Cracking the AP Chemistry Exam, 2012 Edition

1. B At this point, sometimes called the half-equivalence point, enough base has been added to neutralize half of the weak acid. That means that at this point, the concentration of the acid—let’s call it HA—will be equal to the concentration of the conjugate base, A^–.

Now let’s look at the equilibrium expression.

If [HA] = [A^–], then they cancel and K_a = [H⁺].

So pH = pK_a.

2. D The point in the middle of the steep rise in a titration curve is the equivalence point. That’s the point when exactly enough base has been added to neutralize all of the acid that was originally in the solution.

3. E This is the point with the highest pH, so at this point, the solution is most basic.

4. B A buffered solution resists changes to its pH, so the flat part of the titration curve is the buffer region because the pH of the solution is changing very little even when base is being added.

At (B), the solution contains a large quantity of both the acid HA, which absorbs added base, and its conjugate base, A^–, which absorbs added acid.

5. B In any aqueous solution at 25°C, pH + pOH = 14.

So pH + 3 = 14 and pH = 3.

6. A HCl is a strong acid and dissociates completely, so the hydrogen ion concentration is equal to the molarity.

So [H⁺] = 0.1 M

pH = –log[H⁺] = –log(0.1) = 1

7. B HNO₃ is a strong acid and dissociates completely, so the hydrogen ion concentration is equal to the molarity.

So [H⁺] = 0.001 M

pH = –log[H⁺] = –log(0.001) = 3

8. E NaOH is a strong acid and dissociates completely, so the hydroxide ion concentration [OH^–] is equal to the molarity.

So [OH^–] = 0.1 M

pOH = –log[OH^–] = –log(0.1) = 1

Now, pH + pOH = 14

So pH + 1 = 14 and pH = 13

9. A A strong electrolyte is a substance that dissociates to form a lot of ions. HNO₃ is the only substance listed that ionizes completely in solution.

10. E Water can act as a Lewis base by donating an electron pair to a hydrogen ion to form H₃O⁺.

11. A HNO₃ is the only strong acid listed, meaning that it is the only acid listed that ionizes completely, so it will have the lowest pH. By the way, in a 0.1 M solution of a strong acid, the hydrogen ion concentration will be 0.1 M, so the pH will be equal to 1.

12. C H₂CO₃, carbonic acid, is formed in the reaction between CO₂ and H₂O.

13. E NaOH is a strong base, so it can be assumed to dissociate completely. That means that the OH^– concentration will also be 0.01 M.

pOH = –log[OH^–], so pOH = 2, but we’re looking for the pH.

In an aqueous solution, pH + pOH = 14, so pH = 12.

14. B Every mole of Mg(OH)₂ molecules dissociates to produce 2 moles of OH^– ions, so a 0.10 M Mg(OH)₂ solution will be a 0.20 M OH^– solution.

The solution will be neutralized when the number of moles of H⁺ ions added is equal to the number of OH^– ions originally in the solution.

Moles = (molarity)(volume)

Moles of OH^– = (0.20 M)(50 ml) = 10 millimoles = moles of H⁺ added

Volume =

Volume of HCl = = 200 ml

15. B You can eliminate (C), (D), and (E) by using common sense. HBrO is a weak acid, so an HBrO solution will be acidic, with a pH of less than 7.

To choose between (A) and (B) you have to remember that HBrO is a weak acid. If HBrO were a strong acid, it would dissociate completely and [H⁺] would be equal to 0.01-molar, for a pH of exactly 2. Because HBrO is a weak acid, it will not dissociate completely and [H⁺] will be less than 0.01-molar, which means that the pH will be greater than 2. So by POE, (B) is the answer.

16. E To find the lowest pH, we should look for the salt that produces an acidic solution.

The salt composed of the conjugate of a weak base (NH₄⁺ is the conjugate of NH₃) and the conjugate of a strong acid (Cl^– is the conjugate of HCl) will produce an acidic solution.

As for the other choices, (A), (C), and (D) are composed of conjugates of strong acids and bases and will produce neutral solutions.

(B) is composed of the conjugate of a strong base (Na⁺) and the conjugate of a weak acid (C₂H₃O₂^–) and will produce a basic solution.

17. A A pH of greater than 7 means that the solution is basic.

The salt composed of the conjugate of a strong base (Na⁺ is the conjugate of NaOH) and the conjugate of a weak acid (CN^– is the conjugate of HCN) will produce a basic solution.

As for the other choices, (B), (C), and (E) are composed of conjugates of strong acids and bases and will produce neutral solutions.

(D) is composed of the conjugate of a weak base (NH₄⁺) and the conjugate of a strong acid (NO₃^–) and will produce an acidic solution.

18. C The best buffered solution occurs when pH = pK_a. That happens when the solution contains equal amounts of acid and conjugate base. If you want to create a buffer with a pH of 5, the best choice would be an acid with a pK_a that is as close to 5 as possible. You shouldn’t have to do a calculation to see that the pK_a for choice (C) is much closer to 5 than that of any of the others.

19. C An amphoteric species can act either as an acid or a base, gaining or losing a proton.

HCO₃^– can act as an acid, losing a proton to become CO₃^2–, or it can act as a base, gaining a proton to become H₂CO₃.

20. D We want to change the hydrogen ion concentration from 0.1 M (pH of 1) to 0.01 M (pH of 2).

The HCl is completely dissociated, so the number of moles of H⁺ will remain constant as we dilute the solution.

Moles = (molarity)(volume) = Constant

(M₁)(V₁) = (M₂)(V₂)

(0.1 M)(1 L) = (0.01 M)(V₂)

So, V₂ = 10 L, which means that 9 L must be added.

21. A A pH of 5 means that [H⁺] = 1 × 10^–5

For every HA that dissociates, we get one H⁺ and one A^–, so [H⁺] = [A^–] = 1 × 10^–5.

The acid is weak, so we can assume that very little HA dissociates and that the concentration of HA remains 1-molar.

22. A For conjugates, (K_a)(K_b) = K_w = 1 × 10^–14

23. D First let’s find out how many moles of H⁺ ions we need to neutralize.

Every H₂SO₃ will produce 2 H⁺ ions, so for our purposes, we can think of the solution as a
0.4-molar H⁺ solution.

Moles = (molarity)(volume)

Moles of H⁺ = (molarity)(volume) = (0.4 M)(100 ml) = 40 millimoles = moles of OH^– required

Volume of NaOH = = = 400 ml

24. C Let’s look at the equilibrium expression for the dissociation of HC₂H₃O₂.

When HCl is added to the solution, H⁺ ions are added and [H⁺] will increase, so (I) is correct.

Because of Le Châtelier’s law, the equilibrium will shift to consume some of the added H⁺ ions, so some of the H⁺ ions will combine with C₂H₃O₂^– ions to form more HC₂H₃O₂.

So [C₂H₃O₂^–] will decrease, making (II) wrong, and [HC₂H₃O₂] will increase, making (III) correct.

25. C An amphoteric species can act either as a base and gain an H⁺ or as an acid and lose an H⁺.

HSO₄^– can lose an H⁺ to become SO₄^2– or it can gain an H⁺ to become H₂SO₄.

HNO₃ (A), HC₂H₃O₂ (B), and H₃PO₄ (D) can only lose an H⁺.

ClO₄^– (E) can only gain an H⁺.

26. D Every unit of HNO₃ added to the solution will place 1 unit of H⁺ ions in the solution. So first find the moles of HNO₃ added.

Moles =

Moles of H⁺ = = 0.01 moles

Now it’s easy to find the H⁺ concentration.

Molarity =

[H⁺] = = 0.01 M

pH = –log [H⁺] = – log(0.01) = 2

27. D A buffered solution can be prepared by mixing a weak acid with an equal amount of its conjugate or by adding enough strong base to neutralize half of the weak acid present in a solution.

In (I), equal amounts of a strong acid and base are mixed. They’ll neutralize each other completely to produce salt water, which is not a buffer.

In (II), enough strong base is added to neutralize half of the weak acid. This will leave equal amounts of weak acid and its conjugate base, producing a buffered solution.

In (III), equal amounts of a weak acid and its conjugate base are mixed. This will produce a buffered solution.

28. A A Lewis acid must be able to accept a pair of electrons. By drawing the Lewis structure of AlBr₃, shown below, one can see that Al lacks a full octet and therefore is a very strong Lewis acid.

NH₃ is a good Lewis base, since it has a lone pair of electrons to donate. Both CH₄ and Xe have full octets, and will not accept more electrons. The compound PCl₃ looks a lot like NH₃, and has a lone pair of electrons and acts as a Lewis base.

29. E The half equivalence point of this titration, shown in the plot below, is around pH = 9.8.

However, this is not the pK_b of the weak base. Remember that the Henderson-Hasselbalch equation relates pH with pK_a, not pK_b. At the half equivalence point shown, the amount of base (which we can call [A^–] as we do when talking about acids and pH values) is equal to the neutralized base [HA]. When these two quantities are equal, the H-H equation tells us that the pH = the pK_a of the acid form, which when subtracted from 14 gives the pK_b of the base. So, the pK_a of the acid form is 9.8, so the pK_b of the base is 4.2.

30. E Use the equilibrium expression

For every HA that dissociates, we get one H⁺ and one A^–, so [H⁺] = [A^–] = y.

The acid is weak, so we can assume that very little HA dissociates and that the concentration of HA remains 1-molar.

So [H⁺] = y =

PROBLEMS

1. (a) First let’s find out how many moles of NaOH were added.

Moles = (molarity)(volume)

Moles of NaOH = (0.100 M)(.040 L) = 0.004 moles

Every OH^– ion neutralizes one H⁺ ion, so at the equivalence point, the number of moles of NaOH added is equal to the number of moles of HOCl originally present.

(moles of NaOH) = (moles of HOCl) = 0.004 moles

Now we can find the original concentration of the HOCl solution.

Molarity =

[HOCl] = = 0.040 M

(b) Use the equilibrium expression.

[H⁺] = 10^–pH = 10^–4.46 = 3.47 × 10^–5 M

[H⁺] = [OCl^–] = 3.47 × 10^–5 M

[HOCl] = 0.040 M

0.040 is much larger than 3.47 × 10^–5, so we can simplify the expression.

K_a= = 3.00 × 10^–8M

% ionized = = 0.087%

(d) At the equivalence point, all of the HOCl initially present has been converted into OCl^– ions.

From (a), we know that there were initially 0.004 moles of HOCl, so at equivalence, there are 0.004 moles of OCl^–.

At equivalence, we have added 40 ml to the 100 ml of solution originally present, so we must take this into account in our concentration calculation.

Molarity =

[OCl^–] = = 0.029 M

(e) We know the K_a for HOCl, so we can use it to find the K_b for OCl^–.

K_b = = 3.33 × 10^–7

Now use the K_b expression.

[HOCl] = [OH^–] = x

[OCl^–] = 0.029 M

3.33 × 10^–7 =

Let’s assume that 0.029 M is much larger than x. This simplifies the expression.

3.33 × 10^–7 =

x² = (3.33 × 10^–7)(0.029 M) = 9.67 × 10^–9

x = 9.83 × 10^–5 = [OH^–]

pOH = –log[OH^–] = 4

pH = 14 – pOH = 14 – 4 = 10

2. (a) Use the equilibrium expression for H₂S to find [H⁺].

[H⁺] = [HS^–] = x

[H₂S] = 0.100 M

K₁ = 1.0 × 10^–7

1.0 × 10^–7 =

Let’s assume that 0.100 is much larger than x. That simplifies the expression.

1.0 × 10^–7 =

x² = (1.0 × 10^–7)(0.100) M² = 1.0 × 10^–8 M²

x = 1.0 × 10^–4 M = [H⁺]

pH = –log[H⁺] = –log(1.0 × 10^–4) = 4.0

(b) At the midpoint, pH = pK₁

Let’s look at the Henderson-Hasselbalch expression.

pH = pK + log

When pH = pK, [HS^–] must be equal to [H₂S], making equal to one and log equal to zero.

For every molecule of H₂S neutralized, one unit of HS^– is generated, so we must add enough NaOH to neutralize half of the HS^– initially present. (We are assuming that the further dissociation of HS^– into H⁺ and S_2– is negligible and can be ignored.)

Let’s find out how many moles of H₂S were initially present.

Moles = (molarity)(volume)

Moles of H₂S = (0.100 M)(0.500 L) = 0.050 moles

We need to neutralize half of that, or 0.025 moles, so we need 0.025 moles of NaOH.

Volume =

Volume of NaOH = = 0.250 L = 250 ml

(c) The concentrations of the H₂S solution and NaOH solution are both 0.100 M, so the first 500 ml of NaOH solution will completely neutralize the H₂S. The final 300 ml of NaOH solution will neutralize HS^– molecules.

Initially, there were 0.050 moles of H₂S, so after 500 ml of NaOH solution was added, there were 0.050 moles of HS^–.

By adding 300 ml more of NaOH solution, we added (0.100 M)(0.300 L) = 0.030 moles of NaOH. [Moles = (molarity)(volume)]

Every unit of NaOH added neutralizes one unit of HS^–, so when all the NaOH has been added, we have 0.030 moles of S^2– and (0.050 moles) – (0.030 moles) = 0.020 moles of HS^–.

Now we can use the Henderson-Hasselbalch expression to find the pH. We can use the number of moles we just calculated (0.030 for S^2– and 0.020 for HS^–) instead of concentrations because in a solution the concentrations are proportional to the number of moles.

pH = pK + log

pK = –log(1.3 × 10^–13) = 12.9

pH = 12.9 + 0.18 = 13.1.

(d) The reaction (H₂S 2 H⁺ + S^2–) is the sum of the two acid dissociation reactions below.

H₂S H⁺ + HS^–

HS^– H⁺ + S^2–

If one reaction is the sum of two other reactions, then its equilibrium constant will be the product of the equilibrium constants for the other two reactions.

So K_eq = K₁K₂ = (1.0 × 10^–7)(1.3 × 10^–13) = 1.3 × 10^–20

3. (a) pK_b = –log K_b = –log(1.79 × 10^–5) = 4.75

(b) This is a buffered solution, so we’ll use the Henderson-Hasselbalch expression.

First let’s find [NH₄⁺] and [NH₃].

Moles = (molarity)(volume)

Moles of NH₄⁺ = (0.100 M)(0.100 L) = 0.010 moles

Moles of NH₃ = (0.200)(0.080 L) = 0.016 moles

When we mix the solutions, the volume becomes (0.100 L) + (0.080 L) = 0.180 L.

Molarity =

[NH₄⁺] = = 0.056 M

[NH₃] = = 0.089 M

Now we can use the Henderson-Hasselbalch expression for bases.

pOH = pK + log

pK = 4.75

pOH = 4.75 + (–0.20) = 4.55

pH = 14 – pOH = 14 – 4.55 = 9.45

Moles =

Moles of NaOH = = 0.005 mol

When NaOH is added to the solution, the following reaction occurs:

NH₄⁺ + OH^– → NH₃ + H₂O

So for every unit of NaOH added, one ion of NH₄⁺ disappears and one molecule of NH₃ appears. We can use the results of the molar calculations we did in part (b).

Moles of NH₄⁺ = (0.010) – (0.005) = 0.005 moles

Moles of NH₃ = (0.016) + (0.005) = 0.021 moles

Now we can use the Henderson-Hasselbalch expression. We can use the number of moles we just calculated (0.005 for NH₄⁺ and 0.021 for NH₃) instead of concentrations because in a solution the concentrations will be proportional to the number of moles.

pOH = pK + log

pK = 4.75

pOH = 4.75 + (–0.62) = 4.13

pH = 14 – pOH = 14 – 4.13 = 9.87

(d) When equal quantities of a base (NH₃) and its conjugate acid (NH₄⁺) are mixed in a solution, the pOH will be equal to the pK_b.

From (a), pOH = pK_b = 4.75

pH = 14 – pOH = 14 – 4.75 = 9.25

ESSAYS

4. (a) An amphoteric species can act either as an acid or base.

H2PO is amphoteric.

Conjugate base: HPO

Conjugate acid: H₃PO₄

HPO is amphoteric.

Conjugate base: PO

Conjugate acid: H₂PO

(b) The smaller the dissociation constant, the more tightly the acid holds its hydrogen ions and the weaker the acid becomes.

After a hydrogen ion has been removed, the remaining species has a negative charge that attracts the remaining hydrogen ions more strongly.

For H₂PO₄^–, pK_a = –log(6.2 × 10^–8) = 7.2

The best buffer solution is made with an acid whose pK_a is approximately equal to the desired pH.

(d)

The axes should have the correct labeling (pH and NaOH added).

The curve should show rising pH, and it should have three bumps.

5. (a) The solution will be basic. A salt composed of the conjugate of a strong base (Na⁺) and the conjugate of a weak acid (C₂H₃O₂^–) will create a basic solution in water.

NaC₂H₃O₂ dissociates into Na⁺ and C₂H₃O₂^–.

Na⁺ does not react with water but C₂H₃O₂^– does.

C₂H₃O₂^– + H₂O HC₂H₃O₂ + OH^–

(b) The solution will be acidic. A salt composed of the conjugate of a strong acid (Cl^–) and the conjugate of a weak base (NH₄⁺) will create an acidic solution in water.

NH₄Cl dissociates into NH₄⁺ and Cl^–.

Cl^– does not react with water but NH₄⁺ does.

NH₄⁺ + H₂O NH₃ + H₃O⁺

(c) For a buffer solution, it is necessary for both an acid (HA) and its conjugate base (A^–) to be present in the solution.

By definition, strong acids dissociate completely, so the conjugate base will be too weak to accept any hydrogen atoms that are added to the solution.

Weak acids do not dissociate completely, so it is possible to have a solution that contains both undissociated acid (HA) and conjugate base (A^–). The undissociated acid can react with added hydroxide, and the conjugate base can react with added hydrogen ions.

(d) KOH separates into K⁺ and OH^– in water because the ionic bond between O and K in the unit can be broken by water, which is a polar solvent.

HBrO separates into H⁺ and BrO^– because the covalent O–Br bond pulls electrons away from the O–H bond. This makes the O–H bond more polar and more easily broken by water, a polar solvent.

6. (a) HCl is the stronger acid. The bond between H and Cl is weaker than the bond between H and F, so HCl dissociates more easily.

(b) H₂SO₃ is stronger. As a polyprotic acid gives up hydrogen ions, it becomes negatively charged, causing it to hang onto its remaining hydrogen ions more tightly, which weakens it as an acid.

(c)

HNO₃ is stronger. As more oxygens are added to the central atom of an oxyacid, the oxidation number of the central atom increases.

In HNO₂, the central atom, N, has an oxidation state of +3.

In HNO₃, N has an oxidation state of +5.

The increased oxidation state of N makes the N–O bond stronger. This draws electrons away from the O–H bond, making it more polar or weaker. The weakness of the O–H bond makes the acid stronger.

(d)

HClO₄ is stronger. Cl has greater electronegativity than Br, so Cl draws electrons away from O in the Cl–O bond; this makes the O–H bond more polar or weaker. A weak O–H bond makes for a strong acid.

11

Acids and Bases

HOW OFTEN DO ACIDS AND BASES APPEAR ON THE EXAM?

DEFINITIONS

ARRHENIUS

BRØNSTED-LOWRY

LEWIS

pH

WEAK ACIDS

WEAK ACIDS AND CALCULATIONS

STRONG ACIDS

STRONG ACIDS AND CALCULATIONS

K w

ACID AND BASE SALTS

ACID-BASE SALTS AND CALCULATIONS

BUFFERS

POLYPROTIC ACIDS AND AMPHOTERIC SUBSTANCES

ANHYDRIDES

TITRATION

CHAPTER 11 QUESTIONS

MULTIPLE-CHOICE QUESTIONS

PROBLEMS

ESSAYS

CHAPTER 11 ANSWERS AND EXPLANATIONS

MULTIPLE-CHOICE QUESTIONS

PROBLEMS

ESSAYS