CHEM 213
Chemical Analysis
Practice Final
1 __20___ (of 20)
2 __10___ (of 10)
3 __20___ (of 20)
4 __10___ (of 10)
5 __10___ (of 10)
6 __10___ (of 10)
7 __10___ (of 10)
8 __10___ (of 10)
Σ_100_ (of 100)
%
KEY
Name:___________________________________________
(please print)
1. The following are relative peak areas for chromatograms of standard solutions
of methyl vinyl ketone (MVK).
x, MVK,
yi, Relative
xi2
yi2
xiyi
mmol/L
Peak Area
0.500
3.76
0.25
14.1376
1.88
1.50
9.16
2.25
83.9056
13.74
2.50
15.03
6.25
225.9009
37.575
3.50
20.42
12.25
416.9764
71.47
4.50
25.33
20.25
641.6089
113.985
5.50
31.97
30.25
1022.0809
175.835
Σxi = 18.00
Σyi = 105.67 Σxi2 = 71.5 Σyi2 = 2404.6103 Σxiyi = 414.485
a. Calculate by the least squares method the equation of the best straight line for
the calibration curve. Report you result in the form y = [m(±sm)]x + [b(±sb)].
(15 points)
xi = 3.00, ỹ i = 17.612
Sxx = Σxi2 – (Σxi)2/N = 71.5 – 324/6 = 17.50
Syy = Σyi2 – (Σyi)2/N = 2404.6103 – 11166.1489/6 = 543.5855
Sxy = Σxiyi – (Σxi)(Σyi)/N = 414.485 – (18.00)(105.67)/6 = 97.475
m = Sxy/Sxx = 5.57
b = y – mx = 17.612 – (5.57)(3.00) = 0.902
sr =
sm =
sb = sr
Syy - m2Sxx
N-2
sr2
(0.40304)
=
Sxx
543.5855 - (5.57)2(17.5)
=
6-2
2
= 0.40304
= 0.09634
17.5
Σxi2
=
NΣxi2 - (Σxi)2
0.40304
71.5
(6)(71.5) - 324
= 0.332588
Equation of best line: y = [5.57(±0.10)]x + 0.902(±0.33)
b. A sample containing MVK yielded a relative peak area of 6.3. Calculate the
concentration of MVK in the solution, assuming that the result represents a
single measurement as well as the mean of four measurements. (5 points)
x = (y – b)/m = (6.3 – 0.902)/5.57 = 0.96912 = 0.97
Uncertainty in x, single measurement (M = 1):
sr
sc =
m
1
M
+
1
N
+
(yc - y)2
2
m Sxx
=
0.40304
1
5.57
1
+
1
6
+
(6.3 - 17.612)
2
(5.57)217.5
= 0.086 mmol/L
Measurement median of four, M = 4, sc = 0.058
The final answer is 0.97 ± 0.09 mmol/L (1 measurement), or 0.97 ± 0.06
mmol/L (median of 4 measurements)
2
2. Suppose that 0.010 M Mn2+ is titrated with 0.005 M EDTA at pH 7.00.
Kf(MnY2-) = 1013.87.
pH
α4
pH
α4
pH
α4
-23
-7
0
1.3•10
5
3.7•10
10
0.36
1
1.9•10-18
6
2.3•10-5
11
0.85
-14
-4
2
3.3•10
7
5.0•10
12
0.98
3
2.6•10-11
8
5.6•10-3
13
1.00
-9
-2
4
3.8•10
9
5.4•10
14
1.00
a. What is the concentration of free Mn2+ (pMn2+) at the equivalence point. (5
points)
One volume of Mn2+ will require two volumes of EDTA to reach the equivalence
point. The formal concentration of MnY2- at the equivalence point is (1/3)(0.010)
= 0.00333 M
Mn2+ + EDTA ↔ MnY2x
x
0.00333 - x
(0.00333 – x)/x2 = αY4-Kf = (5.0•10-4)1013.87 = 3.7•1010
x = [Mn2+] = 3.0•10-7
b. What is the quotient [H3Y-]/[H2Y-] in the solution when the titration is just
63.7% of the way to the equivalence point? Use the equilibrium given below
for your calculation: (5 points)
= 10-2.66
Because the pH is 7.00, the ratio [H3Y-]/[H2Y-] is constant throughout the entire
titration.
[H2Y2-][H+]/[H3Y-] = K4
[H3Y-]/[H2Y-] = [H+]/K4 = 10-7.00/10-2.66 = 4.6•10-5
3
3. Consider the titration of 50.0 mL of 0.050 M malonic acid (Ka1 = 1.42•10-3, Ka2
= 2.01•10-6) with 0.100 M NaOH. The titration reaction occurring is:
HOCCH2CO2H + OH- → -O2CCH2CO2H + H2O
OCCH2CO2H + OH- → -O2CCH2CO2- + H2O
Designate malonic acid as H2M and use the following values:.
a. At 0.0 mL (3 points)
H2M ↔ H+ + HM0.050 – x
x
x
2
-3
x /(0.050 – x) = K1 → x = 7.75•10
pH = 2.11
b. At 8.0 mL (3 points)
H2M +
OH- →
HMInitial:
25
8
Final:
17
o
8
pH = pKa1 + log([HM-]/[H2M]) = 2.487 + log(8/17) = 2.52
+
H2O
-
c. At 12.5 mL (4 points)
Vb = ½ Ve → pH = pKa1 = 2.85
d. At 25.0 mL (4 points)
At the first equivalence point, H2M has been converted to HM-.
+
[H3O ] =
Ka1Ka2F + Ka1Kw
Ka1 + F
where F = (50/75)(0.050) = 0.0333 M
[H3O+] = 5.23•10-5 M → pH = 4.28
e. At 50.0 mL (3 points)
At the second equivalence point, H2M has been converted to M2-:
M2- + H2O ↔ HM- + OH(50/100)(0.050) –x
x
x
x2/(0.025 – x) = Kb1 = Kw/Ka2
→ x = 1.12•10-5 M
pH = -log(Kw/x) = 9.05
f. At 56.3 mL (3 points)
There are 6.3 mL of excess NaOH
[OH-] = (6.3/106.3)(0.100) = 5.93•10-3 M
→pH = 11.77
4
4. A 20.0 mL solution of 0.005 M Sn2+ in 1 M HCl was titrated with 0.02 M Ce4+
to give Sn4+ and Ce3+. Calculate the potential at the following volumes of
Ce4+, using the half-cell potentials: EºSn = 0.139 V, EºCe = 1.47 V.
a. At 0.100 mL (3 points)
E = 0.139 – (0.0592/2)•log([Sn2+]/[Sn4+])
= 0.139 – (0.0592/2)•log(9.90/0.100) = 0.080 V
b. at 10.00 mL (4 points)
2E = 2(0.139) – 0.0592•log([Sn2+]/[Sn4+])
E = 1.47 – 0.0592•log([Ce3+]/[Ce4+])
3E = 1.748 – 0.0592•log([Sn2+][Ce3+]/[Sn4+][Ce4+])
At the equivalence point, [Sn4+] = ½[Ce3+] and [Sn2+] = ½[Ce4+] wich makes
the log term 0. Therefore, 3E = 1.748, and E = 0.583 V
c. at 10.10 mL (3 points)
E = 1.47 – 0.0592•log([Ce3+]/[Ce4+])
= 1.47 – 0.0592•log(10.0/0.10) = 1.35 V
5
5. Using activities from the table below, calculate the pH and concentration of
the hydronium ion, H3O+ in 0.050 M LiBr at 25 ºC. (10 points)
At an ionic strength of 0.050 M:
γH3O+ = 0.85 and γOH- = 0.81
[H3O+]γH3O+[OH-]γOH- = (x)(0.85)(x)(0.81) = 1.0•10-14
x = [H3O+] = 1.2•10-7 M
pH = -log(1.2•10-7)(0.85)] = 6.99
6
6. a. Fe(III) precipitates from acidic solution by addition of OH- to form Fe(OH)3
(s). At what pH (i.e., concentration of OH-) will the concentration of Fe(III) be
reduced to 1.0•10-10 M? Ksp(Fe(OH)3) = 1.6•10-39 (5 points)
[Fe3+][OH-]3 = (10-10)[OH-]3 = 1.6•10-39
→ [OH-] = 2.5•10-10
pH = 14 – pOH = 4.40
b. If Fe(II) is used instead, at what pH would the Fe(II) concetration be reduced
to 1.0•10-10? Ksp(Fe(OH)2)= 7.9•10-16 (5 points)
[Fe2+][OH-]3 = (10-10)[OH-]3 = 7.9•10-16
→ [OH-] = 2.8•10-3
pH = 14 – pOH = 11.45
7
7. A 48.0 wt % solution of HBr in water has a density of 1.50 g/mL.
a. Find the formal concentration of HBr. (3 points)
(48.0 g HBr)/100.0 g solution)(1.50 g solution/mL solution)
= 0.720 g HBr/mL solution = 720 g HBr/L solution = 8.90 M
b. What mass of solution contains 36.0 g of HBr? (2 points)
36.0 g HBr/(0.480 g HBr/g solution) = 75.0 g solution
c. What volume of solution contains 233 mmol of HBr? (2 points)
233 mmol = 0.233 mol
0.233 mol/8.90 mol/L = 0.0262 L = 26.2 mL
d. How much solution is required to prepare 0.250 L of 0.160 M HBr? (3 points)
Mconc•Vconc = Mdil•Vdil
(8.90 M)•(x mL) = 0.160 M)•(250 mL)
→ x = 4.49 mL
8
8. Calculate the voltage of the following cell (10 points):
Fe(s)│FeBr2(0.010 M)││NaBr (0.050 M)│Br2(l)│Pt(s)
using the following standard half-cell potentials:
Br2(l) + 2 e- ↔ 2Br- Eº = 1.078 V
Fe2+ + 2e- ↔ Fe(s) Eº = -0.44 V
E = {1.078 – (0.0592/2)•log(0.050)2} – {-0.44 – (0.0592/2)•log(1/0.010)} = 1.65 V
9