1 / 40

Entry Task: Feb 25 th Monday

Entry Task: Feb 25 th Monday. Sign off on Ch. 17 Review ws Break out Practice multiple choice test Review Ch. 16-17 part 1. I can…. Explain and calculate the common ion effect. Describe how a buffer system works and calculate the pH changes. Explain and calculate acid-base titrations

betha
Download Presentation

Entry Task: Feb 25 th Monday

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. Entry Task: Feb 25th Monday Sign off on Ch. 17 Review ws Break out Practice multiple choice test Review Ch. 16-17 part 1

  2. I can… • Explain and calculate the common ion effect. • Describe how a buffer system works and calculate the pH changes. • Explain and calculate acid-base titrations • Plot data to provide a titration curve and identify what types of acid-base solutions they are.

  3. 1. A 0.10 M solution of lactic acid (HC3H5O3) has a pH of 2.44. Calculate Ka for lactic acid. 10-2.44 = [H+] = 3.6 x 10-3 = [OH-] (3.6 x 10-3)2 = 0.10 1.3 x 10-4 = Ka

  4. 2. A 0.200 M solution of a weak acid HX is 9.4% ionized. Calculate the pH and Ka for this acid. = 0.019 M 0.200M=[H+] X 0.094 -log = pH=1.72 (0.019)2 = 0.200-0.019M = 2.0 x 10-3 = Ka

  5. 3. Calculate the pH of a 0.050 M solution of ethylamine (C2H5NH2, Kb= 6.4 x10–4). Kb= 6.4 x 10-4 (x)2 = 0.050 (0.050)(6.4 x 10-4)=x2 x = 5.7 x 10-3 of OH pOH = 2.25 – 14 = pH= 11.75

  6. 4. The Ka for hydrocyanic acid, HCN, is 5.0 x10–10. What is the Kb for CN–? 1.0 x 10-14 = 5.0 x 10-10 2.0 x 10-5

  7. 5. Hydrosulfuric acid is a polyprotic acid with the following equilibria: a) Calculate the pH of a 0.100 M H2S solution. b) Calculate the [S2–] for the solution above. (x)2 = 0.100 1.1 x 10-7 pH= 4.00 [H+]= 1.05 x 10-4 Ka2 = 1.2 x 10-13 is the concentration of S2- ions

  8. 6. Sodium benzoate, C6H5CO2Na, is the salt of the weak acid, benzoic acid (C6H5CO2H). A 0.10 M solution of sodium benzoate has a pH of 8.60 at room temperature. a) Calculate the Kb value for benzoate ion (C6H5CO2–). b) Calculate the Ka value for benzoic acid. 8.60 – 14 = pOH= 5.4 10-5.4 = 4.0 x10-6 = [OH-] (4.0 x 10-6)2 0.10 Kb= 1.6 x 10-10 1.0 x 10-14 1.6 x 10-10 Ka= 6.3 x 10-5

  9. 7. Potassium sorbate (KC6H7O2) is the salt of the weak acid, sorbic acid (HC6H7O2, Ka = 1.7x10–5), and is commonly added to cheese to prevent mold. What is the pH of a solution containing 4.93 g of potassium sorbate in 500 mL of solution? = 0.0657 M / 0.5 L 4.93 g/150 g = 0.0328 1.0 x 10-14 / 1.7 10-5 = Kb = 5.9 10-10 (x)2 0.0657 = 5.9 10-10 x2= 3.88 10-10 x= 6.2 x10-6 pOH= 5.2 – 14 pH= 8.8

  10. 8. A buffer is prepared by adding 20.0 g of acetic acid (HC2H3O2) and 20.0 g of sodium acetate (NaC2H3O2) in enough water to prepare 2.00 L of solution. Calculate the pH of this buffer? (Ka = 1.8x10–5) = 1.67 x 10-1 M Acetic acid 20.0g/60 g = 0.33 mole / 2 L = 1.22 x 10-1 M Na2C2H3O2 20.0g/82 g = 0.244 mole / 2 L 1.22 x 10-1 1.67 x 10-1 = 0.73 = -log = -0.136 pH = 4.6 + (-0.136) pH= 4.6

  11. 9. What is the ratio of HCO3– to H2CO3 in blood of pH 7.4? (Kafor H2CO3 = 4.3x10–7) [HCO3] [H2CO3] 7.4 = 6.37 10-1.03 = 11 : 1

  12. 11. Acetylsalicylic acid (aspirin, HC9H7O4) is a weak acid with Ka = 2.75x10–5 at 25°C. 3.00 g of sodium acetylsalicylate (NaC9H7O4) is added to 200.0 mL of 0.100 M solution of this acid. Calculate the pH of the resulting solution at 25°C. 3.00g / 201.99g = 1.49 x 10-2 0.200 = 7.4 x 10-2 7.4 x 10-2(x) 0.100 = 2.75 x 10-5 x = 3.7 x 10-5 pH= 4.42

  13. 12. The equations and dissociation constants for three different acids are given below: Identify the conjugate pair that is best for preparing a buffer with a pH of 7.2. Clearly explain your choice. pH = 6.4 pH = 7.2 ** pH = 1.9 Because it’s the closest to the Ka expression

  14. 13. A buffer solution is prepared by adding 0.10 L of 2.0 M acetic acid solution to 0.10 L of 1.0 M NaOH solution. Ka= 1.7 x10-5 Calculate the pH of this buffer solution. (10 ml)(2.0) = 20 mmol acetic acid (10 ml)(1.0) = 10 mmolNaOH 10 mmol 20 ml = 0.50 mmol of [H+] left over AND 0.50 mmol of [C2H3O2-] pH = 4.77 + (0) pH= 4.7

  15. 14. A 10.0 mL solution of 0.100 M NH3 (Kb = 1.8x10–5) is titrated with a 0.100 M HCl solution. Calculate the pH of this solution at equivalence point. (10 ml)(0.100M) = 1.0 mmol NH3 (10 ml)(0.100M) = 1.0 mmolHCl 1.0 mmol / 20.0 ml = 0.05 mmol of NH3 1.0 mmol / 20.0 ml = 0.05 mmol of HCl Assume all the HCl and NH3react to form NH4Cl, then some of the NH4+ hydrolyzes back to ammonia. 1.0 x 10-14 / 1.8 x 10-5 = Kb = 5.6 x 10-10

  16. 14. A 10.0 mL solution of 0.100 M NH3 (Kb = 1.8x10–5) is titrated with a 0.100 M HCl solution. Calculate the pH of this solution at equivalence point. Assume all the HCl and NH3react to form NH4Cl, then some of the NH4+ hydrolyzes back to ammonia. 1.0 x 10-14 / 1.8 x 10-5 = Kb = 5.6 x 10-10 (x)2 0.050 = 5.6 10-10 x2= 2.8 10-11 x= 5.3 x10-6 pH= 5.28

  17. 15. A sample of 25.0 mL of 0.100 M solution of HBr is titrated with 0.200 M NaOH. Calculate the pH of solution after 10.0 mL of the base is added. (25 ml)(0.100 M) = 2.5 mmol of HBr (10 ml)(0.200 M) = 2.0 mmol of NaOH This means there is 0.50 mmol of [H+] left over 0.50 mmol/ 35 mls of solution [H+]= 1.4 10-2 pH= 1.85

  18. 16. A 10.0­mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HCl solution: Kb=1.8 x10-5 (a) 0.0 mL (x)2 0.300 = 1.8 10-5 x2= 5.4 10-6 x= 2.3 x10-3 pOH= 2.6 – 14 = pH 11.4

  19. 16. A 10.0­mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HClsolution: Kb=1.8 x10-5 (b) 10.0 mL (10 ml)(0.300 M) = 3.0 mmol of NH3 (10 ml)(0.100 M) = 1.0 mmol of HCl This means there is 2.0 mmol of [NH3] left over in 20 mls of solution [NH3]= 1.0 10-1 M This means there is 1.0 mmol of [NH4+] left over in 20 mls of solution [NH4+]= 5.0 10-2 M

  20. 16. A 10.0­mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HClsolution: Kb=1.8 x10-5 (b) 10.0 mL [NH3]= 1.0 10-1 M [NH4+]= 5.0 10-2 M We need the Ka value from the conjugate acid 1.0 x 10-14 / 1.8 x 10-5 = 5.6 x 10-10 [1.0 10-1] [5.0 10-2] pH= 9.25 + log pH= 9.55

  21. 16. A 10.0­mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HCl solution: (c) 30.0 mL (10 ml)(0.300 M) = 3.0 mmol of NH3 (30 ml)(0.100 M) = 3.0 mmol of HCl This means there is 3.0 mmol of [NH4+] left over in 40 mls of solution [NH4+]= 7.5 10-2 M

  22. 16. A 10.0­mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HClsolution: Kb=1.8 x10-5 (c) 30.0 mL [NH4+]= 7.5 10-2 M We need the Ka value from the conjugate acid 1.0 x 10-14 / 1.8 x 10-5 = 5.6 x 10-10 [x]2 [7.5 10-2] = 5.6 x 10-10 x2= 4.2 10-11 x= 6.48 x10-6 pH= 5.19

  23. 17. A 45.0­mL sample of 0.200 M acetic acid is titrated with 0.180 M NaOH. Calculate the pH of the solution. Ka= 1.7 x 10-5 (a) before addition of NaOH, (x)2 0.200 = 1.7 10-5 x2= 3.4 10-6 x= 1.8 x10-3 = pH 2.73

  24. 17. A 45.0­mL sample of 0.200 M acetic acid is titrated with 0.180 M NaOH. Calculate the pH of the solution. Ka= 1.7 x 10-5 (b) after addition of 20.0 mL of NaOH (45.0 ml)(0.200 M) = 9.0 mmol of acetic acid (20 ml)(0.180 M) = 3.6 mmol of NaOH This means there is 5.4 mmol of [Acetic acid] left over in 65 mls of solution [Acetic acid]= 8.3 10-2 M This means there is 3.6 mmol of [C2H3O2-] left over in 65 mls of solution [C2H3O2-]= 5.5 10-2 M

  25. 17. A 45.0­mL sample of 0.200 M acetic acid is titrated with 0.180 M NaOH. Calculate the pH of the solution. Ka= 1.7 x 10-5 (b) after addition of 20.0 mL of NaOH [Acetic acid]= 8.3 10-2 M [C2H3O2-]= 5.5 10-2 M Ka value 1.7 x 10-5 [5.5 10-2] [8.3 10-2] pH= 4.77 + log pH= 4.59

  26. 17. A 45.0­mL sample of 0.200 M acetic acid is titrated with 0.180 M NaOH. Calculate the pH of the solution (c) at the equivalence point.  (45 ml)(0.200 M) = (x)(0.180M) This means that there is 50 mls of base needed (45.0 ml)(0.200 M) = 9.0 mmol of acetic acid (50 ml)(0.180 M) = 9.0 mmol of NaOH This means there is 9.0 mmol of [C2H3O2-] left over in 95 mls of solution [C2H3O2-]= 9.47 x 10-2 M

  27. 17. A 45.0­mL sample of 0.200 M acetic acid is titrated with 0.180 M NaOH. Calculate the pH of the solution  (c) at the equivalence point.  [C2H3O2-]= 9.47 x 10-2 M We need the Kb value from the conjugate base 1.0 x 10-14 / 1.7 x 10-5 = 5.88 x 10-10 [x]2 ___ [9.47 x 10-2 ] x2= 5.57 10-11 = 5.88 x 10-10 x= 7.46 x10-6 pOH= 5.13 – 14 = pH = 8.87

  28. Practice Test 1Answers

  29. 19. What is the pH of a sodium formate (NaCHO2) solution prepared by adding 0.680 grams of sodium formate to 100.0 ml of water at 25.0 oC? The Ka at 25ºC for formic acid 1.8 x 10-4 First you should realize that NaCHO2 is a salt that dissolves in water to give Na+1 ions that have no effect on pH and CHO2-1ions that are a weak base. Write an equation for the ionization of CHO2-1ions. CHO2-1+ H2O ←→ HCHO2 + OH-1 Calculate the concentration of NaCHO2 (0.68 g * 1mole/68 g = 0.01 moles/0.10 L = 0.10M) which is equal to the concentration of CHO2-1 ions. Then make an ICE-box. Then write an equilibrium expression Ka = [CHO2-1] / [HCHO2] [OH-1] Unfortunately you are only given the Kafor HCHO2 = 1.8 x 10-4 so you must solve for Kb of CHO2-1 with the equation Ka * Kb = Kw Thus Kb = 5.56 x 10-11 So 5.56 x10-11 = x2 / 0.1 Solve for x = [OH-1] = 2.36 x 10-6 M. So pOH = 5.63 and pH = 8.37

More Related