University:
North Central College
Course:
CHEM 121 | General Chemistry I
Academic year:

2019
Views:

195

Pages:

14
Author:

5CelestialChronicle

Week Acid and Weak Base Questions and Problems - Set 1 Because the answers are on this sheet, I expect to see all of your work if you want credit for completing this assignment. 1 1. 1. Formic acid, HCOOH, is a monoprotic acid. In a 0.100 M solution of formic acid, the pH is 2.38 at 25°C. Calculate the Ka for formic acid at this temperature. (answer: 1.8 x 10⁻⁴) HCOOH ⇌ H⁺ + HCOO⁻ 1.00M - x x x Ka = (x²)/(0.100 - x) ≈ x² / 0.100 10⁻².³⁸ = 0.0042 = x = [H⁺] Ka = (0.0042)² / 0.100 = 1.8 x 10⁻⁴ Calculate the pH and [H⁺] of a 0.100 M solution of acetic acid, CH₃COOH, Ka = 1.8 x 10⁻⁵. (answer: 2.87, 0.00134 M) 1.8 x 10⁻⁵ = x² / 0.100 x = 0.0013 = [H⁺] pH = -log(0.0013) = 2.87 Which of the following solutions has the highest pH? (a) 0.40 M HNO₂ (b) 0.40 M HClO (c) 0.40 M CH₃COOH Ka = 4.5 x 10⁻⁴ Ka = 3.5 x 10⁻⁸ Ka = 1.8 x 10⁻⁵ Lowest [H⁺] takes highest pH The Kₐ for benzoic acid is 6.5 x 10⁻⁵. Calculate the concentrations of all the species (C₆H₅COOH, C₆H₅COO⁻, and H⁺) in a 0.10 M benzoic acid solution. (answer: 0.10 M, 0.0025 M, 0.0025 M, and 4 x 10⁻¹²) 6.5 x 10⁻⁵ = x² / 0.10 x = 0.0025 = [C₆H₅COO⁻] = [H⁺] [C₆H₅COOH] = 0.10 - 0.0025 ≈ 0.10 [OH⁻] = 1 x 10⁻¹⁴ / [H⁺] = 4 x 10⁻¹² Calculate the percent ionization in a 0.20 M solution of hydrofluoric acid, HF. Ka = 6.7 x 10⁻⁴. (answer: 5.7%) 6.7 x 10⁻⁴ = x² / 0.20 x = 0.012 = [H⁺] (0.012 / 0.20) x 100% = 6% The pH of a 0.10 M HCN solution is found to be 5.2. What is the Ka of HCN? (answer: 4 x 10⁻¹⁰) 10⁻⁵.² = 6.31 x 10⁻⁶ = [H⁺] Ka = (6.31 x 10⁻⁶)² / 1 = 3.98 x 10⁻¹¹ Calculate the pH of a 16.0 M NH₃ solution. The Kb of ammonia is 1.8 x 10⁻⁵. (answer: 12.2) 1.8 x 10⁻⁵ = x² / 16 x = 0.016 = [OH⁻] pOH = -log(0.016) = 1.8 pH = 14 - 1.8 = 12.2 The pH of a 1.0 M CH₃NH₂ solution is 12.32. What is the Kb of methylamine? (answer: 4.5 x 10⁻⁴) pOH = 14 - 12.32 = 1.68 10⁻¹.⁶⁸ = 0.021 = [OH⁻] Kb = (0.021)² / 1 = 4.41 x 10⁻⁴ 9. What is the initial molarity of formic acid (HCOOH; Ka = 6.9 x 10⁻⁴) whose pH is 3.26 at equilibrium? (answer: 5.5 x 10⁻⁴) 10⁻³².⁶ = 5.5 x 10⁻⁴ = [H⁺] 6.9 x 10⁻⁴ = (5.5 x 10⁻⁴)² / x x = 4.37 x 10⁻⁴ M 10. Calculate the percent ionization of a weak acid (HA, Ka = 6.5 x 10⁻⁹) solution at the following concentrations: (a) 0.20 M (b) 0.00020 M (answer: 1.80%; 43%) a) 6.5 x 10⁻⁹ = x² / 0.20 x = 0.0011 % ionization = (0.0011 / 0.20) x 100 = 1.8% b) 6.5 x 10⁻⁹ = x² / 0.00020 x² + 6.5 x 10⁻⁵x - 1.3 x 10⁻¹¹ = 0 x = 0.0001 % ionization = (0.0001 / 0.00020) x 100 = 43% 11. A 0.04 M solution of a weak acid, HA, in water is 14% ionized. Calculate the Ka of this acid. (answer: 6.1 x 10⁻⁴) HA ⇌ H⁺ + A⁻ 0.04 0 0 -0.0056 +0.0056 +0.0056 0.0344 0.0056 0.0056 Ka = (0.0056)² / 0.0344 = 9.1 x 10⁻⁴ 12. What is the pH of a 1.0 M NH₃ solution? Start by writing the appropriate ionization reaction and a Kb expression. The Kb of ammonia is 1.8 x 10⁻⁵. (answer: 11.6) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ 1.8 x 10⁻⁵ = x² / 1 x = 0.0042 = [OH⁻] pOH = -log(0.0042) = 2.4 pH = 14 - 2.4 = 11.6 13. What is the H⁺ ion concentration in a 0.82 M HOCl solution? Ka = 3.2 x 10⁻⁸. (answer: 1.62 x 10⁻⁴ M) HOCl ⇌ H⁺ + OCl⁻ 3.2 x 10⁻⁸ = x² / 0.82 x = 1.62 x 10⁻⁴ = [H⁺] 14. A 1.0 M HF solution is only 2.6% ionized. What is the Ka value for HF? (answer: 6.8 x 10⁻⁴) [H⁺] / 1.0 x 100 = 2.6 [H⁺] = 0.026 Ka = (0.026)² / 1 = 6.8 x 10⁻⁴ 15. Calculate the pH and the % ionization in a 1.0 M HNO₂ solution. Ka = 5.1 x 10⁻⁴. (answer: 1.6; 2.3%) HNO₂ ⇌ H⁺ + NO₂⁻ 5.1 x 10⁻⁴ = x² / 1 x = 0.023 = [H⁺] pH = -log(0.023) = 1.6 % ionization = (0.023 / 1) x 100 = 2.3% Titrations Questions and Problems Set 1 1. What determines the pH found at the equivalence point of a titration? When only products are there 2. Predict the pH found at the equivalence point for each of the following titrations, and choose an appropriate indicator: (a) strong acid + strong base: pH ~ 7, indicator: bromothymol blue (b) strong acid + weak base: pH < 7, indicator: methyl red (c) weak acid + strong base: pH > 7, indicator: phenolphthalein (d) weak acid and weak base: bromothymol blue 3. Write a balanced neutralization reaction for each of the examples listed in question 6. (a) ... (b) ... (c) ... (d) ... 4. a) Write a balanced equation for the titration of a 25mL sample of 0.2M HF with 0.2M KOH. HF + OH⁻ → H₂O + F⁻ b) List the ions or molecules present at the equivalence point of the titration, and predict the pH of the solution at the equivalence point. At the equivalence point: F⁻ + H₂O ⇌ HF + OH⁻ c) Calculate the pH at the equivalence point. [F⁻] = 0.1 M Ka = 3.5 x 10⁻⁴ Kb = 2.86 x 10⁻¹¹ 2.86 x 10⁻¹¹ = x² / 0.1 x = 5.35 x 10⁻⁷ = [OH⁻] pOH = 6.27 pH = 14 - 6.27 = 7.73 5. Consider the titration of a 20 mL sample of 0.10M formic acid, HCOOH, with 0.10M NaOH. a) Write the balanced equation for the reaction between HCOOH and NaOH. HCOOH + NaOH → H₂O + NaCOOH b) Calculate the pH of the formic acid solution before the start of the titration. 1.8 x 10⁻⁴ = x² / 0.1 x = 0.0042 = [H⁺] pH = -log(0.0042) = 2.37 c) List the ions or molecules present in your dish when 10mL of the NaOH have been added to the solution of HCOOH. What kind of solution is this? HCOOH, HCOO⁻, Na⁺, H₂O Acidic d) Calculate the pH of the resulting solution when 10mL of the NaOH have been added to the 20mL sample of HCOOH. [HCOOH] = 0.033 M [HCOO⁻] = 0.033 M pH = 3.74 + log(1) = 3.74 e) List the ions or molecules present when the equivalence point of the titration has been reached. H₂O, HCOO⁻, Na⁺ f) Calculate the Kb of formate 1.8 x 10⁻⁴ * Kb = 1 x 10⁻¹⁴ Kb = 5.56 x 10⁻¹¹ g) Calculate the volume of the solution at the equivalence point of the titration of HCOOH with NaOH. 40 mL h) Calculate the pH at the equivalence point of the titration of the HCOOH with NaOH. 5.56 x 10⁻¹¹ = x² / 0.05 x = 1.67 x 10⁻⁶ pH = 8.2 i) List the ions or molecules present when the 25mL of the NaOH have been added to the 20mL HCOOH sample. OH⁻, HCOO⁻, Na⁺, H₂O j) Calculate the pH of the solution formed when 25mL of the NaOH have been added to the 20mL HCOOH sample. [OH⁻] = 0.05 M pOH = -log(0.05) = 1.3 pH = 14 - 1.3 = 12.7

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