1 / 77

Acids and Bases

Acids and Bases. Chapter 7 and 8 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/. Acids and Bases – ch. 7 . 1 . Calculate the pH for the following solutions at 25 °C and determine if the solution is acidic, basic or neutral. a. [H + ] = 5 x 10 -4 M

sauda
Download Presentation

Acids and Bases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Acids and Bases Chapter 7 and 8 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/

  2. Acids and Bases – ch. 7 1. Calculate the pH for the following solutions at 25 °C and determine if the solution is acidic, basic or neutral. a. [H+] = 5 x 10-4 M b. [OH-] = 2 x 10-9 M

  3. Acids and Bases – ch. 7 pH = - log[H+] pOH = - log[OH-] [H+] = 10 –pH [OH-] = 10 –pOH Kw = [H+][OH-] = 1 x 10-14 (at 25°C) pKw = pH + pOH = 14 (at 25°C)

  4. Acids and Bases – ch. 7 2. Calculate the [H+] and [OH-] for the following solutions at 25 °C. a. pH = 2 b. pOH = 4.2

  5. Acids and Bases – ch. 7 3. At 20 °C, the water ionization constant, Kw, is 6.8 x 10–15. What is the pH and H3O+ concentration in neutral water at this temperature?

  6. Acids and Bases – ch. 7 4. a. Which of the following is a stronger acid? HNO2 (Ka= 4.0 x 10–4) or HCN (Ka = 6.2 x 10–10) b. Which is a stronger base? NO2–or CN–

  7. Acids and Bases – ch. 7 As acid strength ↑ % ionization ↑ Ka↑ As base strength ↑ % ionization ↑ Kb ↑ Conjugate pairs are inversely related As acid strength ↑ conjugate base strength ↓ Kw = (Ka)(Kb)

  8. Acids and Bases – ch. 7 5. Bicarbonate ion is an amphoteric substance. Will an aqueous solution of HCO3- be acidic, basic or neutral?

  9. Acids and Bases – ch. 7 6. Predict if the following salts are acidic, basic orneutral. a. NaClO4 b. LiF c. NH4I d. KNO3 e. NH4F

  10. Acids and Bases – ch. 7 Salts ⇒ the ionic compound that is the result of an acid base reaction ⇒ when a salt is dissolved in water each ion can potentially affect the pH of the solution ⇒ you need to analyze the ions individually Salts Cations Group 1 metal ions ⇒ neutral Ca2+, Sr2+ and Ba2+⇒ neutral All else ⇒ acids Anions Cl-, Br-, I-, NO3-, ClO3-, ClO4-⇒ neutral Starts with H ⇒ amphoteric All else ⇒ bases

  11. Acids and Bases – ch. 7 7. Calculate the pH of the following solutions: a. 0.004 M HBr b. 10 g of H2SO4 and 10 g of HCl in 4.2 Lof solution c. 2.5 g of Ba(OH)2 in 750 mL of solution

  12. Acids and Bases – ch. 7 8. Calculate the pH and the % ionization of the following solutions a. 0.05 M HNO2 (Ka= 4.0 x 10–4) b. 0.33 M CH3COOH (Ka= 1.8 x 10–5) mixed with 0.85 M HCN (Ka= 6.2 x 10–10) c. 0.04 M NH4I (for NH3Kb = 1.8 x 10–5)

  13. Acids and Bases – ch. 7 9. What concentration of HCOOH (Ka = 1.77x10-4) solution will have a pH of 2.2?

  14. Acids and Bases – ch. 7 10. If a solution with 0.5 M of unknown weak acid ionizes 0.62% identify the weak acid.

  15. Acids and Bases – ch. 7 11. What concentration of a weak acid that ionizes 0.1% will have a pH of 3.8?

  16. Acids and Bases – ch. 7 12. What concentration of HNO2(Ka= 4.0 x 10–4) will have the same pH as 0.004M HNO3?

  17. Acids and Bases – ch. 7 13. If 0.1M of a weak acid has a pH of 3 what will be the pH of a 0.001M solution of the weak acid?

  18. Acids and Bases – ch. 7 14. Calculate the pH and % ionization for the following: a. 0.01M NH3 (Ka of NH4+ = 5.6 x 10-10) b. 0.05 M KClO (Ka of HClO = 3 x 10-8)

  19. Acids and Bases – ch. 7 15. How many grams of KF must be dissolved in 500 mL of water in order to get a pH of 8.4? (Ka of HF = 7.2 x 10-4)

  20. Acids and Bases – ch. 7 16. If a 1.2 M solution of NaA has a pH of 10.4, what is the ionization constant (Ka) for the acid HA?

  21. Acids and Bases – ch. 8 17. What is the pH at the equivalence point for the following titrations? a. NaOH with HBr pH = 7 pH > 7 pH < 7 b. HCl with NaCH3COO pH = 7 pH > 7 pH < 7 c. KOH with HCN pH = 7 pH > 7 pH < 7

  22. Acids and Bases – ch. 8 18. Which of the following will result in a buffer solution upon mixing? a. 0.1 mol HClO3 and 0.1 mol NaClO3 is put into 1L ofwater b. 0.5 mol H2S and 0.8 molNaHS is put into 1L of water c. 0.03 mol KHC2O4 and 0.02 mol K2C2O4 is put into 1L of water d. 0.1 molLiOH and 0.2 mol H3PO4 is put into 1L of water e. 0.1 mol HNO3 and 0.04 mol NH3 is put into 1 L of water

  23. Acids and Bases – ch. 8 19. Which of the following 50 mL solutions can absorb the most acid without changing the pH? a. 0.4 M HF /0.5 M NaF b. 0.8 M HF /0.7 M NaF c. 0.3 M HF/0.7 M NaF

  24. Acids and Bases – ch. 8 20. Calculate the pH of the following: a. A 35 mL solution with 0.2 M HN3 (Ka = 1.9 x 10-5) and 0.1 M NaN3 b. A solution prepared by mixing 30 mL of 0.1 M HN3 with 40 mL of 0.2 M NaN3

  25. Acids and Bases – ch. 8 21. Calculate the pH for the following titrations: a. 25 mL of 0.1 M HCl is mixed with 25 mL of 0.2 M KNO2 (Kb = 2.5 x 10-11) b. 50 mL of 0.1 M HCl is mixed with 25 mL of 0.2 M KNO2 c. 60 mL of 0.1 M HClis mixed with 25 mL of 0.2 M KNO2 d. Draw a titration curve and label the solutions in a, b and c

  26. Acids and Bases – ch. 8 3 scenarios when titrating a weak acid or weak base with a strong acid or a strong base 1. nstrong < nweak=> buffer zone 2. nstrong = nweak=> equivalence point 3. nstrong > nweak=> excess titrant

  27. Acids and Bases – ch. 8 22. Calculate the pH of the following titrations: a. 40 mL of 0.20 M KOH is mixed with 40 mL of 0.5 M HClO (Ka = 3 x 10-8) b. 80 mL of 0.25 M KOH is mixed with 40 mL of 0.5 M HClO c. 100 mL of 0.25 M KOH is mixed with 40 mL of 0.5 M HClO d. Draw a titration curve and label the solutions in a, b and c

  28. Acids and Bases – ch. 8 23. Calculate the pH of the following: a. 10 mL of 1 M HCl is added to a 100 mL buffer with 0.5 M CH3CH2COOH (Ka = 1.34 x 10-5) and 0.6 M CH3CH2COONa. b. 20 mL of 0.2 M KOH is added to a 100 mL buffer with 0.5 M CH3CH2COOH and 0.6 M CH3CH2COONa.

  29. Acids and Bases – ch. 8 24. How would you prepare 1.0 L of a buffer at pH = 9.0 from 1.0 M HCN (Ka = 6.2 x 10-10) and 1.5 M NaCN?

  30. Acids and Bases – ch. 8 25. How many grams of NaOH need to be added to a 50 mL of 0.3 M HNO2 (Ka = 4.0 x 10-4) to have a solution with a pH = 4?

  31. Acids and Bases – ch. 8 26. Determine the solubility in mol/L and g/L for the following compounds: a. BaCO3 (Ksp = 1.6 x 10-9) b. Ag2S (Ksp = 2.8 x 10-49)

  32. Acids and Bases – ch. 8 27. Determine the Ksp values for the following compounds: a. Al(OH)3 (solubility = 5 x 10-9mol/L) b. MgF2 (solubility = 0.0735 g/L)

  33. Acids and Bases – ch. 8 28. What is the solubility of Ca3(PO4)2 (Ksp = 1 x 10-54) in 0.02M solution of Na3PO4?

  34. Acids and Bases – ch. 8 29. What is the solubility of Zn(OH)2 (Ksp = 4.5 x 10-17) in a solution with pH of 11?

  35. Acids and Bases – ch. 8 30. Will BaCrO4 (Ksp = 8.5 x 10-11) precipitate when 200 mL of 1 x 10-5 M Ba(NO3)2 is mixed with 350 mL of 3 x 10-5 M KCrO4?

  36. Answer Key – ch. 7 & 8 1. Calculate the pH and pOH for the following solutions and determine if the solution is acidic, basic or neutral. a. [H+] = 5 x 10-4 M pH = -log[H+] ⇒ pH = -log(5 x 10-4)⇒ pH = 3.3 pH + pOH = 14 ⇒ pOH = 14 – 3.3 ⇒ pOH = 10.7 acidic b. [OH-] = 2.5 x 10-9 M pOH = -log[OH-] ⇒ pOH = -log(2 x 10-9 ) ⇒ pOH = 8.6 pH = 14 – 8.6 ⇒ pH= 5.4 acidic

  37. Answer Key – ch. 7 & 8 2. Calculate the [H+] and [OH-] for the following solutions. a. pH = 2 [H+] = 10 -pH⇒ [H+] = 10 -2 M or 0.01 M pOH = 14 – 2 = 12 ⇒ [OH-] = 10 -pOH⇒ [OH-] = 10 -12 M b. pOH = 4.2[OH-] = 10 -4.2 M or 6.3 x 10 -5 M pH = 14 – 4.2 = 10.8 ⇒ [H+] = 10 -10.8 or 1.6 x 10 -11 M

  38. Answer Key – ch. 7 & 8 3. At 20 °C, the water ionization constant, Kw, is 6.8 x 10–15. What is the pH and H3O+ concentration in neutral water at this temperature? If a solution is neutral => [H3O+] = [OH-] For all solutions => [H3O+]x[OH-] = Kw x2 = 6.8 x 10–15 => x = 8.25x10-8M [H3O+]=8.25x10-8M pH = -log(8.25x10-8M) pH = 7.08

  39. Answer Key – ch. 7 & 8 4. a. Which of the following is a stronger acid? HNO2 (Ka= 4.0 x 10–4) or HCN (Ka = 6.2 x 10–10) as Ka↑ % ionization ↑ or acid strength ↑ ⇒HNO2 is a stronger acid b. Which is a stronger base? NO2–or CN– as acid strength ↑ conjugate base strength ↓ since HNO2 is a stronger acid ⇒CN– is a stronger base

  40. Answer Key – ch. 7 & 8 5. Bicarbonate ion is an amphoteric substance. Will an aqueous solution of HCO3- be acidic, basic or neutral? Ka tells you a relative acid strength and Kb tells you a relative base strength ⇒ so by comparing Ka to Kb we can determine what it prefers to be Ka = 4.8 x 10-11 Kb = Kw/Ka of H2CO3⇒ Kb = (1 x 10-14)/(4.3 x 10-7) = 2.3x10-8 Since Kb > Ka bicarbonate will be a base

  41. Answer Key – ch. 7 & 8 6. Predict if the following salts are acidic, basic orneutral. a. NaClO4 Na+ (neutral) + ClO4– (neutral) ⇒ Neutral salt b. LiF Li+ (neutral) + F– (weak base) ⇒ Basic salt c. NH4I  NH4+ (weak acid) + I–(neutral) ⇒ Acidic salt d. KNO3  K+ (neutral) + NO3– (neutral ⇒ Neutral salt e. NH4F  NH4+ (weak acid) + F– (weak base) since there’s an acid and a base we need to compare their ionization constants: NH4+⇒ Ka = 5.6 x 10-10 verses F–⇒ Kb = Kw/Ka of HF ⇒ Kb = (1 x 10–14)/(7.2 x 10–4) ⇒ Kb = 1.4 x 10–11 since Ka > Kb it’s an Acidic salt

  42. Answer Key – ch. 7 & 8 7. Calculate the pH of the following solutions a. 0.004 M HBr⇒ Strong acid [H+] = 0.004 M ⇒ pH = - log(0.004) ⇒ pH = 2.4 b.10 g of H2SO4 and 10 g of HCl in 4.2 Lof solution ⇒ Since both acids are strong they both contribute significantly to the [H+] [H2SO4] = (10g/98.09g/mol)/(4.2L) = 0.024M => [H+] = 0.024M [HCl] = (10g/36.46g/mol)/(4.2L) = 0.065 M => [H+] = 0.065M [H+]total = 0.089M => pH = -log(0.089) = 1.05 c. 2.5 g of Ba(OH)2 in 750 mL of solution => strong base [Ba(OH)2] = (2.5g/171.32g/mol)/(0.75L) = 0.019M [OH-] = 2(0.019M) = 0.038M pOH = -log(0.038) = 1.4 pH = 14-1.4 = 12.6

  43. Answer Key – ch. 7 & 8 8. Calculate the pH and the % ionization of the following solutions a. 0.04 M HNO2(Ka= 4.0 x 10–4) ⇒ weak acid Use Ka to solve for x 4 x 10–4 = (x)(x)/0.04-x x = 0.0038 M [H+] = 0.0038 pH = - log(0.0038) = 2.42 % ionized = ([H+]/[HNO2])x100 %ionized = (0.0038/0.04)x100 = 9.5%

  44. Answer Key – ch. 7 & 8 8. …continued b. 0.33 M CH3COOH (Ka= 1.8 x 10–5) mixed with 0.85 M HCN (Ka= 6.2 x 10–10) If you have 2 or more acids only the strongest acid will contribute significantly to the pH Since acetic acid is stronger we can ignore the HCN 1.8 x 10–5= x2/0.33 x = 0.00244M = [H+] pH = -log(0.00244) = 2.61 % ionized = (0.00244/0.33)100 = 0.74% Insignificantly small

  45. Answer Key – ch. 7 & 8 8. …continued c. 0.04 M NH4I (for NH3 Kb = 1.8 x 10–5) =>if NH3 is a base then NH4+ must be an acid Ka is necessary to solve for x => since NH4+ and NH3 are conjugates Ka = Kw/Kb Ka = 1x10-14/1.8x10-5 = 5.6x10-10 5.6x10-10 = x2/0.002 x = 1.06x10-6 pH = -log(1.06x10) = 5.98 %ionized = (1.06x10-6/0.002)100 = 0.053% Insignificantly small

  46. Answer Key – ch. 7 & 8 8. …continued d. 0.04 M NH4I ⇒salt with weak acid (NH4+)⇒ Ka = 5.6 x 10–10 Use Ka to solve for x 5.6 x 10–10 = (x)(x)/0.04 x = 4.73 x 10–6 [H+] = 4.73 x 10–6 pH = - log(4.73 x 10–6) pH = 5.3 Insignificantly small

  47. Answer Key – ch. 7 & 8 9. An aqueous solution of Ca(OH)2 was found to have a pH of 11. a. Determine the concentration of calcium ions. Since the pH = 11 ⇒ pOH = 3 ⇒ [OH–] = 10–3 M or 0.001 M According to ⇒ Ca(OH)2 Ca2+ + 2 OH– If [OH–] = 0.001 M ⇒ [Ca2+] = 0.0005 M b. How many grams of Ca(OH)2 would you have to dissolve in 2.6 L in order to make the above solution? mol Ca(OH)2 = (0.0005 mol/L)(2.6L) = 0.0013 mol 0.0013 mol Ca(OH)2 x 74.12 g/mol = 0.096 g Ca(OH)2

  48. Answer Key – ch. 7 & 8 10. What concentration of HCOOH (pKa = 3.75) solution will have a pH of 2.2? Since the pH = 2.2 ⇒ [H+] = 10–2.2 M or 0.0063 M ⇒ since the molar ratio is 1:1 ⇒ [H+] = [HCOO–] We need Ka to solve for x since the pKa = 3.75 Ka = 10–3.75or 1.78 x 10-4 1.78 x 10-4 = (0.0063)(0.0063)/x x = 0.223 [HCOOH] = 0.223 M

  49. Answer Key – ch. 7 & 8 11. How many grams of KF must be dissolved in 500 mL of water in order to get a pH of 8.4? KF is a salt with a weak base F–⇒ since the pH = 8.4 ⇒ pOH = 5.6 ⇒ [OH–] = 10–5.6 M or 2.51 x 10–6 M We need Kb to solve for x ⇒ the Ka for HF = 7.2 x 10–4 ⇒ Kb for F– = (1 x10–14)/(7.2 x 10–4) = 1.4 x 10–11 1.4 x 10–11 = (2.51 x 10–6)(2.51 x 10–6)/x ⇒ x = 0.45 ⇒[F–] = 0.45 M (0.45 mol/L)(0.5 L) = 0.225 mol KF (0.225 mol KF)(56.11 g/mol) = 12.6 g of KF

  50. Answer Key – ch. 7 & 8 12. If a 0.0286 M of unknown weak acid (HA) solution has a pH of 4.5, what is the ionization constant and identification of the acid? If we know Ka it can be used to identify the acid ⇒ since the pH = 4.5 ⇒ [H+] = 10–4.5 M or 3.16 x 10–5 Ka = (3.16 x 10–5 )(3.16 x 10–5 )/(0.0286) Ka = 3.5 x 10–8 which corresponds with HOCl

More Related