How to Use This Worksheet
Work in order. Section A checks the ideas: writing Ka and Kb expressions, ranking by constant size, conjugate conversions through Kw, and reading salt solutions. Section B practices the working skills: ICE tables for a weak acid and a weak base, a constant recovered from a pH in each direction, a percent ionization that breaks the approximation, and a dilution pair. Section C is exam style: a basic salt solution built from a conjugate Kb, two acids whose constants differ a hundredfold, and an acidic salt whose constant must be earned through Kw first.
Reference facts. Every temperature is 25 °C, Kw = 1.0 × 10⁻¹⁴, concentrations are in mol L⁻¹, and pH and pOH values are quoted to two decimal places. Each question supplies the constant it needs. Two conjugate pairs recur throughout: Ka of CH₃COOH = 1.8 × 10⁻⁵ with Kb of CH₃COO⁻ = 5.6 × 10⁻¹⁰, and Kb of NH₃ = 1.8 × 10⁻⁵ with Ka of NH₄⁺ = 5.6 × 10⁻¹⁰.
One working habit ties the sheet together: never let an approximation leave the page unchecked. Every ICE calculation ends by computing percent ionization, (x ÷ C) × 100 percent. At 5 percent or below the small-x route stands; above it the quadratic replaces it, and the answer key states which route each question needed and why.
Two checks run through the whole sheet. The pair check: the Ka and Kb of a conjugate pair multiply to 1.0 × 10⁻¹⁴, and their pKa and pKb add to 14.00. The round-trip check: the x you report, substituted back into x² ÷ (C − x), returns the constant you were given. If either check fails, the error is in the working, never in the check.
The sheet also assumes that the ionization being studied is the only equilibrium that matters. Water’s own autoionization is ignored beside the x supplied by the acid or base, which is safe whenever x is far above 10⁻⁷ mol L⁻¹, as it is in every question here. Square brackets always mean equilibrium concentrations.
The answer key shows every ICE row, every logarithm step and every check in full. Do not look at the answer key until you have attempted every question in Sections A, B and C.
Section A: Recall and Concepts
A1. (a) Write the equation for HF ionizing in water, and the expression for its Ka. (b) Write the equation for NH₃ reacting with water, and the expression for its Kb. (c) State what a small Ka value tells you about where the ionization equilibrium lies and about the fraction of molecules ionized.
A2. Five acids and their Ka values: HF 6.8 × 10⁻⁴, HCOOH 1.8 × 10⁻⁴, CH₃COOH 1.8 × 10⁻⁵, H₂CO₃ 4.3 × 10⁻⁷, HCN 4.9 × 10⁻¹⁰. (a) Rank the acids from strongest to weakest. (b) State which acid has the largest pKa, and calculate that pKa. (c) State the general link between the size of pKa and acid strength.
A3. (a) The Ka of an acid HA is 1.0 × 10⁻⁵. Calculate the Kb of its conjugate base A⁻. (b) The Kb of NH₃ is 1.8 × 10⁻⁵. Calculate the Ka of NH₄⁺. (c) For CH₃COOH (Ka = 1.8 × 10⁻⁵), calculate the pKa, and then the pKb of CH₃COO⁻ using the pair relation.
A4. (a) Define percent ionization for a weak acid of concentration C that produces [H₃O⁺] = x at equilibrium. (b) A 0.10 mol L⁻¹ weak acid has [H₃O⁺] = 1.3 × 10⁻³ mol L⁻¹ at equilibrium. Calculate its percent ionization. (c) The same acid is diluted to 0.0010 mol L⁻¹. State what happens to the percent ionization and what happens to [H₃O⁺], and name the principle that explains the change in percent ionization.
A5. Predict whether each solution is acidic, basic or neutral, justifying each answer by naming the ion that reacts with water, if any: (a) NaCH₃COO, (b) NH₄Cl, (c) KNO₃, (d) Na₂CO₃.
Section B: Core Practice
B1. Formic acid, HCOOH, has Ka = 1.8 × 10⁻⁴. Calculate the pH of a 0.20 mol L⁻¹ solution, showing the ICE rows, the small-x approximation, its percent check, and the quadratic value of x.
B2. Calculate the pH of a 0.050 mol L⁻¹ solution of NH₃ (Kb = 1.8 × 10⁻⁵), showing the pOH step explicitly.
B3. A 0.10 mol L⁻¹ solution of a weak acid has pH 3.30. Calculate the Ka of the acid and its percent ionization.
B4. A weak acid HA has Ka = 2.0 × 10⁻⁴. Calculate the percent ionization of HA in a 0.050 mol L⁻¹ solution. Show the approximation result as well as the value you actually report, and state why they differ.
B5. A 0.10 mol L⁻¹ solution of a weak base B has pH 9.80. Calculate the Kb of B.
B6. CH₃COOH has Ka = 1.8 × 10⁻⁵. (a) Calculate the pH and the percent ionization of a 0.10 mol L⁻¹ solution, using the quadratic. (b) Repeat the calculation for a 0.0010 mol L⁻¹ solution. (c) Using your numbers, state why the small-x approximation is acceptable in one case and not in the other.
Section C: Exam Style
C1. Sodium acetate, NaCH₃COO, dissolves completely to give Na⁺ and CH₃COO⁻ ions. Ka of CH₃COOH is 1.8 × 10⁻⁵. (a) Write the equation for the ion that reacts with water, and explain why Na⁺ plays no part. (b) Calculate the Kb of that ion. (c) Calculate the pH of a 0.010 mol L⁻¹ solution of NaCH₃COO. (d) Calculate the fraction of the ion that has reacted at equilibrium, and comment on the approximation used in part (c).
C2. Two weak acids, P (Ka = 1.0 × 10⁻⁴) and Q (Ka = 1.0 × 10⁻⁶), are both made up at 0.10 mol L⁻¹. (a) Calculate [H₃O⁺] and the pH of each solution. (b) Calculate the ratio of the two [H₃O⁺] values. (c) Explain why a hundredfold difference in Ka does not produce a hundredfold difference in [H₃O⁺].
C3. Ammonium chloride, NH₄Cl, dissolves completely to give NH₄⁺ and Cl⁻ ions. Kb of NH₃ is 1.8 × 10⁻⁵. (a) Calculate the Ka of NH₄⁺. (b) Calculate the pH of a 0.10 mol L⁻¹ solution of NH₄Cl. (c) Explain why this solution is acidic even though the salt contains the anion of a strong acid.
Show the answer key and full working (only after you have attempted every question)
A1. (a) HF + H₂O ⇌ H₃O⁺ + F⁻, with Ka = [H₃O⁺][F⁻] ÷ [HF]. (b) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, with Kb = [NH₄⁺][OH⁻] ÷ [NH₃]. Water appears in neither expression: it is the solvent, and its concentration is folded into the constant. Every concentration in both expressions is raised to its stoichiometric coefficient, which is 1 throughout this topic. (c) The equilibrium lies far to the left: only a small fraction of the molecules are ionized at any moment, and most of the acid remains as intact HA.
A2. (a) HF > HCOOH > CH₃COOH > H₂CO₃ > HCN. At equal concentration the larger Ka sits further right and gives the stronger acid. (b) HCN, the weakest acid: pKa = −log(4.9 × 10⁻¹⁰) = 9.31. (c) pKa runs backwards against strength: the larger the pKa, the smaller the Ka and the weaker the acid.
A3. (a) Kb = Kw ÷ Ka = (1.0 × 10⁻¹⁴) ÷ (1.0 × 10⁻⁵) = 1.0 × 10⁻⁹. (b) Ka = Kw ÷ Kb = (1.0 × 10⁻¹⁴) ÷ (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. (c) pKa = −log(1.8 × 10⁻⁵) = 4.74. Then pKb = 14.00 − 4.74 = 9.26 by the pair relation, matching Kb of CH₃COO⁻ = 5.6 × 10⁻¹⁰. The sum 4.74 + 9.26 = 14.00 is the pair check on the two values.
A4. (a) Percent ionization = (x ÷ C) × 100 percent: the concentration that has ionized, expressed as a percentage of the starting concentration. (b) (1.3 × 10⁻³ ÷ 0.10) × 100 = 1.3 percent. (c) The percent ionization increases, while [H₃O⁺] decreases, so the pH rises toward 7. By Le Chatelier’s principle, dilution shifts the equilibrium toward the side with more dissolved particles, the ions, so a greater fraction of the remaining molecules ionizes even though there are fewer of them per liter. The two changes pull in opposite directions, which is exactly why percent ionization and acidity must never be treated as the same quantity.
A5. (a) Basic. CH₃COO⁻ is the conjugate base of the weak acid CH₃COOH and reacts with water to produce OH⁻; Na⁺ is a spectator. (b) Acidic. NH₄⁺ is the conjugate acid of the weak base NH₃ and donates a proton to water; Cl⁻ is a spectator. (c) Neutral. K⁺ comes from the strong base KOH and NO₃⁻ from the strong acid HNO₃, so neither ion reacts with water. (d) Basic. CO₃²⁻ is the conjugate base of the weak acid HCO₃⁻ and reacts: CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻; Na⁺ is a spectator.
B1. ICE rows for HCOOH + H₂O ⇌ H₃O⁺ + HCOO⁻: initial 0.20, 0, 0; change −x, +x, +x; equilibrium 0.20 − x, x, x, with x = [H₃O⁺]. The equilibrium condition is Ka = x² ÷ (0.20 − x) = 1.8 × 10⁻⁴. Approximation: x ≈ √(1.8 × 10⁻⁴ × 0.20) = √(3.6 × 10⁻⁵) = 6.0 × 10⁻³ mol L⁻¹. Percent ionization = 6.0 × 10⁻³ ÷ 0.20 × 100 = 3.0 percent, inside the check, with an error of about 1.5 percent in x. Quadratic: x = 5.9 × 10⁻³ mol L⁻¹. Round trip: (5.9 × 10⁻³)² ÷ (0.20 − 0.0059) returns 1.8 × 10⁻⁴. pH = −log(5.9 × 10⁻³) = 2.23.
B2. ICE rows for NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ give x = [OH⁻] ≈ √(1.8 × 10⁻⁵ × 0.050) = √(9.0 × 10⁻⁷) = 9.5 × 10⁻⁴ mol L⁻¹. Percent ionization = 1.9 percent, so the check passes; the quadratic gives x = 9.4 × 10⁻⁴ mol L⁻¹. Then pOH = −log(9.4 × 10⁻⁴) = 3.03 and pH = 14.00 − 3.03 = 10.97. Round trip: (9.4 × 10⁻⁴)² ÷ (0.050 − 0.00094) returns 1.8 × 10⁻⁵. The pH is reached only through the pOH: taking −log of [OH⁻] and reporting it as a pH is the standing error in base problems.
B3. The pH fixes x directly: x = [H₃O⁺] = 10⁻³.³⁰ = 5.0 × 10⁻⁴ mol L⁻¹. Then Ka = x² ÷ (C − x) = (5.0 × 10⁻⁴)² ÷ (0.10 − 0.0005) = 2.5 × 10⁻⁶. Percent ionization = 5.0 × 10⁻⁴ ÷ 0.10 × 100 = 0.50 percent. No quadratic is needed in this direction: a measured pH is already an equilibrium value.
B4. Approximation: x ≈ √(2.0 × 10⁻⁴ × 0.050) = √(1.0 × 10⁻⁵) = 3.2 × 10⁻³ mol L⁻¹, which would give a percent ionization of 6.3 percent. That is above the 5 percent line, so the approximation is not accepted. Quadratic: x = 3.1 × 10⁻³ mol L⁻¹, and the percent ionization to report is 3.1 × 10⁻³ ÷ 0.050 × 100 = 6.1 percent. The approximation overstates x by about 3 percent because the amount ionized is a real share of 0.050 mol L⁻¹, so replacing C − x by C is no longer harmless.
B5. First the pOH: pOH = 14.00 − 9.80 = 4.20. Then x = [OH⁻] = 10⁻⁴.²⁰ = 6.3 × 10⁻⁵ mol L⁻¹. So Kb = x² ÷ (C − x) = (6.3 × 10⁻⁵)² ÷ (0.10 − 0.000063) = 4.0 × 10⁻⁸. The percent ionization is 6.3 × 10⁻⁵ ÷ 0.10 × 100 = 0.063 percent. The subtraction in the denominator changes nothing at this precision, but the route matters: pH to pOH to [OH⁻] to Kb, never pH straight into the Kb expression.
B6. (a) Quadratic: x² + (1.8 × 10⁻⁵)x − 1.8 × 10⁻⁶ = 0 gives x = 1.33 × 10⁻³ mol L⁻¹, so pH = 2.88 and percent ionization = 1.33 percent. Round trip: (1.33 × 10⁻³)² ÷ (0.10 − 0.00133) returns 1.8 × 10⁻⁵. The approximation gives x = 1.34 × 10⁻³ mol L⁻¹, an error of 0.7 percent. (b) Quadratic: x = 1.25 × 10⁻⁴ mol L⁻¹, so pH = 3.90 and percent ionization = 12.5 percent. (c) At 0.10 mol L⁻¹ only 1.33 percent ionizes, below the 5 percent line, so dropping x beside C costs under 1 percent in x and the approximation is acceptable. At 0.0010 mol L⁻¹ the ionized share is 12.5 percent, x is a real part of C, and the approximation returns 1.34 × 10⁻⁴ mol L⁻¹, about 7 percent too large: only the quadratic is honest there. Dilution raised the percent ionized roughly tenfold while [H₃O⁺] fell roughly tenfold.
C1. (a) CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Na⁺ is the cation of the strong base NaOH; it has no proton to give and no tendency to take one, so it is a spectator and the acetate ion alone sets the pH. (b) Kb = Kw ÷ Ka = (1.0 × 10⁻¹⁴) ÷ (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. (c) x = [OH⁻] ≈ √(5.6 × 10⁻¹⁰ × 0.010) = √(5.6 × 10⁻¹²) = 2.4 × 10⁻⁶ mol L⁻¹. pOH = −log(2.4 × 10⁻⁶) = 5.63, and pH = 14.00 − 5.63 = 8.37. (d) Fraction reacted = 2.4 × 10⁻⁶ ÷ 0.010 = 2.4 × 10⁻⁴, about 0.024 percent. So little of the ion reacts that neglecting x beside 0.010 in the denominator changes nothing at the precision quoted: the approximation in part (c) is fully justified.
C2. (a) Acid P: approximation x ≈ √(1.0 × 10⁻⁴ × 0.10) = 3.2 × 10⁻³ mol L⁻¹, a percent ionization of 3.2 percent, which passes the check; the quadratic gives x = 3.1 × 10⁻³ mol L⁻¹ and pH = 2.51. Acid Q: x ≈ √(1.0 × 10⁻⁶ × 0.10) = 3.2 × 10⁻⁴ mol L⁻¹, 0.32 percent ionized, and the quadratic agrees at this precision; pH = 3.50. (b) Ratio = 3.1 × 10⁻³ ÷ 3.2 × 10⁻⁴ ≈ 10 (9.9 from the unrounded roots). (c) At the same concentration, x ≈ √(Ka × C), so [H₃O⁺] scales with the square root of Ka. A factor of 100 in Ka becomes √100 = 10 in [H₃O⁺], which is why the pH values sit about one unit apart, 2.51 against 3.50, rather than two. The square root is already visible in Ka = x² ÷ (C − x): doubling [H₃O⁺] at fixed C would quadruple the numerator, so the constant, not the concentration, carries the full hundredfold factor.
C3. (a) Ka of NH₄⁺ = Kw ÷ Kb of NH₃ = (1.0 × 10⁻¹⁴) ÷ (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. (b) For NH₄⁺ + H₂O ⇌ H₃O⁺ + NH₃, x = [H₃O⁺] ≈ √(5.6 × 10⁻¹⁰ × 0.10) = √(5.6 × 10⁻¹¹) = 7.5 × 10⁻⁶ mol L⁻¹. The percent ionization is 0.0075 percent, so the approximation is safe. pH = −log(7.5 × 10⁻⁶) = 5.13. Round trip: (7.5 × 10⁻⁶)² ÷ (0.10 − 0.0000075) returns 5.6 × 10⁻¹⁰, the Ka earned in part (a). (c) The acidity comes from the cation, not the anion. NH₄⁺ is the conjugate acid of the weak base NH₃, so it donates protons to water and produces H₃O⁺; Cl⁻, the conjugate base of strong HCl, does not react with water at all. A salt solution is judged ion by ion, never by the salt’s name.