Chemistry 10123 and 10125, Exam 3 Answer Key March 3, 2014

Answer Key
Chemistry 10123 and 10125, Exam 3
March 3, 2014
In the "SHOW ALL WORK" questions, include balanced, net-ionic equations for all
relevant chemical reactions. Clearly state and justify any assumptions that you may make.
Also, if you think you need to use the quadratic formula at any point, think again!
1. (10 points) SHOW ALL WORK. For H2CO3, pKa1 = 6.35 and pKa2 = 10.33. A solution is
prepared by dissolving 53.0 g of Na2CO3 and 21.0 g of NaHCO3 in water and bringing the total
volume to 1.00 L. Determine the pH of this solution after equilibrium is established.
(molar masses: Na2CO3 = 106.0, NaHCO3 = 84.0, H2CO3 = 62.0)
(53.0 g Na2CO3) (1 mole / 106 g) = 0.500 mole CO32(21.0 g NaHCO3) (1 mole / 84.0 g) = 0.250 mole HCO3
R
HCO3
H+ + CO32Ka2 = 10-10.33 = 4.7 x 10-11
I
C
0.25 M
-x
0
+x
0.50 M
+x
E
0.25 - x
x
0.50 + x
Ka2 = 4.7 x 10-11 = x (0.50 + x) / (0.25 - x)
assume x << 0.25, 0.50
4.7 x 10-11 ≈ x (0.50) / (0.25)
x = [H+] ≈ 2.35 x 10-11 M
(assumption OK!)
pH = - log (2.35 x 10-11) = 10.63
2. (10 points) In the following reaction, write complete Lewis electron dot formulas for all reactants and
products. Clearly indicate which reactant is the Lewis acid and which is the Lewis base. Use
arrows to illustrate the formation and breaking of any bonds as the reaction occurs from left to right.
CH3S- + BrCN
CH3SBr + CN-
-
H
H
C
H
Lewis
base
..
S:
..
H
+
..
: Br
..
C
Lewis
acid
N:
H
C
..
S
..
..
Br :
..
H
+
:C
N:
-
3. (10 points) SHOW ALL WORK. For NH3, pKb = 4.74. For HC2H3O2, pKa = 4.74. Determine
the pH of a solution that is 0.25 M HC2H3O2 and 0.50 M NH4NO3.
Compare the two acid dissociation equilibria. The HC2H3O2 reaction has a much
larger Ka. Therefore, it determines the H+ concentration so we can ignore the
NH4+ dissociation reaction.
NH4+
H+ + NH3
Ka = Kw / Kb = 10-14 / 10-4.74 = 5.5 x 10-10
R HC2H3O2
H+ + C2H3O2
Ka = 10-4.74 = 1.8 x 10-5
I
0.25 M
0
0
C
-x
+x
+x
E
0.25 - x
x
x
Ka = 1.8 x 10-5 = x2 / (0.25 - x)
assume x << 0.25
1.8 x 10-5 ≈ x2 / (0.25)
x = [H+] ≈ 2.12 x 10-3 M
(assumption OK!)
pH = - log (2.12 x 10-3) = 2.67
4. (12 points) SHOW ALL WORK. Nitrophenol, O2N(C6H4)OH (molar mass = 139.1), is a weak,
monoprotic acid. A certain aqueous solution of nitrophenol has a pH of 4.152. When a 50.00 mL
portion of the same nitrophenol solution was titrated with aqueous NaOH, it required 29.17 mL of
0.1200 M NaOH for complete neutralization. Determine the pKa value for nitrophenol.
Let NpOH = nitrophenol
Equilibrium:
Titration:
NpOH
NpO-
+
H+
Ka = [H+] [NpO-] / [NpOH]
moles NaOH = moles NpOH = (0.02917 L) (0.1200 mole/L)
= 0.003500 moles
[NpOH] = 0.00350 mole / 0.0500 L = 0.0700 M
[H+] = [NpO-] = 10-4.152 = 7.05 x 10-5
Ka = [H+] [NpO-] / [NpOH] = (7.05 x 10-5)2 / (0.0700) = 7.09 x 10-8
pKa = - log (7.09 x 10-8) = 7.15
5. (18 points) Indicate whether an aqueous solution of each of the following substances is acidic (A),
basic (B), or neutral (N). For each solution, write the net ionic equation for the major equilibrium
reaction that is occurring.
(a) B KOCN
OCN + H2O
HOCN + OH
(c) A Cr(NO3)3
Cr(H2O)63+ + H2O
Cr(H2O)5(OH)2+ + H3O+
(c) A (CH3)2NH2Br
(CH3)2NH2+ + H2O
(d) N NaClO4
2 H2O
(e) B Ba(HCO2)2
HCO2- + H2O
(CH3)2NH + H3O+
H3O+ + OH-
HCO2H + OH-
6. (4 points) The pH of 0.035 M Ba(OH)2 is 12.85. The pH of 12.0 M HNO3 is -1.08.
7. (3 points) Among the following, circle the weakest Brønsted-Lowry acid.
H3PO4
H2SO4
H3AsO3
H3PO3
H2SO3
H3AsO4
8. (2 points) The conjugate acid of HPO42- is H2PO4-. The conjugate base of HSO4- is SO42-.
9. (3 points) Among the following, circle the weakest Brønsted-Lowry base.
AsH2FOHHSeBr-
NH2-
10. (8 points) SHOW ALL WORK. At 75 °C, the pH of a 0.0300 M KOH solution is found to be 11.18.
Determine the value of pKw at 75 °C.
[H+] = 10-11.18 = 6.61 x 10-12 M
Kw = [H+] [OH-] = (6.61 x 10-12) (0.0300) = 1.98 x 10-13
pKw = - log Kw = - log (1.98 x 10-13) = 12.70
11. (10 points) SHOW ALL WORK. For CH3NH2, pKb = 3.36. For H2SO3, pKa1 = 1.80 and pKa2 =
7.19. Determine the equilibrium constant (Kc) for the following reaction. Your answer must contain
the appropriate balanced chemical equation(s), not just calculations.
HSO3 (aq) + CH3NH2(aq)
CH3NH3+(aq) + SO32-(aq)
HSO3
SO32- + H+
Ka2
CH3NH2 + H+
CH3NH3+
1/Ka = 1 / (Kw/Kb) = Kb/Kw
HSO3
-
+ CH3NH2
CH3NH3+ + SO32-
Kc = Ka2 Kb / Kw
Kc = (10-7.19) (10-3.36) / (1.0 x 10-14) = 2.8 x 103
12. (10 points) SHOW ALL WORK. A stock solution of NaOH is 32.0 % by mass NaOH and has a
density of 1.349 g/mL. Determine the volume (in mL) of this stock solution that is needed to prepare
6.00 L of an NaOH solution with a pH of 12.50? (molar mass: NaOH = 40.0)
Molarity of stock (concentrated) solution:
32 % by mass → 32.0 g NaOH in 100 g solution
(100 g) (1 mL / 1.349 g) = 74.1 mL = 0.0741 L
(32.0 g NaOH) (1 mole / 40.0 g) = 0.800 mole NaOH
0.800 mole / 0.0741 L = 10.80 M
Molarity of final (dilute) solution:
pOH = pKw - pH = 14.00 - 12.50 = 1.50
[OH-] = 10-1.50 = 0.0316 M
Volume of stock (concentrated) solution:
moles solute = McVc = MdVd
(10.80 moles/L) Vc = (0.0316 moles/L) (6.00 L)
Vc = 0.0176 L = 17.6 mL

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