Question 1. A typical air sample in the lungs contains
oxygen at 100 mm Hg, nitrogen at 573 mm Hg, carbon dioxide at 40 mm Hg, and
water vapor at 47 mm Hg. What is the
total pressure of air in the lungs?
Poxygen = 100 mm Hg
Pnitrogen = 573 mm Hg
Pcarbon dioxide = 40 mm Hg
Pwater vapor = 47 mm Hg
Ptotal = P1 + P2 + …
Ptotal = Poxygen
+ Pnitrogen + Pcarbon dioxide + Pwater vapor
Ptotal = 100 mm
Hg + 573 mm Hg + 40 mm Hg + 47 mm Hg = 760 mm Hg
Question 2. A
35.20 g sample of metal is heated to 100.0 °C and is then placed in a
calorimeter containing 42.5 g of water at 19.2 °C. If the final temperature of the metal and the
water is 29.5 °C, what is the specific heat of the solid? Assume no heat loss; the specific heat of
water is 4.184 J/g °C.
mmetal = 35.20 g mwater
= 42.5 g
Ti metal =
100.0 °C Ti water
= 19.2 °C
Tf metal = 29.5
°C Tf water
= 29.5 °C
cmetal = ? cwater
= 4.184 J/g °C
qin = - qout q = mc∆T
(mmetal)(cmetal)(∆Tmetal)
= (mwater)(cwater)(∆Twater)
(35.20 g)(cmetal)(29.5
°C – 100.0 °C) = - (42.5 g)(4.184 J/g °C)(29.5 °C – 19.2 °C)
(35.20 g)(cmetal)(-70.5
° C) = - (42.5 g)(4.184 J/g °C)(10.3 °C)
(-2481.6 g °C)(cmetal)
= - 1831.546 J
cmetal = 0.738
J/g °C
Question 3. Calculate the
amount of heat in calories required to vaporize 2.00 g of acetone, C3H6O,
at 25 °C. The heat of vaporization is
31.0 kJ/mol; the heat of fusion is 5.69 kJ/mol.
m = 2.00 g
T = 25 °C
∆Hvap = 31.0 kJ/mol
∆Hfus = 5.69 kJ/mol
q = m ∆H

Question 4. Identify each as
an ionic, molecular, network or metallic solid.
a.
silicon dioxide, SiO2 network
solid
b.
titanium tetrachloride, TiCl4 molecular
solid
c.
cesium chloride ionic
solid
Question
5. Name the types of intermolecular
forces for each molecule:
a. ethanol, CH3CH2CH2OH London-dispersion forces, dipole
forces,
hydrogen bonding
b. silicon dioxide, SiO2 London-dispersion forces
c. carbon monoxide, CO London-dispersion forces, dipole forces
Question 6. In
hexane, C6H14, the intermolecular attraction is less than
that in isopropyl alcohol, C3H7OH. Predict which liquid has the higher value for
each of the following:
- vapor pressure hexane
- viscosity isopropyl alcohol
- boiling point isopropyl alcohol
- surface tension isopropyl alcohol
Question 7. A tincture of
iodine is prepared with 0.10 g I2 and 10.0 mL ethyl alcohol, density
of ethyl alcohol solution is 1.064 g/mL.
What is the solute, and what is the solvent?
Solute: I2 Solvent: ethyl
alcohol
Question 8. Would the
following pairs of liquids be miscible or immiscible?
a. carbon tetrafluoride, CF4 and
chloroform CHCl3 miscible
b. carbon tetrafluoride, CF4, and
water immiscible
Question 9. What mass of
solution containing 6.50% sodium sulfate, by mass contains 1.50g of sodium
sulfate?
m/m% = 6.50 % Na2SO4
msolute = 1.50 g Na2SO4
msolution = ?


10. How
many grams of Na2SO4 are required to prepare 25.00 mL of
0.200 M Na2SO4 solution?

11. How
many mL of 0.150 M CaCl2 contain 6.00 g of CaCl2?

12. In
33.00 mL of 0.22 M (NH4)3PO4, there are how
many
a. Moles
of (NH4)3PO4?

b. Grams
of phosphorous?

c. Moles
of phosphate ions?


d. NH4+
ions?

13. A
solution is prepared by dissolving 81.15 grams of Na3PO4·12H2O
in enough water to prepare 500.0 mL of solution. The density of the solution is 1.077 g/mL. Calculate
a. The
% anhydrous salt

b. The
% hydrated salt

c. The
molarity of Na3PO4

14. If 15.00 g of a mixture of 25.00% CaCl2 and
75.00% inert material are dissolved in 1.00 liter of water, how many mL of
0.500 M silver nitrate solution would be required to precipitate all of the
chloride according to the equation:
2
AgNO3(aq) + CaCl2(aq) à Ca(NO3)2(aq) + 2
AgCl(s)

15. A 1.00 gram
sample of (NH4)2SO4 is added to 50.0 mL of
0.450 M (NH4)3PO4 solution. Assuming no volume change, what are the
resultant molarities of all ions in solution?
Mol (NH4)2SO4
= 1.00 g x 1 mol (NH4)2SO4 /132.15g = 7.56e-3
mol (NH4)2SO4
Mol (NH4)3PO4 = 0.0500 L x
0.450 mol (NH4)3PO4/L = 0.0225 mol (NH4)3PO4
[NH4+1 ] =
[2(7.56e-3 mol (NH4)2SO4) + 3(0.0225 mol(NH4)3PO4)]/0.0500
L = 1.65 M NH4+1
[SO4-2 ] = 7.56e-3 mol SO4-2
/0.0500 L = 0.151 M SO4-2
[PO4-3 ] = 0.0225 mol PO4-3/0.0500
L = 0.450 M PO4-3
16. 2.500
g of K3PO4 are dissolved in 100.0 mL of deionized
water. To this solution are added 75.00
mL of 0.300 M AgNO3 solution.
K3PO4(aq) + 3
AgNO3(aq) à 3 KNO3(aq) + Ag3PO4(s)
Initial 0.01178 mol
0.0225 mol
Change -X -3X +3X +X
X =
0.00750 mol
Final 0.00428 mol 0
mol 0.0225 mol 0.00750 mol
a. How
many grams of Ag3PO4 will precipitate?

b. How
many grams of which reactant remains unreacted?

c. What
is the final concentration of K+?

17. When
35.00 mL of 0.100 M iron(III) chloride are added to 20.00 mL of 0.200 M
potassium phosphate, a precipitate of iron(III) phosphate is observed to
form. Inventory the concentrations of
all ions in solution and calculate the mass of the precipitate.
Fe(Cl)3 + K3PO4 à FePO4 +
3 KCl
I
|
3.50
e-3 mol
|
4.00e-3
mol
|
|
0
mol
|
0
mol
|
||
|
-x
mol
|
-x
mol
|
|
+x
mol
|
+3x
mol
|
||
E
|
0
mol
|
5.0e-4
mol
|
|
3.5
e-3 mol
|
1.05
e-2 mol
|
[Fe+3] = 0M
[PO4-3] = 5.0 e-4/0.055 = 0.0091
M
[K+] = 3(4.00 e-3)/0.055 L = 0.218 M
[Cl-] = 1.05 e-2/0.055 L = 0.191 M
3.5 e-3 mol FePO4 x 150.82 g FePO4/mol
= 5.27 g FePO4
18. A 35.00 mL
aliquot of 0.1861 M sulfuric acid is mixed with 20.00 mL of a 0.3016 M barium
hydroxide solution. Write the balanced
chemical equation for the reaction and determine [H3O+],
[OH- ],
[Ba+2], [SO4-2], pH, and pOH. Find the mass of Barium sulfate that
precipitates.
H2SO4 +
Ba(OH)2 à 2 H2O + BaSO4(s)
|
I
|
6.514 e -3 mol
|
6.032 e-3 mol
|
|
|
0 mol
|
D
|
-x mol
|
-x mol
|
|
+ 2x mol
|
+ 1x mol
|
E
|
4.82 e-4 mol
|
0 mol
|
|
|
6.032 e-3 mol
|
[Ba+2] = 0 M Ba+2
[SO4-1] = 4.82 e-4mol /0.05500 L =
8.18 e-3 M SO4-1
[H3O+1] = 2(4.82 e-4)/0.05500 L = 0.0175
M pH = 1.756
[OH-1] = kw/[H3O+1] = 1.0
e-14/0.0175 = 5.7 e -13 pOH
= 12.24
mass BaSO4 = 6.032 e-3 mol * 233.4 g/mol = 1.408
g BaSO4
19. Fill in the
table below
Conjugate acid
|
Conjugate base
|
HC2H3O2
|
C2H3O2-1
|
HCN
|
CN-1
|
NH4+1
|
NH3
|
H3O+1
|
H2O
|
H2O
|
OH-1
|
CH3NH3+
|
CH3NH2
|
20. Calculate
the hydroxide ion concentration in a solution if 35.27 mL of it to exactly
neutralize 24.30 mL of a 0.07243 M solution of sulfuric acid.

21. A solution
is prepared by dissolving 0.1162 grams of KOH in 250.0 mL of solution. Calculate
a. [H+]
= 1.2107e-12
b. [OH- ] =
8.284e-3
c. pH
= 11.9182
d. pOH
= 2.0818
22. Fill
in the table below
[H3O+]
|
[
|
pH
|
pOH
|
3.582 x 10-6
|
2.792e-9
|
5.4459
|
8.5541
|
1.539e-6
|
6.498 x 10-9
|
5.8128
|
8.1872
|
1.705e-4
|
5.864e-11
|
3.7682
|
10.2318
|
2.986e-5
|
3.349e-10
|
4.5249
|
9.4751
|
23. A primary
standard is made by dissolving 3.7627 grams of oxalic acid dihydrate in enough
water to make 500.0 mL of solution. A
25.00 mL aliquot of the oxalic acid solution requires 41.87 mL of potassium
hydroxide solution to titrate it to a phenophalein end point. A 25.00 mL sample of vinegar is titrated with
15.97 mL of the sodium hydroxide solution.
What is the concentration of acetic acid in the vinegar?
Concentration of oxalic acid standard

Concentration of potassium hydroxide solution

Concentration of Acetic acid

24. Solution A
is a solution of HNO3 with a pH of 2.80. Solution B is a solution of NaOH with a pH of
11.30. Calculate the pH of the following
solutions:
Solution A pH = 2.80 [H3O+]
= 10-2.80 = 1.6e-3 M
Solution B pH = 11.30 pOH
= 2.70 [OH-1] =
10-2.70 = 2.0e-3 M
a. 10.00
mL of A and 5.00 mL of B

I
|
1.6e-5 mol
|
1.0e-5 mol
|
|
|
D
|
-x mol
|
-x mol
|
|
|
E
|
0.6e-5 mol
|
0 mol
|
|
|
[H3O+] = 0.6 e-5mol/0.015 L = 4e-4 M
pH = 3.40
b. 10.00
mL of B and 5.00 mL of A
HNO3
+ NaOH à H2O +
NaNO3
I
|
0.8e-5 mol
|
2.0e-5 mol
|
|
|
D
|
-x mol
|
-x mol
|
|
|
E
|
0mol
|
1.2e-5 mol
|
|
|
[OH-1] = 1.2 e-5mol/0.015 L = 8e-4 M
pOH = 3.10
pH = 10.90
c. How
many mL of A must be added to 25.00 mL of B to make a solution of pH = 7.00?
pH = 7 when {H3O+] = {OH-1]
at the equivalence point of a titration
We need to find out how many mL of A are required to
completely react with 25.00 mL of B

25. What
is the pH and pOH of a solution prepared by dissolving 2.5 g hydrochloric acid,
HCl, in water to make 425 mL of solution?

pH = - log[H+] = - log(0.16) = 0.792
pH + pOH = 14.000 à pOH = 14.000 – pH =
14.000 – 0.792 = 13.208