Chem Hydrolysis Questions Answers

Hydrolysis Questions and Problems 1. What requirement must an ion meet in order to react with water? strong enough to ionize water molecule 2. Write the formula for the salts containing the following ions: (a) SO₄²⁻: CaSO₄ (b) NH₄⁺: NH₄Cl (c) CH₃COO⁻: NaCH₃COO (d) PO₄³⁻: AlPO₄ (e) Ca²⁺: CaCO₃ (f) NO₃⁻: NaNO₃ 3. Which of the ions listed in question 2 could react with water rendering it acidic? B + D Which of the ions listed in question 2 could react with water rendering it basic? C + E Which of the ions listed in question 2 could not react with water? A + F 4. Predict the pH of the following aqueous salt solutions. Write hydrolysis reactions to support your predictions. (a) LiCl: pH = 7 (b) CaSO₃: pH > 7 (c) NaHCO₃: pH > 7 (d) Fe(NO₃)₃: pH < 7 5. What determines the pH found at the equivalence point of a titration? The pH of the products 6. Predict the pH found at the equivalence point for each of the following titrations: (a) strong acid + strong base: pH = 7 (b) strong acid + weak base: pH < 7 (c) weak acid + strong base: pH > 7 (d) weak acid and weak base: ? 7. Write a balanced neutralization reaction for each of the examples listed in question 6. a) HCl + NaOH → NaCl + H₂O b) H₂SO₄ + 2B → HB⁺ + H₂O c) HA + OH⁻ → A⁻ + H₂O 8. Calculate the pH of the following aqueous salt solutions (hint: treat these solutions as either a weak acid solution or weak base solution) 1   Buffer Problems Do the following questions and problems on a separate sheet of paper. 1. What are the essential ingredients of a buffer solution? An acid and its conjugate base 2. Identify which of the following result in buffer solutions when equal volumes of the two solutions are mixed. 0.1 M HNO₃ and 0.1 M HNO₂ 0.1 M NaNO₂ and 0.1 M HNO₂ 0.1 M HCl and 0.1 M NH₃ 0.2 M HF and 0.1 M NH₃ 0.1 M HCOOH and 0.2 M NH₃ 3. Calculate the pH of a solution containing 0.150 M acetic acid and 0.300 M sodium acetate. 4. Calculate the ratio [NH₃]/[NH₄⁺] needed in an ammonia/ammonium chloride buffer to produce a pH of 8.55. 5. If you wanted to increase the capacity of a buffer and maintain the pH, what would you do? What is the advantage of a larger capacity buffer? 1   1. 6. A buffer solution is prepared by mixing 500 mL of 0.2 M HNO₂ with 500 mL of 0.40 M NaNO₂. (a) Calculate the pH of the resulting solution. (b) What is the new pH after 2.0 mL of 2.0 M HCl is added to this buffer? 7. The Kₐ values for H₃PO₄, H₂PO₄⁻, and HPO₄²⁻ are 7.5 x 10⁻³, 6.2 x 10⁻⁸, and 4.8 x 10⁻¹³, respectively. (a) You are told to make a buffer having a pH of 7.4, what acid would you start with? (b) What salts could you use to supply H₂PO₄⁻ and HPO₄²⁻ ions? (c) How would you prepare such a buffer? Carry out the appropriate calculations. Solutions: 3. pKa = -log(1.76 x 10⁻⁵) = 4.75 pH = 4.75 + log(0.3/0.15) = 5.05 4. pKa = -log(5.7 x 10⁻¹⁰) = 9.24 8.55 = 9.24 + log([NH₃]/[NH₄⁺]) -0.69 = log([NH₃]/[NH₄⁺]) [NH₃]/[NH₄⁺] = 0.204 : 1 5. Increase both [HA] and [A⁻] and neutralize since larger capacity buffers can neutralize more. 6. [HNO₂] = (0.5 L)(0.2 M) = x (1 L) = 0.1 M [NO₂⁻] = (0.5 L)(0.4 M) = x (1 L) = 0.2 M a) pH = 3.29 + log(0.2/0.1) = 3.59 b) 0.004 mol HCl + 0.004 mol NO₂⁻ → 0.004 mol HNO₂ 0.2 mol NO₂⁻ - 0.004 mol = 0.196 mol NO₂⁻ 0.002 mol HNO₂ + 0.004 mol = 0.006 mol HNO₂ pH = 3.29 + log(0.196/0.006) = 4.57 7. a) H₂PO₄⁻ b) NaH₂PO₄ and Na₂HPO₄ c) 7.4 = 7.2 + log(x) 0.2 = log(x) x = 1.52 M Na₂HPO₄ More Buffer Questions and Problems 1. (a) Calculate the pH of a buffer solution that is 0.25 M HF and 0.50 M NaF. (Ka for HF = 3.5 x 10^-4) pH = 3.46 + log(0.5/0.25) = 3.76 (b) Calculate the % ionization of HF in this buffer solution. (0) HF ⇌ H⁺ + F⁻ 0.25 - x x x 3.5 x 10⁻⁴ = x²/0.25 x = 0.009 M % ionization = (0.009 / 0.25) x 100 = 3.6% (c) Suppose 3.0 mL of 2.0 M HCl is added to 100 mL of the original buffer. What is the new pH of the buffer after neutralization? (3) 0.003 L (2 M) = 0.006 mol H⁺ added H⁺ + F⁻ → HF 0.006 mol 0.025 mol → 0.031 mol / 0.103 L = 0.3 M HF: (0.1 L)(0.25 M) = 0.025 mol + 0.006 mol = 0.031 mol / 0.103 L = 0.3 M pH = 3.46 + log(0.43/0.3) = 3.61 (d) Suppose 3.0 mL of 2.0 M NaOH is added to 100 mL of the original buffer. What is the new pH of the buffer after neutralization? (3.6) 0.003 L (2 M) = 0.006 mol OH⁻ added OH⁻ + HF → H₂O + F⁻ 0.006 mol 0.025 mol → 0.019 mol / 0.103 L = 0.18 M F⁻: 0.05 mol + 0.006 mol = 0.056 mol / 0.103 L = 0.54 M pH = 3.46 + log(0.54/0.18) = 3.94 2. In lab, you are told to make 500 mL of a buffer solution having a pH of 3.94. Using your Kₐ table and your vast storehouse of buffer knowledge, how exactly would you prepare the assigned buffer solution? Specify the chemicals and their concentrations that you would use to prepare this buffer.   ANSWERS WILL VARY pH = pKa + log(x) 3.94 = 3.12 + log(x) 0.82 = log(x) x = 6.6 [H₂PO₄⁻] / [H₃PO₄] = 6.6 [H₂PO₄⁻] = 6.6 M [H₃PO₄] = 1 M 250 mL of each: 6.6 M H₂PO₄⁻ 1 M H₃PO₄ 3. In lab, you are told to make 100 mL of a buffer solution having a pH of 7.8. Using your Kₐ table and your vast storehouse of buffer knowledge, how exactly would you prepare the assigned buffer solution? Specify the chemicals and their concentrations that you would use to prepare this buffer.   ANSWERS WILL VARY pKa H₂PO₄ = 7.21 7.8 = 7.21 + log(x) 0.59 = log(x) x = 3.89 [HPO₄²⁻] / [H₂PO₄⁻] = 3.89 [HPO₄²⁻] = 3.89 M [H₂PO₄⁻] = 1 M 50 mL of each: 3.89 M HPO₄²⁻ 1 M H₂PO₄ Weak 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⁻ 0.100 0 0 -x +x +x 0.100-x x x Ka = (x^2) / (0.100-x) Ka = 1.8 x 10^-4 2. 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) CH₃COOH ⇌ H⁺ + CH₃COO⁻ 0.100 0 0 -x +x +x 0.100-x x x 1.8 x 10⁻⁵ = x² / 0.100 x = 0.00134 = [H⁺] pH = 2.87 3. Which of the following solutions has the highest pH? (a) 0.40 M HNO₂ (b) 0.40 M HCIO (c) 0.40 M CH₃COOH Answer: (c) 0.40 M CH₃COOH *because they are the same concentration, the one with the smallest Ka (CH₃COOH) has the highest pH.* 4. 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. What is the [OH⁻] in this solution? (answer: 0.0025 M, 0.0025 M, 4 x 10⁻¹²) C₆H₅COOH ⇌ C₆H₅COO⁻ + H⁺ 0.10 0 0 -x +x +x 0.10-x x x 6.5 x 10⁻⁵ = x² / (0.10-x) x ≈ 0.0025 = [H⁺] = [C₆H₅COO⁻] [OH⁻] = 4 x 10⁻¹² 5. Calculate the percent ionization in a 0.20 M solution of hydrofluoric acid, HF. Ka = 6.7 x 10⁻⁴. (answer: 5.8%) HF ⇌ H⁺ + F⁻ 0.20 0 0 -x +x +x 0.20-x x x 6.7 x 10⁻⁴ = x² / (0.20-x) x ≈ 0.0116 = [H⁺] % ionization = (0.0116 / 0.20) x 100 = 5.8% 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) HCN ⇌ H⁺ + CN⁻ 0.10 0 0 -x +x +x 0.10-x x x [H⁺] = 10⁻⁵.² Ka = (6.31 x 10⁻⁶)² / (0.10) ≈ 3.98 x 10⁻¹⁰ 7. Calculate the pH of a 1.50 M NH₃ solution. The Kb of ammonia is 1.8 x 10⁻⁵. (answer: 12.2) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ 1.50 0 0 -x +x +x 1.50-x x x 1.8 x 10⁻⁵ = x² / 1.50 x ≈ 0.0164 = [OH⁻] pOH = 1.8 pH = 12.2 8. The pH of a 1.0 M CH₃NH₂ solution is 12.32. What is the Kb of methylamine? (answer: 4.5 x 10⁻⁴) CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH⁻ 1.0 0 0 -x +x +x 1.0-x x x [OH⁻] = 10⁻¹².³² Kb = (0.021)² / (1.0) ≈ 4.5 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? HCOOH ⇌ H⁺ + HCOO⁻ x x x 6.9 x 10^-4 = (x)(x) / (x) x = 6.9 x 10^-4 M (answer: 6.9093 x 10^-4 M) 10. Calculate the percent ionization of a weak acid (HA, Ka = 3.8 x 10^-4) at the following concentrations: a) 0.20 M b) 0.0020 M a) 3.8 x 10^-4 = x² / 0.20 x = 0.0087 M % ionization = (0.0087 / 0.20) x 100 = 4.35% b) 3.8 x 10^-4 = x² / 0.0020 x = 1.14 x 10^-4 M % ionization = (1.14 x 10^-4 / 0.0020) x 100 = 5.7% 11. A 0.04 M solution of a weak acid, HA, in water is 14% ionized. Calculate the Ka of this acid. (answer: 9.1 x 10^-4) 14 / 100 = x / 0.04 x = 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 Kь expression. The Kb of ammonia is 1.8 x 10⁻⁵. (answer: 11.62) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ 1 0 0 -x +x +x 1-x x x 1.8 x 10⁻⁵ = x² / 1 x = 0.0042 = [OH⁻] pOH = 2.38 pH = 11.62 13. What is the H⁺ ion concentration in a 0.82 M HOCl solution? Kₐ = 3.2 x 10⁻⁸. (answer: 1.6 x 10⁻⁴ M) 3.2 x 10⁻⁸ = x² / 0.82 x = [H⁺] = 1.62 x 10⁻⁴ 14. A 0.10 M HF solution is only ionized 2.6%. What is the Kₐ value for HF? (answer: 6.9 x 10⁻⁴) 0.26 / 100 = x / 0.1 x = 0.026 = [H⁺] Ka = (0.026)² / (0.1 - 0.026) = 6.9 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%) 5.1 x 10⁻⁴ = x² / 1 x = 0.023 M pH = -log(0.023) = 1.6 % ionization = (0.023 / 1) x 100 = 2.3%