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

11/17 in-class activity

1.  What is the molar mass of HBr in SI units?

Molar mass of HBr = 80.908 g/mol, but THIS IS NOT IN SI UNITS!

In SI units, molar mass of HBR =  0.080908 kg/mol

2.  What are the SI units for the following quantities?

Pressure	pascal (Pa)
Volume	liter (L)
Mass	kg
Temperature	Kelvin (K)
Energy	Joule (J)

(J  and Pa are actually an SI-derived units, J is equal to kg*m²/s², and Pa is N/m²)

3. The mole fraction of argon in dry air is 0.00934.  How many liters of air at STP will contain enough argon to fill a 35.4-L cylinder to a pressure of 150. atm at 20^oC?

How many moles of Ar must there be to fill the cylinder?
PV=nRT

150. atm * 35.4 L = n * 0.08206 atm*L/mol*K * 293 K

n 220.8 mol Ar

STP: 0^oC = 273 K and 1 bar = 0.9869 atm

P_i = 0.00934 bar, P_tot = 1 bar (note we are using bar, not atm, because we are told we are at a total pressure of 1 bar).  P_i is the partial pressure of Ar in the atmosphere in bar.

We use the P_i to find the volume of air needed to obtain 220.8 mol:

P_iV=nRT
Note: WE DO NOT USE THE SAME R.  WE USE R IN UNITS OF BAR.  Conversely, we could convert P_i to atm and use the same R.  We must be sure that units cancel!

R = 0.08314 L*bar/mol*K

(Also, not the temperature change.)

0.00934 bar * V = 220.8 mol * 0.08314 L*bar/mol*K * 273 K

V = 536,665 L of dry air at STP

4. Which of these will change the position of equilibrium (circle all that apply)?

            allow more time to pass

            add a catalyst

            remove some products

Adding a catalyst does not shift the equilibrium (it only lowers the activation energy – more to follow on this later in the class).  Allowing more time to pass will not change a system at equilibrium from equilibrium.

5.  For the following reaction, predict the effect of each change to the system.

N₂ (g) + 6HCl (g) + energy ßà 2NH₃ (g) + 3 Cl₂ (g)

Triple the volume of the system:

Shifts left

The amount of nitrogen is doubled:

Shifts right

Heat is added to the system:

Shifts right

6.  Fill in the following table:

pH	pOH	[H+]	[OH-]	Acid, base, or neutral?
6.88	7.12	1.3 x 10-7	7.6 x 10-8	Acid
0.92	13.08	0.12	8.4 x 10-14	Acid
10.89	3.11	1.3 x 10-11	7.8 x 10-4	Base
7.00	7.00	1.0 x 10-7	1.0 x 10-7	Neutral

7.  What is the conjugate base of acetic acid?  Would you expect it to be a strong or weak base?  What is the conjugate acid of potassium hydroxide?  Would you expect it to be a strong or a weak acid?

Acetic acid (pK_a = 4.75): CH₃COOH ßà CH₃COO^-  (acetate anion)+ H⁺

Acetate anion is the conjugate base.  You would expect it to be a strong base.

KOH is a strong base, so you would expect its conjugate acid, K^-, to be a weak acid.

8.  Would you expect a sodium acetate solution to be acidic or basic?  Would you expect an ammonium chloride solution to be acidic or basic?

You would expect a sodium acetate solution to be basic (acetate is a strong base and sodium cation is a weak acid).

You would expect ammonium chloride solution to be acidic (ammonium is a strong acid and chloride anion is a weak base).

9.  The value for K_a is 7.45 x 10^-4 for citric acid (C₆H₁₀O₈) (a monoprotic acid).  Calculate the pOH for a 0.200 M citric acid solution.

We can write this equilibrium in terms of a “general acid” HA:

HA ßà A^- + H⁺

We also have the autoiniozation of water:

H₂O ßà H⁺ + OH^-

We know that the concentration of H⁺ in water is 1.0 x 10^-7 M, so we start out with 1.0 x 10^-7 M H⁺ concentration (I in the ICE diagram).  Generally, we won’t need to account for this, but sometimes we do.  We will include it just in case.

Concentrations (M)	HA	A-	H+
I	0.200	0	1.0 x 10-7
C	-x	+x	+x
E	0.200 - x	x	1.0 x 10-7 + x

We can write: K_a = [A^-][H⁺]/[HA]

Plugging in the equilibrium concentrations, we get:

7.45 x 10^-4 = x*(1.0 x 10^-7 + x)/(0.200 – x)

x² + 7.451 x 10^-4x – 1.49 x 10^-4 = 0

Plugging this into the quadratic equation, we get two values for x, but only one of them makes physical sense.  We get x = 0.01184 M.

Therefore, equilibrium concentration of H⁺ = 0.01184 M

pH = -log[H⁺] = 1.937

pOH = 14 – pH = 12.073

We can check our answer by plugging the equilibrium values into the equation for K_a.  If we get back the correct value, we know we have the correct answer.

Original worksheet:

1.  What is the molar mass of HBr in SI units?

2.  What are the SI units for the following quantities?

Pressure
Volume
Mass
Temperature
Energy

3. The mole fraction of argon in dry air is 0.00934.  How many liters of air at STP will contain enough argon to fill a 35.4-L cylinder to a pressure of 150. atm at 20^oC?

4. Which of these will change the position of equilibrium (circle all that apply)?

            allow more time to pass

            add a catalyst

            remove some products

5.  For the following reaction, predict the effect of each change to the system.

N₂ (g) + 6HCl (g) + energy ßà 2NH₃ (g) + 3 Cl₂ (g)

Triple the volume of the system:

The amount of nitrogen is doubled:

Heat is added to the system:

6.  Fill in the following table:

pH	pOH	[H+]	[OH-]	Acid, base, or neutral?
6.88
			8.4 x 10-14
	3.11
		1.0 x 10-7

7.  What is the conjugate base of acetic acid?  Would you expect it to be a strong or weak base?  What is the conjugate acid of potassium hydroxide?  Would you expect it to be a strong or a weak acid?

8.  Would you expect a sodium acetate solution to be acidic or basic?  Would you expect an ammonium chloride solution to be acidic or basic?

9.  The value for K_a is 7.45 x 10^-4 for citric acid (C₆H₁₀O₈) (a monoprotic acid).  Calculate the pOH for a 0.200 M citric acid solution.

11/10 in-class activity

For the reaction of solid carbon (C_(s)) and gaseous carbon dioxide (with the product gaseous carbon monoxide), K_p=1.50.

1.) Write the balanced chemical equation (remember this is an equilibrium).

                                          k₁

            C_(s) + CO_2,(g)    ßà       2 CO_(g)

                                     k_-1

2.) Write the forward and reverse rate laws in terms of partial pressure.  Come up with an expression for equilibrium.

Forward rate = k₁ (Pco₂)

Reverse rate = k_-1 (Pco)²

k₁ is the forward reaction rate constant and k_-1 is the reverse reaction rate constant.

Equilibrium constant, K_p = (Pco)²/(Pco₂)

Remember that solids and pure liquids are not included in ICE diagrams of equilibrium expressions (it doesn’t make sense to have a concentration or a partial pressure of a solid or pure liquid).

3.) For excess C_(s) and initial partial pressures Pco₂ = 0.56 atm and Pco = 0.14 atm, find Q_p, the reaction quotient.  Which way will the reaction proceed to equilibrium?

Q_p = (Pco)²/(Pco₂) = 0.035

Q<K so the reaction will proceed to the right (products) to reach equilibrium.

4.) Make an ICE diagram for this reaction using the partial pressures.

	CO2            à	2CO
I (initial)	0.56 atm	0.14 atm
C (change)	-x	+2x
E (equilibrium)	0.56 - x	0.14 + 2x

5.) Solve for x.  You will get two values for x here.  Which one makes physical sense?

K_p = 1.50 (given in the problem)

Given the equilibrium constant expression K_p = (Pco)²/(Pco₂), we can use the new equilibrium concentration expressions (found in the ICE table under equilibrium).  Plugging this in, we get:

1.50 = (0.14 + 2x)²/(0.56 – x)

multiplying both sides by (0.56 – x):

1.5*(0.56 – x) = (0.14 + 2x)²

0.84 – 1.50x = 0.0196 + 0.56x + 4x²

Therefore, 0 = 4x² + 2.06x – 0.8204

Plugging into the quadratic equation (see video), we get:

x = 0.263 atm and x = -0.778 atm

We know that the partial pressure of CO₂ will decrease, so x must be positive.  Also, for x = -0.778, we get an equilibrium pressure of CO of 0.14 + 2*-0.778, which doesn’t make physical sense (we can’t have a negative partial pressure).  Therefore, we get x = 0.263 atm.

6.) If you add 0.042 mol CO to the 2L system at 293 K at equilibrium, what is Q_p?  According to Le Châtelier’s principle, to which direction will the reaction proceed?

For x = 0.263 atm, we get equilibrium partial pressures of: Pco = 0.666 atm and Pco₂ = 0.297 atm.  Putting the added number of moles into the equation for the ideal gas law (PV=nRT), we get a partial pressure of 0.505 atm.  We add this to the partial pressure of CO we already have and plug it into the expression for Q_p:

Q_p = (0.505 + 0.666)²/(0.297) = 4.62

Q>K, so the reaction will proceed to the left (toward reactants).  Le Châtelier’s principle also tells us that when we add more products to a system at equilibrium, the system will shift toward reactants.