8.1 Introduction to Acids and Bases

1. ABoth are acceptable, but they give different K_w values, as H⁺ and H₃O⁺ differ
   A student who thinks H⁺(aq) and H₃O⁺(aq) are different ions picks this. They are the same ion written two ways, so [H⁺] = [H₃O⁺] and both equations give K_w = 1.0 × 10⁻¹⁴ at 25°C.
2. BBoth are acceptable, as H⁺(aq) and H₃O⁺(aq) stand for the same aqueous ion Correct
   H⁺(aq) and H₃O⁺(aq) are two symbols for the same aqueous ion of hydrogen. Both equations describe the autoionization of water and give the same K_w; H₃O⁺ is preferred, and H⁺ is accepted on the AP Exam.
3. CNeither is acceptable, as pure water consists only of H₂O molecules, not ions
   A student who thinks pure water contains no ions picks this. Water autoionizes to a small extent, so pure water contains H₃O⁺ and OH⁻, and both equations describe that process.
4. DThe first alone is acceptable, as H₃O⁺(aq) is made by acids and not by pure water
   A student who thinks H₃O⁺ forms only when an acid is added picks this. Pure water produces H₃O⁺ by its own autoionization, so the second equation is correct and is the preferred form.

Working No calculation. H⁺(aq) and H₃O⁺(aq) are interchangeable symbols for the aqueous hydrogen ion (H₃O⁺ preferred, H⁺ accepted). Both equations describe autoionization and give K_w = [H⁺][OH⁻] = [H₃O⁺][OH⁻].

CED 8.1.A.1 · Read this in Fix

1. ADiagram 1 Correct
   Water autoionizes, 2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq), to a very small extent, giving equal numbers of H₃O⁺ and OH⁻ ions among a great majority of H₂O molecules. The box with eight H₂O molecules, one H₃O⁺ and one OH⁻ shows the right kinds of particles, with H₂O the most common.
2. BDiagram 2
   A student who thinks pure water contains no ions picks the box with only H₂O molecules. Water autoionizes, so pure water always contains equal, small concentrations of H₃O⁺ and OH⁻.
3. CDiagram 3
   A student who thinks equilibrium means comparable amounts of reactant and products picks the box in which most of the particles are ions. K_w is tiny, so water molecules vastly outnumber the ions.
4. DDiagram 4
   A student who thinks ionization breaks water into hydrogen and oxygen picks the box with H₂ and O₂. Autoionization is a proton transfer between water molecules, producing H₃O⁺ and OH⁻.

Working No calculation. 2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) with K_w = 1.0 × 10⁻¹⁴: pure water contains equal, very small numbers of H₃O⁺ and OH⁻, and mostly H₂O. The box with eight H₂O, one H₃O⁺ and one OH⁻ fits.

CED 8.1.A.2 · Read this in Fix

1. AH₃O⁺ and OH⁻ have equal concentrations Correct
   A neutral solution is one in which [H₃O⁺] = [OH⁻], so pH = pOH. Their common value, √K_w, depends on temperature, but their equality defines neutrality at any temperature.
2. BThe pH of the solution is equal to 7.00
   A student who thinks neutral always means pH 7.00 picks this. pH 7.00 is neutral only at 25°C; at other temperatures K_w differs and so does the neutral pH.
3. CBoth [H₃O⁺] and [OH⁻] equal 1.0 × 10⁻⁷ M
   A student who thinks K_w is 1.0 × 10⁻¹⁴ at every temperature picks this. In a neutral solution [H₃O⁺] = [OH⁻] = √K_w, which equals 1.0 × 10⁻⁷ M only at 25°C.
4. DThe solution contains no H₃O⁺ or OH⁻ ions
   A student who thinks pure water and neutral solutions contain no ions picks this. Water autoionizes, so every aqueous solution contains both ions; in a neutral solution their concentrations are equal.

Working No calculation. Neutral means [H₃O⁺] = [OH⁻] (pH = pOH). The pH of a neutral solution equals ½pK_w, which is 7.00 only at 25°C, and [H₃O⁺] = [OH⁻] = √K_w, which is 1.0 × 10⁻⁷ M only at 25°C.

CED 8.1.A.3 · Read this in Fix

1. A8.80 Correct
   Use K_w at 50°C: [H₃O⁺] = 5.5 × 10⁻¹⁴/3.5 × 10⁻⁵ = 1.57 × 10⁻⁹ M, so pH = −log(1.57 × 10⁻⁹) = 8.80.
2. B9.54
   A student who uses K_w = 1.0 × 10⁻¹⁴ at every temperature picks this: [H₃O⁺] = 1.0 × 10⁻¹⁴/3.5 × 10⁻⁵ = 2.9 × 10⁻¹⁰ M, pH 9.54. At 50°C, K_w = 5.5 × 10⁻¹⁴, giving [H₃O⁺] = 1.57 × 10⁻⁹ M and pH 8.80.
3. C4.46
   A student who takes −log[OH⁻] as the pH picks this. −log(3.5 × 10⁻⁵) = 4.46 is the pOH; the pH comes from [H₃O⁺] = K_w/[OH⁻] = 1.57 × 10⁻⁹ M.
4. D7.00
   A student who thinks [H₃O⁺] stays at 1.0 × 10⁻⁷ M whatever [OH⁻] is picks this. [H₃O⁺] and [OH⁻] are linked by K_w, so [H₃O⁺] = 1.57 × 10⁻⁹ M and pH = 8.80.

Working [H₃O⁺] = K_w/[OH⁻] = 5.5 × 10⁻¹⁴/3.5 × 10⁻⁵ = 1.57 × 10⁻⁹ M. pH = −log(1.57 × 10⁻⁹) = 8.80.

CED 8.1.A.4 · Read this in Fix

1. ASolution W
   A student who thinks a higher [H₃O⁺] means a higher pH picks W, which has the most H₃O⁺ (0.10 M). W has pH 1.00, the lowest; X, with pH 11.30, has the highest pH.
2. BSolution Y
   A student who reads the pH or pOH as the exponent of the concentration takes Y as pH 12 and X as pOH 3, pH 11, and picks Y. Taking the full logarithms, Y has pH 11.10 and X has pOH 2.70 and so pH 11.30.
3. CSolution Z
   A student who takes −log[OH⁻] as the pH picks Z, reading its pH as 12.00. −log[OH⁻] is the pOH; Z has pOH 12.00 and pH 2.00, so it is strongly acidic.
4. DSolution X Correct
   Using pH = −log[H₃O⁺] (with [H⁺] = [H₃O⁺]) and pH = 14.00 − pOH at 25°C: W has pH 1.00, X has pOH 2.70 and so pH 11.30, Y has pH 11.10 and Z has pOH 12.00 and so pH 2.00. X has the highest pH.

Working pH = −log[H₃O⁺]; [H⁺] means [H₃O⁺]; where [OH⁻] is given, pOH = −log[OH⁻] and pH = 14.00 − pOH at 25°C. W: pH 1.00. X: pOH 2.70, pH 11.30. Y: pH = −log(8.0 × 10⁻¹²) = 11.10. Z: pOH 12.00, pH 2.00. Highest pH: X.

CED 8.1.A.1 · Read this in Fix

1. A30 times
   A student who adds a factor of 10 for each pH unit picks this: 10 + 10 + 10 = 30. The factors multiply: 10 × 10 × 10 = 1000.
2. B1000 times Correct
   [H₃O⁺] = 10^−pH, so X has 1 × 10⁻² M and Y has 1 × 10⁻⁵ M. Each pH unit is a factor of 10, and a difference of 3.0 units is a factor of 10³ = 1000.
3. C3 times
   A student who treats the pH scale as linear picks this, taking the difference of 3.0 pH units as a factor of 3. The scale is logarithmic, so the factor is 10³ = 1000.
4. D2.5 times
   A student who treats [H₃O⁺] as inversely proportional to pH picks this: 5.0/2.0 = 2.5. [H₃O⁺] = 10^−pH, so the ratio is 10^{5.0 − 2.0} = 1000.

Working [H₃O⁺] = 10^−pH: X: 1 × 10⁻² M; Y: 1 × 10⁻⁵ M. Ratio = 10⁻²/10⁻⁵ = 10³ = 1000.

CED 8.1.A.1 · Read this in Fix

1. A4.40
   A student who takes −log[H₃O⁺] as the pOH picks this. −log[H₃O⁺] = 4.40 is the pH; the pOH is 14.00 − 4.40 = 9.60.
2. B7.00
   A student who thinks [OH⁻] stays at 1.0 × 10⁻⁷ M whatever [H₃O⁺] is picks this. K_w = [H₃O⁺][OH⁻] links the two, so [OH⁻] = 2.5 × 10⁻¹⁰ M and pOH = 9.60.
3. C9.60 Correct
   pH = −log(4.0 × 10⁻⁵) = 4.40, and at 25°C pH + pOH = 14.00, so pOH = 9.60. Equivalently, [OH⁻] = K_w/[H₃O⁺] = 2.5 × 10⁻¹⁰ M and −log(2.5 × 10⁻¹⁰) = 9.60.
4. D9.00
   A student who takes the pH to be the exponent, 5, picks this: 14.00 − 5 = 9.00. The full logarithm is needed: pH = −log(4.0 × 10⁻⁵) = 4.40, so pOH = 9.60.

Working pH = −log(4.0 × 10⁻⁵) = 4.40. At 25°C, pH + pOH = pK_w = 14.00, so pOH = 14.00 − 4.40 = 9.60. (Check: [OH⁻] = 1.0 × 10⁻¹⁴/4.0 × 10⁻⁵ = 2.5 × 10⁻¹⁰ M; −log = 9.60.)

CED 8.1.A.3 · Read this in Fix

1. AIt rises to 100 times its original value
   A student who thinks [H₃O⁺] and [OH⁻] are always equal picks this. They are equal only in a neutral solution; K_w = [H₃O⁺][OH⁻] means that when one rises, the other falls.
2. BIt falls to 1/100 of its original value Correct
   At 25°C, [H₃O⁺][OH⁻] = K_w = 1.0 × 10⁻¹⁴. If [H₃O⁺] becomes 100 times as great, [OH⁻] must become 1/100 as great to keep the product equal to K_w.
3. CIt stays equal to its original value
   A student who thinks [OH⁻] is unaffected by a change in [H₃O⁺] picks this. The two are linked by K_w = [H₃O⁺][OH⁻], so [OH⁻] falls to 1/100 of its value.
4. DIt becomes zero, as the acid uses up the OH⁻
   A student who thinks an acidic solution contains no OH⁻ picks this. Every aqueous solution contains both ions; [OH⁻] = K_w/[H₃O⁺] is small but never zero.

Working K_w = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ is constant at 25°C. If [H₃O⁺] is multiplied by 100, [OH⁻] must be divided by 100: [OH⁻]new = [OH⁻]old/100.

CED 8.1.A.2 · Read this in Fix

1. AWater molecules are polar, so the molecules themselves carry charge through the sample
   A student who confuses partial charges with ionic charges picks this. Polar water molecules are neutral overall and cannot carry a current; the current is carried by H₃O⁺ and OH⁻ ions.
2. BWater itself forms no ions, so the current is carried by traces of dissolved impurity ions
   A student who thinks pure water contains no ions picks this. Even ultrapure water contains H₃O⁺ and OH⁻ from its own autoionization, and these ions account for its conductivity.
3. CA tiny fraction of water molecules transfer protons to each other, forming a few H₃O⁺ and OH⁻ ions Correct
   A current in solution is carried by moving ions. Autoionization, 2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq), gives pure water 1.0 × 10⁻⁷ M of each ion at 25°C, far less than the ions in 0.0100 M KCl, so the conductivity is small but not zero.
4. DElectrons pass from molecule to molecule through the hydrogen bonds between the water molecules
   A student who pictures current in a solution as electrons flowing through it picks this. In solutions the current is carried by moving ions, here H₃O⁺ and OH⁻.

Working No calculation. Conduction in solution needs mobile ions. Ultrapure water conducts about 1/26,000 as well as 0.0100 M KCl: it contains ions, but at very low concentration, from 2 H₂O ⇌ H₃O⁺ + OH⁻ (1.0 × 10⁻⁷ M each at 25°C).

CED 8.1.A.2 · Read this in Fix

1. A7.00
   A student who thinks pure, neutral water has pH 7.00 at every temperature picks this. At 40°C, K_w ≈ 2.9 × 10⁻¹⁴, so the neutral pH is ½pK_w = 6.77.
2. B7.23
   A student who uses pH + pOH = 14.00 at 40°C picks this, finding pOH = 6.77 and then pH = 14.00 − 6.77 = 7.23. At 40°C pK_w = 13.54, and in pure water pH = pOH = 6.77.
3. C6.54
   A student who keeps [OH⁻] at 1.0 × 10⁻⁷ M picks this: [H₃O⁺] = 2.9 × 10⁻¹⁴/1.0 × 10⁻⁷ = 2.9 × 10⁻⁷ M, pH 6.54. Autoionization forms H₃O⁺ and OH⁻ in equal amounts, so both rise to 1.7 × 10⁻⁷ M.
4. D6.77 Correct
   The graph gives K_w ≈ 2.9 × 10⁻¹⁴ at 40°C. In pure water [H₃O⁺] = [OH⁻], so [H₃O⁺] = √(2.9 × 10⁻¹⁴) = 1.7 × 10⁻⁷ M and pH = 6.77. The water is still neutral.

Working From the graph, K_w at 40°C ≈ 2.9 × 10⁻¹⁴. Pure water is neutral: [H₃O⁺] = [OH⁻] = √(2.9 × 10⁻¹⁴) = 1.7 × 10⁻⁷ M, so pH = −log(1.7 × 10⁻⁷) = 6.77 (equivalently ½pK_w = ½ × 13.54).

CED 8.1.A.4 · Read this in Fix

1. AIt is correct: a basic solution contains OH⁻ ions but contains no H₃O⁺ ions
   A student who thinks a basic solution contains no H₃O⁺ picks this. Every aqueous solution contains both ions; here [H₃O⁺] = K_w/[OH⁻] = 1.0 × 10⁻¹² M.
2. BIt is incorrect: the solution also contains H₃O⁺ ions, at 1.0 × 10⁻⁷ M
   A student who thinks [H₃O⁺] stays at its pure-water value picks this. [H₃O⁺] and [OH⁻] are linked by K_w, so with [OH⁻] = 0.010 M, [H₃O⁺] = 1.0 × 10⁻¹² M.
3. CIt is incorrect: the solution also contains H₃O⁺ ions, at 1.0 × 10⁻¹² M Correct
   Water autoionizes in every aqueous solution, and [H₃O⁺][OH⁻] = K_w = 1.0 × 10⁻¹⁴ at 25°C. With [OH⁻] = 0.010 M, [H₃O⁺] = 1.0 × 10⁻¹² M: far too little to draw to scale, but H₃O⁺ is a kind of ion present, so the claim is wrong.
4. DIt is incorrect: the solution also contains H₃O⁺ ions, at 0.010 M
   A student who thinks [H₃O⁺] and [OH⁻] are always equal picks this. They are equal only in a neutral solution; here [H₃O⁺] = K_w/[OH⁻] = 1.0 × 10⁻¹² M.

Working K_w = [H₃O⁺][OH⁻] applies to every aqueous solution: [H₃O⁺] = 1.0 × 10⁻¹⁴/0.010 = 1.0 × 10⁻¹² M. The solution contains a very small concentration of H₃O⁺, a kind of ion that the diagram omits.

CED 8.1.A.2 · Read this in Fix

1. AHeating pure water from 25°C to 50°C raises its pH from 7.00 to 7.37
   A student who thinks a higher pH means more H₃O⁺ expects more ionization to raise the pH. More H₃O⁺ means a lower pH, so a rise in pH would mean less ionization at higher temperature, an exothermic process.
2. BCooling pure water from 25°C to 0°C lowers its pH from 7.00 to 6.53
   A student who treats heat as a product of an endothermic reaction expects cooling to favor ionization. For an endothermic process cooling shifts the equilibrium toward H₂O, so K_w falls and the pH of pure water rises on cooling.
3. CHeating pure water from 25°C to 50°C increases its pOH from 7.00 to 7.37
   A student who applies pH + pOH = 14.00 at every temperature picks this, reasoning that the pH falls to 6.63 on heating and so the pOH rises to 14.00 − 6.63 = 7.37. In pure water [OH⁻] = [H₃O⁺], so if ionization increases, both pH and pOH fall.
4. DHeating pure water from 25°C to 50°C lowers its pOH from 7.00 to 6.63 Correct
   For an endothermic process, raising the temperature shifts the equilibrium toward products, so K_w increases. In pure water [OH⁻] = [H₃O⁺] = √K_w, so both concentrations rise and both pOH and pH fall, as in a drop from 7.00 at 25°C to 6.63 at 50°C.

Working No calculation. If autoionization is endothermic, raising the temperature shifts it toward the ions, so K_w increases; in pure water [H₃O⁺] = [OH⁻] = √K_w, so both rise and both pH and pOH fall. So heating pure water should lower its pOH (and its pH), for example from 7.00 at 25°C to 6.63 at 50°C.

CED 8.1.A.4 · Read this in Fix

1. A5.0 × 10⁻⁶ M
   A student who treats pH and pOH as interchangeable picks this, taking 10^−5.30 = 5.0 × 10⁻⁶ M as [OH⁻]. That value is [H₃O⁺]; the pOH is 14.00 − 5.30 = 8.70, so [OH⁻] = 2.0 × 10⁻⁹ M.
2. B1.0 × 10⁻⁷ M
   A student who thinks [OH⁻] stays at its pure-water value whatever [H₃O⁺] is picks this. The two concentrations are linked by K_w = [H₃O⁺][OH⁻]; with [H₃O⁺] = 5.0 × 10⁻⁶ M, [OH⁻] = 2.0 × 10⁻⁹ M.
3. C7.0 × 10⁻⁸ M
   A student who finds pOH = 8.70 and then reads 8 as the exponent and .70 as the coefficient picks this. The whole value is the exponent: [OH⁻] = 10^−8.70 = 2.0 × 10⁻⁹ M.
4. D2.0 × 10⁻⁹ M Correct
   At 25°C, pH + pOH = 14.00, so pOH = 8.70 and [OH⁻] = 10^−8.70 = 2.0 × 10⁻⁹ M. Equivalently, [H₃O⁺] = 10^−5.30 = 5.0 × 10⁻⁶ M and [OH⁻] = K_w/[H₃O⁺] = 2.0 × 10⁻⁹ M.

Working At 25°C, pOH = 14.00 − pH = 14.00 − 5.30 = 8.70. [OH⁻] = 10^−8.70 = 2.0 × 10⁻⁹ M. (Check: [H₃O⁺] = 10^−5.30 = 5.0 × 10⁻⁶ M and K_w/[H₃O⁺] = 1.0 × 10⁻¹⁴/5.0 × 10⁻⁶ = 2.0 × 10⁻⁹ M.) Distractors: 10^−pH taken as [OH⁻], 5.0 × 10⁻⁶ M; [OH⁻] left at its pure-water value, 1.0 × 10⁻⁷ M; pOH 8.70 split into exponent 8 and coefficient 7.0, 7.0 × 10⁻⁸ M.

CED 8.1.A.3 · Read this in Fix

1. A2.10
   A student who thinks a larger number on the pH scale means more H₃O⁺ picks this, halving the pH because [H₃O⁺] has been halved: 4.20/2 = 2.10. The scale runs the other way and is logarithmic: less H₃O⁺ means a higher pH, 4.20 + 0.30 = 4.50.
2. B8.40
   A student who treats pH as inversely proportional to [H₃O⁺] picks this, doubling the pH when the concentration is halved: 2 × 4.20 = 8.40. The relationship is logarithmic, so the pH rises by log 2 = 0.30, to 4.50.
3. C4.50 Correct
   pH = −log[H₃O⁺], so halving [H₃O⁺] raises the pH by log 2 = 0.30: 4.20 + 0.30 = 4.50. A lower hydronium ion concentration means a higher pH, and a factor of 2 is much less than one pH unit, which is a factor of 10.
4. D6.20
   A student who treats the pH scale as linear picks this, taking a factor of 2 in [H₃O⁺] as a change of 2 pH units: 4.20 + 2.00 = 6.20. Each pH unit is a factor of 10, so a factor of 2 changes the pH by log 2 = 0.30, to 4.50.

Working pH = −log[H₃O⁺]. Halving [H₃O⁺] changes the pH by −log(½) = +log 2 = +0.30: pH = 4.20 + 0.30 = 4.50. (Check: 10^−4.20 = 6.3 × 10⁻⁵ M; half is 3.2 × 10⁻⁵ M; −log = 4.50.) Distractors: pH taken to fall in step with [H₃O⁺], 4.20/2 = 2.10; pH taken as inversely proportional to [H₃O⁺], 2 × 4.20 = 8.40; a factor of 2 taken as 2 pH units, 4.20 + 2.00 = 6.20.

CED 8.1.A.1 · Read this in Fix