8.2 pH and pOH of Strong Acids and Bases

1. A0.03
   A student who treats pH as proportional to concentration picks this, dividing 3.00 by the dilution factor of 100. It is [H₃O⁺] that is divided by 100; the pH, −log[H₃O⁺], rises by 2 to 5.00.
2. B3.00
   A student who thinks the pH of a strong acid is fixed by its strength, not its concentration, picks this. Dilution lowers [H₃O⁺] to 1.0 × 10⁻⁵ M, so the pH becomes 5.00.
3. C1.00
   A student who thinks dilution lowers the pH picks this, taking 1 unit off for each tenfold dilution. A lower [H₃O⁺] means a higher pH: 3.00 + 2 = 5.00.
4. D5.00 Correct
   The dilution factor is 1000 mL/10.0 mL = 100, so [H₃O⁺] falls from 1.0 × 10⁻³ M to 1.0 × 10⁻⁵ M and the pH rises by 2 units, to 5.00.

Working HCl is a strong acid, so [H₃O⁺] = 1.0 × 10⁻³ M before dilution. The volume increases by a factor of 1000 mL/10.0 mL = 100, so [H₃O⁺] = 1.0 × 10⁻⁵ M and pH = 5.00: each tenfold dilution raises the pH by 1.

CED 8.2.A.1 · Read this in Fix

1. ADiagram 1 Correct
   Sr(OH)₂ is a strong base: it dissociates completely, Sr(OH)₂(s) → Sr²⁺(aq) + 2 OH⁻(aq), so the solution contains separate ions with two OH⁻ ions for every Sr²⁺ ion, as in the box with three Sr²⁺ and six OH⁻.
2. BDiagram 2
   A student who thinks every strong base gives one OH⁻ per formula unit picks the box with equal numbers of Sr²⁺ and OH⁻ ions. Each Sr(OH)₂ releases two OH⁻ ions, and a solution with equal numbers would not be electrically neutral.
3. CDiagram 3
   A student who attaches the 2 : 1 ratio to the wrong ion picks the box with twice as many Sr²⁺ ions as OH⁻ ions. One formula unit of Sr(OH)₂ contains one Sr²⁺ and two OH⁻.
4. DDiagram 4
   A student who thinks a dissolved hydroxide stays as whole formula units picks the box of undissociated Sr(OH)₂ units. A strong base dissociates completely into separate Sr²⁺ and OH⁻ ions.

CED 8.2.A.2 · Read this in Fix

1. A12.40
   A student who takes [OH⁻] as equal to the concentration of the hydroxide picks this: pOH = −log(0.025) = 1.60 and pH = 12.40. Each Ba(OH)₂ releases two OH⁻, so [OH⁻] = 0.050 M and the pH is 12.70.
2. B12.70 Correct
   Ba(OH)₂ is a group 2 hydroxide and dissociates completely, giving [OH⁻] = 2 × 0.025 M = 0.050 M. pOH = −log(0.050) = 1.30, and pH = 14.00 − 1.30 = 12.70.
3. C12.10
   A student who halves the concentration picks this: [OH⁻] = 0.0125 M, pOH = 1.90 and pH = 12.10. One formula unit gives two OH⁻ ions, so the concentration is doubled, not halved: pH = 12.70.
4. D12.00
   A student who reads the pOH as the exponent of [OH⁻] = 5.0 × 10⁻² M picks this: pOH = 2 and pH = 12.00. The full logarithm gives pOH = 1.30 and pH = 12.70.

Working Ba(OH)₂(s) → Ba²⁺(aq) + 2 OH⁻(aq), complete. [OH⁻] = 2 × 0.025 M = 0.050 M. pOH = −log(0.050) = 1.30. pH = 14.00 − 1.30 = 12.70.

CED 8.2.A.2 · Read this in Fix

1. ASolution X
   A student who thinks the pH is fixed by which acid is present, rather than by its concentration, picks X, taking HClO₄ to be the strongest acid listed. All three acids are strong; X has [H₃O⁺] = 0.010 M, one-tenth of that in W.
2. BSolution Y
   A student who takes −log of the concentration as the pH for every solution picks Y: −log(1.0) = 0.00. KOH is a strong base, so 0.00 is the pOH and the pH is 14.00, the highest of the four.
3. CSolution W Correct
   For a strong acid [H₃O⁺] is set by the concentration: W has [H₃O⁺] = 0.10 M (pH 1.00), X has 0.010 M (pH 2.00), and Z has at most 0.020 M (pH no lower than 1.70). Y is a strong base with pH 14.00. W has the highest [H₃O⁺] and the lowest pH.
4. DSolution Z
   A student who judges acidity by the number of H atoms in the formula picks Z. Even if both hydrogens ionized completely, [H₃O⁺] in Z would be only 0.020 M, less than the 0.10 M in W.

Working Lowest pH means highest [H₃O⁺]. W: strong acid, [H₃O⁺] = 0.10 M, pH 1.00. X: strong acid, [H₃O⁺] = 0.010 M, pH 2.00. Y: strong base, [OH⁻] = 1.0 M, pOH 0.00, pH 14.00. Z: each H₂SO₄ can provide at most two H₃O⁺, so [H₃O⁺] is at most 0.020 M and the pH is no lower than 1.70. W has the lowest pH.

CED 8.2.A.1 · Read this in Fix

1. AAll three solutions have the same pH, 1.00
   A student who thinks a strong acid has a pH of about 1 whatever its concentration picks this. The pH depends on [H₃O⁺], which is 0.010 M in each solution, so the pH is 2.00.
2. BAll three solutions have the same pH, 2.00 Correct
   HCl, HBr and HI are strong acids: each ionizes completely in water, so each 0.010 M solution has [H₃O⁺] = 0.010 M and pH = −log(0.010) = 2.00.
3. CThe pH decreases from HCl to HBr and to HI
   A student who thinks the strong acids ionize to different extents in water, with HI ionizing most, picks this. All three ionize completely, so [H₃O⁺] = 0.010 M and the pH is 2.00 in each solution.
4. DThe pH increases from HCl to HBr and to HI
   A student who thinks the acid with the most electronegative halogen gives the most H₃O⁺ picks this. All three acids ionize completely in water, so [H₃O⁺] = 0.010 M and the pH is 2.00 in each solution.

CED 8.2.A.1 · Read this in Fix

1. AIt is right: HCl ionizes completely, so [H₃O⁺] is equal to the HCl concentration
   A student who applies [H₃O⁺] = concentration of the acid at any dilution picks this. HCl does ionize completely, but the H₃O⁺ from water (about 1.0 × 10⁻⁷ M) is larger than that from the HCl, and an acid solution cannot have a pH above 7 at 25°C.
2. BIt is wrong: HCl is a strong acid, so the pH stays near 1 however dilute it is
   A student who thinks the pH of a strong acid is fixed by its strength picks this. The pH depends on [H₃O⁺], which here is only slightly above 1.0 × 10⁻⁷ M, giving a pH just below 7.
3. CIt is wrong: this much water neutralizes the acid, so the pH is exactly 7.00
   A student who thinks water neutralizes an acid picks this. Dilution does not remove the H₃O⁺ supplied by the HCl; it adds to the H₃O⁺ from water, so the solution stays slightly acidic, with a pH just below 7.
4. DIt is wrong: water supplies more H₃O⁺ than the HCl, so the pH is below 7.00 Correct
   The rule [H₃O⁺] = concentration of the strong acid neglects the H₃O⁺ from the autoionization of water. Here water supplies about ten times as much H₃O⁺ as the HCl does, so [H₃O⁺] is slightly greater than 1.0 × 10⁻⁷ M and the pH is slightly below 7.00 (about 6.98).

Working A solution of an acid cannot be basic. The autoionization of water alone gives [H₃O⁺] = 1.0 × 10⁻⁷ M at 25°C, ten times the 1.0 × 10⁻⁸ M supplied by the HCl, so it cannot be neglected. Total [H₃O⁺] is a little more than 1.0 × 10⁻⁷ M (about 1.05 × 10⁻⁷ M), so the pH is just below 7 (about 6.98).

CED 8.2.A.1 · Read this in Fix

1. A1.0 × 10⁻² M Correct
   pOH = 14.00 − 12.30 = 1.70, so [OH⁻] = 10^−1.70 = 2.0 × 10⁻² M. Each formula unit of Ca(OH)₂ that dissolves releases two OH⁻ ions, so the Ca(OH)₂ concentration is half of [OH⁻]: 1.0 × 10⁻² M.
2. B2.0 × 10⁻² M
   A student who takes [OH⁻] as equal to the concentration of the dissolved hydroxide picks this. [OH⁻] is 2.0 × 10⁻² M, but each Ca(OH)₂ supplies two OH⁻ ions, so only 1.0 × 10⁻² M of Ca(OH)₂ is needed.
3. C4.0 × 10⁻² M
   A student who thinks two units of Ca(OH)₂ are needed for each OH⁻ ion picks this, doubling [OH⁻] = 2.0 × 10⁻² M. One formula unit gives two OH⁻ ions, so the Ca(OH)₂ concentration is half of [OH⁻], 1.0 × 10⁻² M.
4. D3.5 × 10⁻¹ M
   A student who reads pOH 1.70 as exponent 1 and coefficient 7.0 picks this: [OH⁻] = 7.0 × 10⁻¹ M, halved to 3.5 × 10⁻¹ M. The whole value is the exponent: [OH⁻] = 10^−1.70 = 2.0 × 10⁻² M, giving 1.0 × 10⁻² M of Ca(OH)₂.

Working pOH = 14.00 − 12.30 = 1.70, so [OH⁻] = 10^−1.70 = 2.0 × 10⁻² M. Ca(OH)₂ is a group 2 hydroxide and dissociates completely, Ca(OH)₂ → Ca²⁺ + 2 OH⁻, so [OH⁻] is double the Ca(OH)₂ concentration: [Ca(OH)₂] = (2.0 × 10⁻² M)/2 = 1.0 × 10⁻² M. Distractors: [OH⁻] taken as the Ca(OH)₂ concentration, 2.0 × 10⁻² M; [OH⁻] taken as half the Ca(OH)₂ concentration, 2 × 2.0 × 10⁻² = 4.0 × 10⁻² M; pOH 1.70 split into exponent 1 and coefficient 7.0, [OH⁻] = 7.0 × 10⁻¹ M and half of it, 3.5 × 10⁻¹ M.

CED 8.2.A.2 · Read this in Fix