Download Homework Solutions Week 5

1
EA02 Analytical Chemistry
Week 4 Workshop/Homework solutions
1) We start with a 0.8040-g sample of iron ore. The ore sample is dissolved and the iron
converted to the Iron(II) (Fe2+) ion. The solution is titrated with a 0.02242 M KMnO4
solution. The initial buret reading is 0.30 mL and the reading at the end point is 47.52
mL. The balanced equation for the reaction is
MnO4- + 5 Fe2+ + 8 H+ == > Mn2+ + 5 Fe3+ + 4 H2O
a) What is the percent iron in the ore?
b) If we have the following uncertainty in the measured quantities:
mass  0.0001 g
volume  0.02 mL
molarity  0.00003 M.
What is the uncertainty in the percent iron in the ore?
The first part of the problem is a stochiometry problem. Let’s look at the volume first.
47.52 – 0.30 mL = 47.22 mL
The uncertainty in an added or subtracted reading is found from the absolute
uncertainties:
.
.
2
2
Uncertainty in answer = √( 0.02) + ( 0.02) =  0.0283 mL
47.22 mL MnO4- x 0.02242 mmol MnO4- x 5 mmol iron x 55.845 mg iron
mL MnO41 mmol MnO4- mmol iron
=
295.608 mg Fe in sample or 0.295608 g Fe
0.295608 g iron x 100% = 36.77 % iron in the ore sample
0.8040 g sample
The uncertainty in multiplication and/or division is found from percent relative errors:
Volume
0.0283/47.22 x 100%
=  0.06%
Concentration
0.00003/0.02242 x 100% =  0.134%
Mass
0.0001/0.8040 x 100% =  0.012%
Atomic Weight
0.002/55.845 x 100%
=  0.004%
.
.
2
2
2
2
Uncertainty in answer = √ ( 0.06) + (0.134) + (0.012) + (0.004) =  0.1474%
2
= 0.001474
36.77 x  0.1474% = 36.77  0.05 % Fe in ore
There is a very important utility to this exercise. Which measurement contributes the
most uncertainty to the answer? If you want to decrease the uncertainty, which
measurement needs to be improved?
Got to here
2) The pKsp at 25oC and zero ionic strength of lanthanum iodate, La(IO3)3 is 10.99.
La(IO3)3 < == > La3+ + 3 IO3Assume that La(III) and iodate are the only ions formed when the salt dissolves. What is
the solubility of lanthanum iodate (FM 663.63) under these conditions?
Ksp = 10-10.99 = 1.02 x 10-11
The molar solubility of lanthanum iodate equals the concentration of lanthanum ion since
one lanthanum ion results for each formula unit dissolved.
Before
After
La(IO3)3 < == > La3+ + 3 IO3solid
0
0
solid
x
3x
Ksp = [La3+][IO3-]3 = 1.02 x 10-11 = x(3x)3 = 27x4  x4 = (1.02 x 10-11)/27
x = 7.85 x 10-4 M x 663.63 g/mol = 0.52 g La(IO3)3 /L
Homework These two problems and the following from Chapter 9: 2, 5, 8
9-2.
Write a charge balance for a solution containing H+, OH-, Ca2+, HCO3-, CO32-,
Ca(HCO3)+, Ca(OH)+, K+, and ClO4-.
For charge balance we want the sum of the positive charges in the solution equal to the
sum of the negative charges. An ion with a charge of 2 has its concentration multiplied
by 2, etc. So
[H+] + 2[Ca2+] + [Ca(HCO3)+] + [Ca(OH)+] + [K+] = [OH-] + [HCO3-] + [CO32-] + [ClO4-]
3
9-5.
(a) Suppose that MgBr2 dissolves to give Mg2+ and Br-. Write a charge balance
equation for this aqueous solution.
We will have Mg2+, Br-, OH- and H+ ions in solution. So the charge balance is
2 [Mg2+] + [H+] = [Br-] + [OH-]
(b) What is the charge balance if, in addition to Mg2+ and Br-, MgBr+ is formed?
This adds another cation to the equation
2 [Mg2+] + [H+] + [MgBr+] = [Br-] + [OH-]
9-8. Suppose that MgBr2 dissolves to give Mg2+ and Br-.
(a) Write the mass balance for Mg2+ for 0.20 M MgBr2.
(b) Write the mass balance for Br- for 0.20 M MgBr2.
The chemical reaction is
MgBr2(s)  Mg2+ + 2 Br(a) Thus one mole of magnesium ion goes into solution for each mole of magnesium
bromide dissolved. Thus [Mg2+] is equal to the concentration of dissolved solid, i.e. 0.20
M.
[Mg2+] = 0.20 M
(b)The concentration of the bromide ion is twice the concentration of dissolved salt.
[Br-] = 0.40 M
Now suppose that MgBr+ is formed in addition to Mg2+ and Br-.
(c) Write the mass balance for Mg2+ for 0.20 M MgBr2.
(d) Write the mass balance for Br- for 0.20 M MgBr2.
c) Some of the Mg2+ has become MgBr+, but the total of all the magnesium species is
still 0.20 M
[Mg2+] + [MgBr+] = 0.20 M
d) The same is true, all bromine species still equals the total amount in solution
which is 0.40M.
[Br-] + [MgBr+] = 0.40 M