12.9 key

Answers:

New Material:

1. Co(OH)₃ ßà Co³⁺ + 3 OH^-

K_sp = (3[OH^-])³*[Co³⁺] = 27x⁴

x = [Co³⁺] = [OH^-]/3

[OH^-] = 2.94 x 10^-11

pOH = 10.53

This answer doesn’t really make sense because we think this compound should be basic.  Here, the [OH^-] increase is so small compared to the autoionization of water that we wouldn’t notice any change in pH.  The autoionization of water is the splitting of water into protons and hydroxide ions, and in a solution of water at 25^oC, concentration of protons = concentration of hydroxide ions = 1 x 10^-7.  Therefore, total concentration of hydroxide would be 2.94 x 10^-11 + 1 x 10^-7, which is approximately 1 x 10^-7.  That means the pOH = pH = 7.  I wouldn’t expect a question like this, where you have to account for the autoionization of water, to come up on the exam, but I want you to know how to do this sort of calculation.

2.

a.) 0.0 mL

pH = 9.11

b.) 8 mL HCl added.  We must determine the amounts of py and Hpy+ present after the added H+ reacts completely with the py.

Initial mmol py present = 25. * 0.100mmol/mL = 2.50mmol py

mmol H+ added = 8.0mL * 0.100mmol/mL = 0.80mmol H+

Added H⁺ reacts with py to completion

(fyi)

pH = 5.56

c.) 28.0 mL

pH = 2.24

3. The Ksp for silver sulfate (Ag2SO4) is 1.2 x 10-5. Calculate the solubility of silver sulfate in each of the following.


(a) water

(b) 0.10 M AgNO3

(c) 0.20 M K2SO4

 

Review:

1. a.) Ag b.) Kr

2. a.) sodium acetate b.) sodium dihydrogen phosphate c.) potassium permanganate d.) tetraphosphorus hexoxide e.) sodium bisulfite/sodium hydrogen sulfite f.) LiAlH₄ g.) Cu₂O h.) K₂CO₃ i.) BaCl₂ j.) N₂O₄

3. Empirical formula: C₃HCl; molecular formula: C₁₂H₄Cl₄

C: 49.67%, Cl: 48.92%, H: 1.39%

If you had 100 g of sample, you would have 49.67 g C, 48.92 g Cl, and 1.39 g H.  Convert to moles = 4.14 mol C, 1.38 mol Cl, 1.38 mol H.  Divide all numbers by 1.38: 3 mol C, 1 mol Cl, 1 mol H.  Write empirical formula: C₃HCl.  Actual molar mass is 289.9 g/mol.  Molar mass of empirical formula is 72.49 g/mol. 289.9/72.49 = ratio of molecular formula to empirical formula = 4.  Therefore we multiply empirical formula by 4 to get moleculr formula.  Molecular formula: C₁₂Cl₄H₄.

4. 21.99 g water vapor

Combustion of ethanol (balanced reaction):

2 CH₃CH₂OH + 7 O₂ (g) à 4 CO₂ (g) + 6 H₂O (g)

18.74 g ethanol = 0.4068 mol ethanol.  0.4068 mol ethanol * (6 mol water/2 mol ethanol) = 1.22 mol H₂O. 1.22 mol water = 21.99 g water

5. 0.114 M ammonia solution

Since ammonia and HCl react in a 1:1 ratio, the number of moles of HCl is the same as the number of moles of ammonia at the equivalence point.  # of moles is the same as molarity * volume so we can use M₁V₁ = M₂V₂.  18.96 mL * 0.150 M = M₂ * 25 mL.  M₂ = 0.114 M

6. a.) Cr₂O₇^2- + 6 Cl^- + 14 H⁺ à 2 Cr³⁺ + 3 Cl₂ + 7 H₂O

b.) HXeO₄^- + 3 Pb + 2 OH^- à Xe + 3 HPbO₂^-

(See book or previous blog post for explanation)

7. 3 x 10¹³ particles

Use ideal gas law: PV=nRT.  Solve for n.  First convert temperature to units of K and convert pressure to units of atm.  Once you have n, multiply by Avogadro’s number for # of particles.

8. PCl₅: 0.023 atm, PCl₃: 0.177 atm, Cl₂: 0.277 atm

For the reaction PCl₅ ßà PCl₃ + Cl₂, set up an ice diagram to solve for x.  Plug back into expressions for equilibrium pressure to solve for the partial pressures.  Total pressure is the sum of all species’ partial pressures.

9. a.) left b.) right c.) no shift

10. pH = 2.42

Set up ice diagram to solve for [H⁺] (see book).  Note that I gave the wrong Ka on the sheet I gave you.  The correct Ka is listed below.  Had you used the previous Ka I gave you, your answer would be pH = 1.91.

11. pH = 4.55

Look up K_a of acetic acid (check Wikipedia).  Solve for “diluted” concentrations of acetic acid and acetate using M1V1=M2V2.  Plug those values into the Henderson Hasselbalch equation to solve for pH.

12.9 Review

New Material:

1. The Ksp of Cobalt (III) Hydroxide at 25^oC is 2.5 x 10^-43.  What is the pOH of Colbalt (III) hydroxide at 25^oC?

2. A 25.0 mL sample of 0.100 M pyridine (Kb = 1.7 x 10-9) is titrated with 0.100 M hydrochloric acid. Calculate the pH after the addition of the following amounts of hydrochloric acid:
a.) 0.00 mL
b.) 8.00 mL
c.) 28.0 mL


3. The K_sp for silver sulfate is 1.2 x 10^-5. Calculate the solubility of silver sulfate in each of the following:
a.) water
b.) 0.10 M silver nitrate (common ion effect)
c.) 0.20 M potassium sulfate (common ion effect)

Review:

1. Identify the following elements:

a.) ¹⁰⁸₄₇X

b.)⁸⁵₃₆ X

2. Name/write formulas for the following compounds:

a.) NaC₂H₃O₂

b.) KH₂PO₄

c.) KMnO₄

d.) P₄O₆

e.) NaHSO₃

f.) lithium aluminum hydride

g.) copper (I) oxide

h.) potassium chlorate

i.) barium chloride

j.) dinitrogen tetroxide

3. A compound is found my mass spectral analysis to contain the following percentages of elements by mass:

C: 49.67%, Cl: 48.92%, H: 1.39%

Find the empirical formula of the compound.  Mass spectrometry shows that the compound has a molar mass of 289.9 g/mol.  What is the compound’s molecular formula?

4. How many grams of water vapor can be generated from the combustion of 18.74 g ethanol (C₂H₆O)?

5. A 25.0 mL sample of ammonia solution is analyzed by titration with hydrochloric acid.  It took 18.96 mL of 0.150 M hydrochloric acid to titrate the ammonia.  What is the concentration of the original ammonia solution?

6. Balance the following redox reactions:

a.) in acid solution: Cr₂O₇^2- + Cl^- à Cr³⁺ + Cl₂

b.) in base solution: HXeO₄^- + Pb à Xe + HPbO₂^-

7. A vacuum line in a research lab has a volume of 1.013 L.  The temperature in the lab is 23.7^oC, and the vacuum line is evacuated to a pressure of 1 x 10^-6 torr.  How many gas particles remain?

8. Given the initial partial pressures of 0.0500 atm for phosphorus pentachloride, 0.150 atm for phosphorus trichloride, and 0.250 atm for molecular chlorine, all at 250^oC, what is the equilibrium partial pressure of each component of the mixture?  What is the total pressure?  (Phosphorus pentachloride reacts to give phosphorus trichloride and molecular chlorine, where K_p = 2.15.)

9. In the combustion of methane, K = 5.67.  Predict the direction the system will shift in order to reach equilibrium.

a.) Q = 11.85

b.) Q = 3.8 x 10^-4

c.) Q = 5.67

10. Calculate the pH of a 0.237 M solution of benzoic acid, a monoprotic acic, with K_a = 6.14 x 10^-5.

11. A buffer is prepared by adding 20.5 g of acetic acid and 17.8 g sodium acetate to water to make a 500. mL solution.  Calculate the pH.

12/1 in-class activity

The pK_a of CH₃NH₃⁺ is 10.62.  A solution is originally 1.072 M in CH₃NH₂ (methylamine) and 0.836 M CH₃NH₃⁺.

1.) What is the initial pH of the buffer?  Hint: use the Henderson-Hasselbalch equation.

pH = pK_a + log([base]/[acid])                     (Henderson-Hasselbalch equation)

[base] = [CH₃NH₂] = 1.072 M

[acid] = [CH₃NH₃⁺] = 0.836 M

pH = 10.73

                         

2.) 3 mL of 1.5 M HBr is added to a 15 mL solution of the buffer above.  Taking the volume change into consideration, what is the pH of the new solution?  What is the pH of a 15 mL solution of the buffer after 3 mL of pure water is added?

Use M₁V₁=M₂V₂ to find the new concentrations of acid and base in solution.

[base] = [CH₃NH₂] = 0.8933 M

[acid] = [CH₃NH₃⁺] = 0.697 M

[HBr] = 0.25 M

Here, we assume that the HBr reacts completely with the base, methylammine.  We therefore subtract 0.25 M from the concentration of the base.  We now get the concentrations:

[base] = [CH₃NH₂] = 0.6433 M

[acid] = [CH₃NH₃⁺] = 0.947 M

We plug these values into the Henderson-Hasselbalch equation.

pH = 10.47

pH decreases.  This makes sense.

For the second question, we use M₁V₁=M₂V₂ to determine the new concentrations.  We plug these new values into the Henderson-Hasselbalch equation and determine the new pH.  However, since the concentrations of acid and base decrease by the same proportion, the ratio of the two stays the same, and the pH is the same as in problem 1.  pH = 10.73

3.) 5 mL of 1.7 M NaOH is added to 12 mL of the buffer solution.  Taking the volume change into consideration, what is the pOH of the new solution?  What is the pOH of a 12 mL solution of the buffer after 5 mL of pure water is added?

Use M₁V₁=M₂V₂ to find the new concentrations of acid and base in solution

[base] = [CH₃NH₂] = 0.7567 M

[acid] = [CH₃NH₃⁺] = 0.590 M

[NaOH] = 0.50 M

Here, we assume that the NaOH reacts completely with the acid, methylammonia.  We therefore subtract 0.50 M from the concentration of the base.  We now get the concentrations:

[base] = [CH₃NH₂] = 1.257 M

[acid] = [CH₃NH₃⁺] = 0.090 M

We plug these values into the Henderson-Hasselbalch equation.

pH = 11.76, pOH = 2.24

pH increases.  This makes sense.

For the second question, we use M₁V₁=M₂V₂ to determine the new concentrations.  We plug these new values into the Henderson-Hasselbalch equation and determine the new pH.  However, since the concentrations of acid and base decrease by the same proportion, the ratio of the two stays the same, and the pH is the same as in problem 1.  pH = 10.73, pOH = 3.27