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

Section 17-11:  pH at Equivalence Point of a 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 K_c for the reaction above is typically replaced by K_w in the equilibrium expression:

 

Equation 3:  K_w = [H⁺][OH^-]

 

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

 

- log (K_w) = -log ([H⁺][OH^-])

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

Equation 4:  14.00 = pH + pOH

 

On the pH scale at 25°C:

 

pH < 7 = acidic

pH 7 = neutral

pH > 7 = basic

 

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

 

 

 

 

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 (M_i) 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:

 

 

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:

 

 

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:

 

 

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:

 

 

 

 

 

 

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 K_c for the ionization reaction, which is typically expressed as K_a, known as the acid ionization constant.

 

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

 

 

(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 K_a = 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 (K_a = 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:

 

 

 

 

 

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 K_a 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 K_a 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:

 

 

 

 

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

 

Relative strengths of weak acids can also be compared using pK_a values:

 

pK_a = -log K_a

K_a = 10^-pKa

lower pK_a = larger K_a = stronger acid

 

 

 

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 K_a value. 

 

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

 

1^st ionization:  H₂A (aq) ⇌ H⁺ (aq) + HA^- (aq)     K_c = K_a1

 

2^nd ionization:  HA^- (aq) ⇌ H⁺ (aq) + A^2- (aq)     K_c = K_a2

 

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

 

1^st ionization:  H₃A (aq) ⇌ H⁺ (aq) + H₂A^- (aq)     K_c = K_a1

 

2^nd ionization:  H₂A^- (aq) ⇌ H⁺ (aq) + HA^2- (aq)     K_c = K_a2

 

3^rd ionization:  HA^2- (aq) ⇌ H⁺ (aq) + A^3- (aq)     K_c = K_a3

 

The K_a values will decrease with each successive ionization:  K_a1 > K_a2 > K_a3.  In general, only the 1^st ionization will produce enough H⁺ to affect the pH, so we will only use the RICE chart and equilibrium calculation from the 1^st 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:

 

K_a1 = 7.5 x 10^-3

K_a2 = 6.2 x 10^-8

K_a3 = 4.2 x 10^-13

 

Solution:

 

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

 

 

 

 

2^nd ionization:  H₂PO₄^- (aq) ⇌ H⁺ (aq) + HPO₄^2- (aq)    

 

3^rd ionization:  HPO₄^2- (aq) ⇌ H⁺ (aq) + PO₄^3- (aq)  

 

 

Sulfuric acid is a unique case where the 1^st 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 2^nd ionization (similar to a diprotic weak acid):

 

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

 

2^nd ionization:  HSO₄^- (aq) ⇌ H⁺ (aq) + SO₄^2- (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 K_a 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:

 

CO₃^2- (aq) + HC₂O₄^- (aq) ⇌ HCO₃^- (aq) + C₂O₄^2- (aq) 

 

Solution:

 

forward reaction:  Bronsted acid = HC₂O₄^- (donates proton), Bronsted base = CO₃^2- (accepts proton)

reverse reaction:  Bronsted acid = HCO₃^-, Bronsted base = C₂O₄^2-

 

 

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 = C₄H₄O₆^2-

 

 

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)

 

K_c for the reaction above can be expressed as K_b, 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 K_b expression.  Also note that the calculated value of x = [OH^-], not the molarity of hydrogen ions:

 

 

 

 

When comparing different weak bases with the same initial molarity:

 

larger K_b = 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 (K_b = 1.8 x 10^-5).

 

Solution:

 

 

 

 

 

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 K_b as follows:

 

Sample Exercise 17I:

 

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

 

 

Solution:

 

 

 

 

 

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

 

Relative strengths of weak bases can also be compared using pK_b values:

 

pK_b = -log K_b

K_b = 10^-pKb

lower pK_b = larger K_b = stronger base

 

 

Section 17-7:  Lewis Structures and Acid Strength

 

The Lewis structure for an oxoacid (or oxyacid) with the formula H_mXO_n 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 K_a 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)     K_a for BH⁺ = ???

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

__________________________________________________________________________________________

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)     K_w = 1.0 x 10^-14

 

(K_a for BH⁺)(K_b for B) = K_w

 

 

(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 K_b for the reaction above as follows:

 

A^- (aq) + H₂O (l) ⇌ HA (aq) + OH^- (aq)     K_b for A^- = ???

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

_________________________________________________________________________________

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)     K_w = 1.0 x 10^-14

 

(K_b for A^-)(K_a for HA) = K_w

 

 

(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 K_a for BH⁺ > K_b for A^-, then solution is acidic

if K_a for BH⁺ < K_b for A^-, then solution is basic

if K_a for BH⁺ = K_b 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:

 

 

 

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)

 

 

K_a > K_b = 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 (K_b for NH₃ = 1.8 x 10^-5)

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

 

Solution:

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

 

 

 

 

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

 

 

 

 

 

 

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) + A^2- (aq)     K_a for HA^- = K_a2 for polyprotic acid H₂A = given in problem or found on K_a data table

weak base hydrolysis reaction:

HA^- (aq) + H₂O (l) ⇌ H₂A (aq) + OH^- (aq)     K_b for HA^- = given in problem or found on K_b 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 K_a > K_b, then solution is acidic

if K_a < K_b, 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, K_a1 = 4.3 x 10^-7 and K_a2 = 5.6 x 10^-11

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

 

Solution:

Na⁺ = spectator ion

 

weak acid ionization reaction: 

HCO₃^- (aq) ⇌ H⁺ (aq) + CO₃^2- (aq)     K_a for HCO₃^- = K_a2 for H₂CO₃ = 5.6 x 10^-11

weak base hydrolysis reaction:

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

 

K_a < K_b = 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.

 

 

 

Section 17-11:  pH at Equivalence Point of a Titration

When a strong acid is titrated with a strong base, the net ionic equation is as follows:

 

H⁺ (aq) + OH^- (aq) → H₂O (l)

 

Since equal moles of hydrogen ion and hydroxide ion have been mixed to reach the equivalence point, the only species present at the equivalence point will be spectator ions that do not affect the pH and water.  Therefore, the pH will be 7 at the equivalence point:

 

strong acid + strong base titration:  pH = 7 at equivalence point

 

When a weak acid is titrated with a strong base, the net ionic equation is as follows:

 

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

 

In this case, after equal moles of HA and hydroxide ion have been mixed to reach the equivalence point, A^- ions will also be present in addition to spectator ions that do not affect the pH and water.  Since A^- can hydrolyze as a weak base, the pH will be greater than 7 at the equivalence point due to the following equilibrium reaction:

 

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

 

weak acid + strong base titration:  pH > 7 at equivalence point

 

Sample Exercise 17Q:

For which titration below will the pH be above 7 at the equivalence point?

 

HCl (aq) + Ba(OH)₂ (aq)  or  HF (aq) + NaOH (aq)?

 

Solution:

HCl (aq) + Ba(OH)₂ (aq) = strong acid + strong base:  pH = 7 at equivalence point

HF (aq) + NaOH (aq) = weak acid + strong base:  pH > 7 at equivalence point

 

 

 

 

Chapter 17 Practice Exercises and Review Quizzes:

 

 

 

 

 

 

 

 

 

 

 

 

 

    

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Click for Review Quiz 1 Answers