 

1. A double equilibrium arrow will be used in the ionization equation to indicate that the reverse reaction is significant.

 

2. The initial molarity of the weak acid will not decrease to zero as the reaction proceeds.  As such, the molarity of H+ present at equilibrium will be less than the initial molarity of the weak acid.

 

3. To determine the magnitude of the change (C) line on the RICE chart, we will need to do an equilibrium calculation.  This will require knowing the value of Kc for the ionization reaction, which is typically expressed as Ka, known as the acid ionization constant.

 

The RICE chart and equilibrium calculation for all monoprotic weak acids will follow the format below:

 

R

        HA (aq)       ⇌        H+ (aq)         +          A- (aq)     

I

Mi

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

(Note that the anion A- may contain more hydrogen atoms, but we can assume that these are not ionizable.)

 

 

 

Once we know the value of x, we can also calculate the percentage of the original sample of acid molecules that have been ionized as follows:

 

 

When comparing different monoprotic weak acids with the same initial molarity:

 

larger Ka = larger x = larger % ionization = larger [H+] = lower pH = more acidic = stronger acid

   

Sample Exercise 17C:


Write the acid ionization equation and calculate the pH and percent ionization of 0.16 M hydrofluoric acid, HF (Ka = 6.8 x 10-4).

 

Solution:

 

We complete the RICE chart and equilibrium calculation for a monoprotic weak acid, then use the value of x = [H+] to calculate the pH and percent ionization:

 

R

       HF (aq)        ⇌          H+ (aq)        +         F- (aq)     

I

0.16

0

0

C

-x

+x

+x

E

0.16 - x

x

x

 

 

 

 

If we know the pH of a solution, we can calculate [H+] as follows:

 

Equation 1a:  [H+] = 10-pH 

 

If we know the initial molarity and the pH of a monoprotic weak acid solution, we can calculate Ka as follows:

 

 Sample Exercise 17D:


A 0.58 M acetic acid solution has a pH of 2.49.  Write the acid ionization equation and calculate Ka for acetic acid, HC2H3O2 (a more condensed version of the formula CH3COOH showing the one ionizable proton at the beginning, separate from the other non-ionizable hydrogens). 

 

Solution:

 

R

  HC2H3O2 (aq)  ⇌      H+ (aq)        +   C2H3O2- (aq)     

I

0.58

0

0

C

-x

+x

+x

E

0.58 - x

x

x

 

 

 

If we know the initial molarity and the percent ionization for a monoprotic weak acid, we can calculate pH and Ka, as demonstrated at the end of the chapter in Practice Exercise 17-5.  If we know the pH and Ka for a monoprotic weak acid solution, we can calculate the initial molarity, as demonstrated in Practice Exercise 17-6.

 

 

Section 17-4:  Polyprotic Weak Acids and Sulfuric Acid


A polyprotic weak acid has more than one ionizable proton and will ionize in a stepwise series of reactions, each of which has its own distinct Ka value. 

 

A diprotic weak acid will ionize in two steps as follows:

 

1st ionization:  H2A (aq) H+ (aq) + HA- (aq)     Kc = Ka1

 

2nd ionization:  HA- (aq) H+ (aq) + A2- (aq)     Kc = Ka2

 

A triprotic weak acid will ionize in three steps as follows:

 

1st ionization:  H3A (aq) H+ (aq) + H2A- (aq)     Kc = Ka1

 

2nd ionization:  H2A- (aq) H+ (aq) + HA2- (aq)     Kc = Ka2

 

3rd ionization:  HA2- (aq) H+ (aq) + A3- (aq)     Kc = Ka3

 

The Ka values will decrease with each successive ionization:  Ka1 > Ka2 > Ka3.  In general, only the 1st ionization will produce enough H+ to affect the pH, so we will only use the RICE chart and equilibrium calculation from the 1st ionization to calculate the pH:   

 

Sample Exercise 17E:

 

Write the stepwise acid ionization equations and calculate the pH of 1.6 M phosphoric acid, H3PO4, which has the following acid ionization constants:

 

Ka1 = 7.5 x 10-3

Ka2 = 6.2 x 10-8

Ka3 = 4.2 x 10-13

 

Solution:

 

We obtain the pH by completing the RICE chart and equilibrium calculation for the 1st ionization only, after which we proceed with writing the 2nd and 3rd ionization equations:

 

R

    H3PO4 (aq)         H+ (aq)      +       H2PO4- (aq)     

I

1.6

0

0

C

-x

+x

+x

E

1.6 - x

x

x

 

 

 

2nd ionization:  H2PO4- (aq) H+ (aq) + HPO42- (aq)    

 

3rd ionization:  HPO42- (aq) H+ (aq) + PO43- (aq)  

 

 

Sulfuric acid is a unique case where the 1st ionization of H2SO4 occurs completely with a negligible reverse reaction (similar to a monoprotic strong acid), whereas only a fraction of the HSO4- anions further ionize in the 2nd ionization (similar to a diprotic weak acid):

 

1st ionization:  H2SO4 (aq) H+ (aq) + HSO4- (aq)

 

2nd ionization:  HSO4- (aq) H+ (aq) + SO42- (aq)    

 

 

Section 17-5:  Bronsted Theory


Because ammonia, NH3, does not contain hydroxide ion, Arrhenius theory does not explain how ammonia can function as a base in water. On the other hand, Bronsted theory can explain how a solution of ammonia can be basic by defining an acid and base as follows:

 

Bronsted acid = proton (H+) donor

Bronsted base = proton (H+) acceptor

 

In the case of an ammonia solution, water will act as a Bronsted acid to donate a proton to the Bronsted base ammonia.  Since the Bronsted acid water loses a proton, the formula will decrease by one hydrogen and the charge will decrease by 1, thus producing the hydroxide ion that makes the solution basic.  At the same time, the Bronsted base ammonia will gain a proton, so the formula will increase by one hydrogen and the charge will increase by 1 to produce the ammonium ion:

 

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) 

 

Although we have used Arrhenius theory to write acid ionization equations and will continue to do so, we could have chosen to use Bronsted theory instead.  In the case of solution containing the monoprotic weak acid HA, water can act as a Bronsted base to accept a proton from the Bronsted acid HA:

 

HA (aq) + H2O (l) H3O+ (aq) + A- (aq) 

 

The cation H3O+ is known as the hydronium ion and can replace H+ when calculating pH after writing the acid ionization equation according to Bronsted theory:  pH = - log [H3O+].  Since liquid water will be omitted from the RICE chart and Ka expression, the pH calculated using Bronsted theory will be the same as the pH calculated using Arrhenius theory.

 

As shown above, water acts as a Bronsted acid in an ammonia solution, but a Bronsted base in an acid solution.  A molecule or ion that can act as both a Bronsted acid or a Bronsted base in different situations is said to be amphoteric (or amphiprotic).

 

Bronsted theory also applies to proton transfer reactions in which water is not a reactant or a product:

 

Sample Exercise 17F:

 

Identify the Bronsted acids and bases in the forward and reverse directions for the reaction below:

 

CO32- (aq) + HC2O4- (aq) HCO3- (aq) + C2O42- (aq) 

 

Solution:

 

forward reaction:  Bronsted acid = HC2O4- (donates proton), Bronsted base = CO32- (accepts proton)

reverse reaction:  Bronsted acid = HCO3-, Bronsted base = C2O42-

 

 

After a proton has been donated by a Bronsted acid, the resulting molecule or ion is known as the conjugate base.  After a proton has been accepted by a Bronsted base, the resulting molecule or ion is known as the conjugate acid:  

 

Sample Exercise 17G:

 

Write the formula for:

 

a. the conjugate acid of HS-

b. the conjugate base of HC4H4O6-

 

Solution:

 

a. conjugate acid = add H+ to formula = H2S

b. conjugate base = remove H+ from formula = C4H4O62-

 

 

Section 17-6:  Weak Bases

 

For a solution of a neutral weak base B, the base ionization equation can be represented: 

 

B (aq) + H2O (l) BH+ (aq) + OH- (aq)

(proton is typically added at end of base formula)

 

Kc for the reaction above can be expressed as Kb, known as the base ionization constant.  To calculate the pH and percent ionization in a weak base solution, we complete a RICE chart and equilibrium calculation similar to that for a monoprotic weak acid.  However, the column under liquid water will be omitted from the RICE chart and water will be omitted from the Kb expression.  Also note that the calculated value of x = [OH-], not the molarity of hydrogen ions:

 

R

        B (aq)        +          H2O(l)                  BH+ (aq)        +         OH- (aq)

I

Mi

 

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

 

When comparing different weak bases with the same initial molarity:

 

larger Kb = larger x = larger % ionization = larger [OH-] = higher pH = more basic = stronger base

 

Sample Exercise 17H:

 

Calculate the pH and percent ionization of 0.029 M ammonia (Kb = 1.8 x 10-5).

 

Solution:

 

R

       NH3 (aq)      +        H2O(l)                  NH4+ (aq)      +        OH- (aq)

I

0.029

 

0

0

C

-x

+x

+x

E

0.029- x

x

x

 

 

 

 

If we know the pOH of a solution, we can calculate [OH-] as follows:

 

Equation 2a:  [OH-] = 10-pOH 

 

If we know the initial molarity and the pH of a weak base solution, we can calculate Kb as follows:

 

Sample Exercise 17I:

 

A 0.026 M methylamine solution has a pH of 11.51.  Write the base ionization equation and calculate Kb for methylamine, CH3NH2.

 

 

Solution:

 

R

   CH3NH2 (aq)    +         H2O(l)              CH3NH3+ (aq)    +       OH- (aq)

I

0.026

 

0

0

C

-x

+x

+x

E

0.026- x

x

x

 

 

 

 

If we know the initial molarity and the percent ionization for a weak base, we can calculate pH and Kb, as demonstrated in Practice Exercise 17-12.  If we know the pH and Kb for a weak base solution, we can calculate the initial molarity, as demonstrated in Practice Exercise 17-13.

 


Section 17-7:  Lewis Structures and Acid Strength

 

The Lewis structure for an oxoacid (or oxyacid) with the formula HmXOn will generally show the center atom X bonded to each oxygen atom, and each hydrogen atom will be bonded to a different oxygen atom, as demonstrated below for sulfurous acid, H2SO3:

 

 

(other resonance structures also acceptable)

 

To compare the relative strengths of oxoacids, use the following guidelines:

 

1. More terminal oxygens (= oxygens not bonded to a hydrogen = n-m) = oxoacid donates proton more readily = stronger acid.

2. Higher electronegativity of center atom X = oxoacid donates proton more readily = stronger acid.

 

Sample Exercise 17J:

 

Draw Lewis structures for nitric acid, HNO3, and phosphoric acid, H3PO4.  Which is the stronger acid?  Give two reasons to justify your answer.

 

Solution:

 

 

 

 

Nitric acid is stronger acid because:

 

1. HNO3 has more terminal oxygens (3-1=2) than H3PO4 (4-3=1).

2. Electronegativity for central atom N is higher than for central atom P.

 

 

The Lewis structure for a monoprotic organic acid with the formula RCOOH, where R can either represent a single atom or a group of atoms, is shown below:

 

 

The ionizable proton will be the one single-bonded to the oxygen atom on the right end of the Lewis structure, not in R.  To compare the relative strengths of organic acids, use the following general guidelines:

 

1. Higher electronegativity of atoms in R = organic acid donates proton more readily = stronger acid.

2. Highly electronegative atoms in R are closer in proximity to ionizable hydrogen = organic acid donates proton more readily = stronger acid.

 

Sample Exercise 17K:


Which of the two acids shown below is the stronger acid?  Give two reasons to justify your answer.

 

 

 

 

Solution:


The bottom acid is stronger because:

 

1. Electronegativity of F is higher than Br.

2. F is closer to ionizable proton than Br.

 


Section 17-8:  Hydrolysis – Aqueous Ions as Acids and Bases


When an ionic compound dissolves in water, it is sometimes possible for the ions in solution to function as acids or bases in a process known as hydrolysis.  Use the following guidelines to decide if an ion will hydrolyze and, therefore, affect the pH:

 

1. A cation BH+ containing an ionizable proton (ammonium, NH4+, for example) will generally hydrolyze as a weak acid:

 

BH+ (aq) H+ (aq) + B (aq)

 

We can determine Ka for the reaction above by utilizing the fact that when two reactions are added to obtain a third reaction, the product of the equilibrium constants for the two added reactions will equal the equilibrium constant for the third reaction:

 

BH+ (aq) H+ (aq) + B (aq)     Ka for BH+ = ???

B (aq) + H2O (l) BH+ (aq) + OH- (aq)     Kb for B = given in problem or found on Kb data table

__________________________________________________________________________________________

H2O (l) H+ (aq) + OH- (aq)     Kw = 1.0 x 10-14

 

(Ka for BH+)(Kb for B) = Kw

 

 

(so stronger base B = weaker conjugate acid BH+)

 

2. Metal cations will generally not hydrolyze and, therefore, will be spectator ions that do not affect the pH.  Notable exceptions include Al3+ and Fe3+.

 

3. An anion will generally hydrolyze as a weak base:

 

A- (aq) + H2O (l) HA (aq) + OH- (aq)

 

We can determine Kb for the reaction above as follows:

 

A- (aq) + H2O (l) HA (aq) + OH- (aq)     Kb for A- = ???

HA (aq) H+ (aq) + A- (aq)     Ka for HA = given in problem or found on Ka data table

_________________________________________________________________________________

H2O (l) H+ (aq) + OH- (aq)     Kw = 1.0 x 10-14

 

(Kb for A-)(Ka for HA) = Kw

 

 

(so stronger acid HA = weaker conjugate base A-)

 

4. Anions found in monoprotic strong acids (Cl-, Br-, I-, NO3-, ClO4-) do not hydrolyze as weak bases to any significant extent, so these five anions will be spectator ions that do not affect the pH.

 

Using the guidelines above, we can predict if an ionic compound solution will be acidic, basic, or neutral as follows:

 

A. If both the cation and the anion are spectator ions, the solution will be neutral.

 

B. If only the cation hydrolyzes and the anion is a spectator ion, the solution will be acidic.

 

C. If the cation is a spectator ion and only the anion hydrolyzes, the solution will be basic.

 

D. If both the cation and anion hydrolyze, then we must compare the equilibrium constants for the two hydrolysis reactions:

 

if Ka for BH+ > Kb for A-, then solution is acidic

if Ka for BH+ < Kb for A-, then solution is basic

if Ka for BH+ = Kb for A-, then solution is neutral

 

Sample Exercise 17L:


Predict whether a solution of each compound below will be acidic, basic, or neutral.  For solutions that are not neutral, show all relevant hydrolysis reactions that affect the pH and also calculate the equilibrium constant for each reaction you write using information from the following data tables:

 

Acid

Ka

HC2H3O2

1.8 x 10-5

HNO2

4.5 x 10-4

 

Base

Kb

C5H5N

1.7 x 10-9

CH3NH2

4.4 x 10-4

 

a. Mg(NO3)2

b. CH3NH3Cl [composed of CH3NH3+ and Cl-]

c. NaC2H3O2

d. C5H5NHNO2 [composed of C5H5NH+ and NO2-]

 

Solution:


a. Mg2+ and NO3- = spectator ions = solution is neutral

 

b. Cl- = spectator ion, CH3NH3+ hydrolyzes as weak acid = solution is acidic:

 

CH3NH3+ (aq) H+ (aq) + CH3NH2 (aq) 

 

 

c. Na+ = spectator ion, C2H3O2- hydrolyzes as a weak base = solution is basic:

 

C2H3O2- (aq) + H2O (l) HC2H3O2 (aq) + OH- (aq)

 

 

d. Both ions hydrolyze, so we must compare equilibrium constants:

 

C5H5NH+ (aq) H+ (aq) + C5H5N (aq) 

 

 

NO2- (aq) + H2O (l) HNO2 (aq) + OH- (aq)

 

 

Ka > Kb = solution is acidic

 

 

To calculate the pH of a solution where only the cation hydrolyzes as a weak acid and the anion is a spectator ion, we complete a RICE chart and equilibrium calculation similar to that completed earlier for a neutral monoprotic weak acid.  To calculate the pH of a solution where only the anion hydrolyzes as a weak base and the cation is a spectator ion, we complete a RICE chart and equilibrium calculation similar to that completed earlier for a neutral weak base.

 

Sample Exercise 17M:

 

For each solution below, show any relevant hydrolysis reactions and calculate the pH.

 

a. 1.6 M NH4I (Kb for NH3 = 1.8 x 10-5)

b. 0.84 M KClO (Ka for HClO = 3.0 x 10-8)

 

Solution:


a. I- = spectator ion, NH4+ hydrolyzes as a weak acid:

 

R

     NH4+ (aq)     ⇌       H+ (aq)         +      NH3 (aq)     

I

1.6

0

0

C

-x

+x

+x

E

1.6 - x

x

x

 

 

 

b. K+ = spectator ion, ClO- hydrolyzes as a weak base:

 

R

       ClO- (aq)      +        H2O(l)                 HClO (aq)       +        OH- (aq)

I

0.84

 

0

0

C

-x

+x

+x

E

0.84- x

x

x

 

 

 

 

 

If a solution contains metal cations that are spectator ions along with anions of the type HA- having one ionizable proton, a portion of the HA- ions will ionize as weak acids while the remainder of the HA- ions will hydrolyze as weak bases:

 

weak acid ionization reaction: 

HA- (aq) H+ (aq) + A2- (aq)     Ka for HA- = Ka2 for polyprotic acid H2A = given in problem or found on Ka data table


weak base hydrolysis reaction:

HA- (aq) + H2O (l) H2A (aq) + OH- (aq)     Kb for HA- = given in problem or found on Kb data table

 

To determine if an HA- solution is acidic or basic, we must compare the equilibrium constants for the weak acid ionization reaction and the weak base hydrolysis reaction shown above:

 

if Ka > Kb, then solution is acidic

if Ka < Kb, then solution is basic

 

Sample Exercise 17N:


Predict whether a solution of sodium bicarbonate, NaHCO3, will be acidic or basic.  Show all relevant reactions that affect the pH and also give the value of the equilibrium constant for each reaction you write using some or all of the following information:

 

For carbonic acid, H2CO3, Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11

For the bicarbonate ion, HCO3-, Kb = 2.3 x 10-8

 

Solution:


Na+ = spectator ion

 

weak acid ionization reaction: 

HCO3- (aq) H+ (aq) + CO32- (aq)     Ka for HCO3- = Ka2 for H2CO3 = 5.6 x 10-11


weak base hydrolysis reaction:

HCO3- (aq) + H2O (l) H2CO3 (aq) + OH- (aq)     Kb for HCO3- = 2.3 x 10-8

 

Ka < Kb = solution is basic

 


Section 17-9:  Reactions of Nonmetal Oxides with Water to Create Oxoacids


Recall from Chapter 10 that basic hydroxide ion is produced when a metal oxide reacts with water.  When a nonmetal oxide reacts with water, the product is an oxoacid.  Note that this is not a redox reaction, so the atoms do not undergo a change in oxidation number as the reaction proceeds.

 

nonmetal oxide + water oxoacid with no change in oxidation numbers

 

For example, the reaction of SO2 (oxidation number of S = +4) and water will produce H2SO3 (oxidation number of S = +4) rather than H2SO4 (oxidation number of S = +6).

 

Sample Exercise 17O:

 

 

Will the reaction of N2O5 and water produce HNO2 or HNO3?  Write a balanced equation for the reaction.

 

Solution:

 

The reaction of N2O5 (oxidation number of N = +5) and water will produce HNO3 (oxidation number of N = +5) rather than HNO2 (oxidation number of N = +3).  The balanced equation will be N2O5 + H2O 2 HNO3.

 


Section 17-10:  Experiment – Determining the Molar Mass of an Unknown Monoprotic Weak Acid by Titration


Recall from Chapter 3 that we can determine the empirical formula of a compound if we know the percent composition by mass, after which we can determine the molecular formula using the experimentally determined molar mass. We can obtain the molar mass of an unknown monoprotic weak acid HA using titration as follows:

 

1. Dissolve a known mass of HA in water.

 

2. Titrate the HA solution with a metal hydroxide solution of known molarity from a buret.  Record the volume of metal hydroxide solution required to reach the equivalence point.

 

3.  Use stoichiometry from the net ionic equation HA (aq) + OH- (aq) H2O (l) + A- (aq) to calculate the moles of HA reacted, then divide the mass of HA by moles of HA to obtain the molar mass of HA.

   

Sample Exercise 17P:


a. An unknown monoprotic weak acid was found to be 54.53% carbon and 9.15% hydrogen by mass, with the remainder being oxygen.  Determine the empirical formula of the acid.

b. In a separate experiment, 6.7 grams of the acid was dissolved in 25 mL of water and then titrated with 0.85 M barium hydroxide.  The volume of base required to reach the equivalence point was 45 mL.  Calculate the molar mass and determine the molecular formula of the acid.

 

Solution:

 

a. 100% - 54.53% C - 9.15% H = 36.32%O by mass.

 

Assume one hundred grams of unknown acid:

 

 

4.540 mol C: 9.08 mol H: 2.270 mol O (divide each by 2.270)

= 2 mol C: 4 mol H: 1 mol O

Therefore, empirical formula is C2H4O.

 

b. 0.85 M Ba(OH)2 0.85 M Ba2+ and 2 x 0.85 M = 1.7 M OH-

Use stoichiometry from the net ionic equation HA (aq) + OH- (aq) H2O (l) + A- (aq) to calculate the moles of HA reacted, then divide the mass of HA by moles of HA to obtain the molar mass of HA.  Note that the volume of water used to dissolve the HA is not relevant in this calculation:

 

 

Since the acid is monoprotic, we can rewrite the molecular formula as HC4H7O2.

 

 

 

Chapter 17 Practice Exercises and Review Quizzes:

 

17-1) A 74 mL sample of HI gas, measured at 55C and 694 mmHg, was dissolved in water to yield 48 mL of solution.  Calculate the molarity of hydrogen ion, the molarity of hydroxide ion, pH, and pOH in this solution.

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17-1) 

 

 

HI = strong acid:

R

       HI (aq)               H+ (aq)         +          I- (aq)     

I

0.052

0

0

C

-0.052

+0.052

+0.052

E

0

0.052

0.052

 

 

 

 

17-2) A 0.054 gram sample of barium hydroxide was dissolved in water to create 75 mL of solution.  Calculate the molarity of hydroxide ion, the molarity of hydrogen ion, pOH, and pH in this solution.

 

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17-2)

 

 

Ba(OH)2 = strong base:

 

R

  Ba(OH)2 (aq)  →      Ba2+ (aq)      +       2 OH- (aq)     

I

0.0042

0

0

C

-0.0042

+0.0042

+2(0.0042)

E

0

0.0042

0.0084

 

 

 

 

17-3) Write the acid ionization equation and calculate the pH and percent ionization of 0.87 M hypochlorous acid, HClO (Ka = 3.0 x 10-8).

 

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17-3)

R

      HClO (aq)          H+ (aq)        +        ClO- (aq)     

I

0.87

0

0

C

-x

+x

+x

E

0.87 - x

x

x

 

 

 

 

17-4) A 0.92 M hydrocyanic acid solution has a pH of 4.68.  Write the acid ionization equation and calculate Ka for hydrocyanic acid, HCN.

 

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17-4)

R

       HCN (aq)         H+ (aq)       +          CN- (aq)     

I

0.92

0

0

C

-x

+x

+x

E

0.92 - x

x

x

 

 

 

 

17-5) A 0.046 M solution of benzoic acid is 3.6% ionized.  Write the acid ionization equation and calculate the pH of the solution and Ka for benzoic acid, HC7H5O2.   

 

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17-5)

R

  HC7H5O2 (aq      H+ (aq)      +      C7H5O2- (aq)     

I

0.046

0

0

C

-x

+x

+x

E

0.046 - x

x

x

 

 

 

 

17-6) A nitrous acid, HNO2, solution has a pH of 2.58.  Given that Ka = 4.5 x 10-4 for nitrous acid, write the acid ionization equation and calculate the initial molarity of the nitrous acid solution.

  

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17-6)

R

     HNO2 (aq)         H+ (aq)     +           NO2- (aq)     

I

Mi

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

 

 

17-7) Write the stepwise acid ionization equations and calculate the pH of 0.073 M sulfurous acid, H2SO3, which has the following acid ionization constants:

 

Ka1 = 1.7 x 10-2

Ka2 = 6.4 x 10-8

 

Click for Solution

 

17-7)

R

    H2SO3 (aq)          H+ (aq)       +        HSO3- (aq)     

I

0.073

0

0

C

-x

+x

+x

E

0.073 - x

x

x

 

 

 

2nd ionization:  HSO3- (aq) H+ (aq) + SO32- (aq)

    

 

17-8) Identify the Bronsted acids and bases in the forward and reverse directions for the reaction below:

 

HPO42- (aq) + H2AsO4- (aq) H2PO4- (aq) + HAsO42- (aq) 

 

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17-8)

forward reaction:  Bronsted acid = H2AsO4- (donates proton), Bronsted base = HPO42- (accepts proton)

reverse reaction:  Bronsted acid = H2PO4-, Bronsted base = HAsO42-

 

 

17-9) Write the formula for:

 

a. the conjugate acid of HC6H5O72-

b. the conjugate base of HC3H2O4-

 

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17-9)

a. conjugate acid = add H+ to formula = H2C6H5O7-

b. conjugate base = remove H+ from formula = C3H2O42-

 

 

17-10) Write the base ionization equation and calculate the pH and percent ionization of 0.11 M pyridine, C5H5N (Kb = 1.7 x 10-9).

 

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17-10)

R

     C5H5N (aq)     +         H2O(l)          C5H5NH+ (aq)    +       OH- (aq)

I

0.11

 

0

0

C

-x

+x

+x

E

0.11- x

x

x

 

 

 

 

17-11) A 0.72 M aniline solution has a pH of 9.26.  Write the base ionization equation and calculate Kb for aniline, C6H5NH2.

 

 

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17-11)

R

  C6H5NH2 (aq)    +        H2O(l)        ⇌   C6H5NH3+ (aq)   +       OH- (aq)

I

0.72

 

0

0

C

-x

+x

+x

E

0.72- x

x

x

 

 

 

 

 

17-12) A 0.052 M solution of ethylamine is 10.6% ionized.  Write the base ionization equation and calculate the pH of the solution and Kb for ethylamine, C2H5NH2.    

 

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17-12)

R

  C2H5NH2 (aq)    +        H2O(l)        C2H5NH3+ (aq)    +        OH- (aq)

I

0.052

 

0

0

C

-x

+x

+x

E

0.052- x

x

x

 

 

 

 

 

17-13) A trimethylamine, (CH3)3N, solution has a pH of 11.68.  Given that Kb = 6.4 x 10-5 for trimethylamine, write the base ionization equation and calculate the initial molarity of the trimethylamine solution.

 

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17-13)

R

   (CH3)3N (aq)    +          H2O(l)         (CH3)3NH+ (aq)   +       OH- (aq)

I

Mi

 

0

0

C

-x

+x

+x

E

Mi - x

x

x

 

 

 

 

 

17-14) Draw Lewis structures for arsenic acid, H3AsO4, and sulfuric acid, H2SO4.  Which is the stronger acid?  Give two reasons to justify your answer.

 

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17-14)

 

 

 

Sulfuric acid is stronger acid because:

 

1. H2SO4 has more terminal oxygens (4-2=2) than H3AsO4 (4-3=1).

2. Electronegativity for central atom S is higher than for central atom As.

 

 

17-15) Which of the two acids shown below is the stronger acid?  Give two reasons to justify your answer.

 


 

 

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17-15)

The top acid is stronger because:

 

1. Electronegativity of Cl is higher than I.

2. Cl atoms are closer to ionizable proton than I atoms.

 

 

17-16) Predict whether a solution of each compound below will be acidic, basic, or neutral.  For solutions that are not neutral, show all relevant hydrolysis reactions that affect the pH and also calculate the equilibrium constant for each reaction you write using information from the following data tables:

 

Acid

Ka

HClO

3.0 x 10-8

HF

6.8 x 10-4

 

Base

Kb

NH3

1.8 x 10-5

(CH3)3N

6.4 x 10-5

 

a. KF

b. (CH3)3NHClO [composed of (CH3)3NH+ and ClO-]

c. Ca(ClO4)2

d. NH4Br

 

 

Click for Solution

 

17-16)

a. K+ = spectator ion, F- hydrolyzes as a weak base = solution is basic:

 

F- (aq) + H2O (l) HF (aq) + OH- (aq)

 

 

b. Both ions hydrolyze, so we must compare equilibrium constants:

 

(CH3)3NH+ (aq) H+ (aq) + (CH3)3N (aq) 

 

 

ClO- (aq) + H2O (l) HClO (aq) + OH- (aq)

 

 

Ka < Kb = solution is basic

 

c. Ca2+ and ClO4- = spectator ions = solution is neutral

 

d. Br- = spectator ion, NH4+ hydrolyzes as weak acid = solution is acidic:

 

NH4+ (aq) H+ (aq) + NH3 (aq)

 

 

 

17-17) For each solution below, show any relevant hydrolysis reactions and calculate the pH.

 

a. 1.8 M NaNO2 (Ka for HNO2 = 4.5 x 10-4)

b. 0.66 M CH3NH3NO3 [composed of CH3NH3+ and NO3-] (Kb for CH3NH2 = 4.4 x 10-4)

 

 

Click for Solution

 

17-17)

a. Na+ = spectator ion, NO2- hydrolyzes as a weak base:

 

R

       NO2- (aq)    +          H2O(l)                HNO2 (aq)     +         OH- (aq)

I

1.8

 

0

0

C

-x

+x

+x

E

1.8- x

x

x

 

 

 

b. NO3- = spectator ion, CH3NH3+ hydrolyzes as a weak acid:

 

R

  CH3NH3+ (aq)     H+ (aq)       +      CH3NH2 (aq)     

I

0.66

0

0

C

-x

+x

+x

E

0.66 - x

x

x

 

 

 

 

17-18) Predict whether a solution of potassium bisulfite, KHSO3, will be acidic or basic.  Show all relevant reactions that affect the pH and also give the value of the equilibrium constant for each reaction you write using some or all of the following information:

 

For sulfurous acid, H2SO3, Ka1 = 1.7 x 10-2 and Ka2 = 6.4 x 10-8

For the bisulfite ion, HSO3-, Kb = 5.9 x 10-13

 

Click for Solution

 

17-18)

K+ = spectator ion

 

weak acid ionization reaction: 

HSO3- (aq) H+ (aq) + SO32- (aq)     Ka for HSO3- = Ka2 for H2SO3 = 6.4 x 10-8


weak base hydrolysis reaction:

HSO3- (aq) + H2O (l) H2SO3 (aq) + OH- (aq)     Kb for HSO3- = 5.9 x 10-13

 

Ka > Kb = solution is acidic

 

 

 

17-19) Will the reaction of Cl2O7 and water produce HClO3 or HClO4?  Write a balanced equation for the reaction.

 

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17-19) The reaction of Cl2O7 (oxidation number of Cl = +7) and water will produce HClO4 (oxidation number of Cl = +7) rather than HClO3 (oxidation number of Cl = +5).  The balanced equation will be Cl2O7 + H2O 2 HClO4.

 

 

17-20)

a. An unknown monoprotic weak acid was found to be 14.13% carbon, 1.78% hydrogen, and 74.67% iodine by mass, with the remainder being oxygen.  Determine the empirical formula of the acid.

b. In a separate experiment, 9.03 grams of the acid was dissolved in 35 mL of water and then titrated with 0.414 M calcium hydroxide.  The volume of base required to reach the equivalence point was 32.1 mL.  Calculate the molar mass and determine the molecular formula of the acid.

 

Click for Solution

 

17-20) a. 100% - 14.13% C – 1.78% H - 74.67% I = 9.42% O by mass

 

Assume one hundred grams of unknown acid:

 

 

1.177 mol C: 1.77 mol H: 0.5884 mol I: 0.5888 mol O (divide each by 0.5884)

= 2 mol C: 3 mol H: 1 mol I: 1 mol O

Therefore, empirical formula is C2H3IO.

 

b. 0.414 M Ca(OH)2 0.414 M Ca2+ and 2 x 0.414 M = 0.828 M OH-

HA (aq) + OH- (aq) H2O (l) + A- (aq)

 

 

Since the acid is monoprotic, we can rewrite the molecular formula as HC4H5I2O2.

 

 

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Click for Review Quiz 1 Answers