Practical Project
ANSWERS TO CLASS EXERCISES
Chapter 1
1.1 Find a method for the analysis of iron in aqueous solution. Look for an ashing method for a
similar material to Corn Flakes, eg bread, cake etc. Once the organic matrix is destroyed, the
aqueous solution analysis will be suitable.
1.2
Physical state of the sample
most techniques require solutions, so if the sample is solid,
this will require some sort of dissolving, requiring time
if a possible technique can handle solids, it will be timeefficient
Complexity of sample matrix
certain techniques handle matrix interference better than
others (eg ICP vs AAS) so must be considered if the sample
is likely to contain a complex matrix
Accuracy required
sometimes quick with less accuracy will suit the job
requirements better than slow and spot on
Number of analyses to be performed
Extended sample preparation time becomes an issue when
large sample batches are involves
Time allowable for result
Different situations require different turnaround times for
the results
Multiple analytes
If the technique can do more than one analyte accurately,
there is a clear time saving
Existing laboratory equipment
Clearly this will be preferred where possible
Laboratory budget for purchases
Where equipment isn’t already available, how much
money can spent
Skill of technician assigned to task
While anyone can press the “Start” button, the complexity
of setup and troubleshooting varies widely and needs to
be matched with the person who will do the job
Instrument characteristics
Accuracy
Time Required
Costs
Operator Skill
Multi-component
Volumetric
Medium
Medium
Low
Low
No
IR
Medium
Fast
Medium
Low
Yes
UV/VIS
High
Fast
Medium
Low
Mostly no
Accuracy
Time Required
Costs
Operator Skill
Multi-component
Flame Em.
Medium
Fast
Low
Low
Yes
ISE
Low-medium
V. fast
Very low
Low
No
GC
High
Slow
Medium-high
Medium-High
Yes
FAAS
Medium
Medium
Medium
Low-Medium
Yes (fast sequential
only)
Project Theory Notes
Answers to Exercises
1.3(a) Acid in vinegar by titration or GC
concentration
physical state
matrix
accuracy
number of analyses
time
multiple analytes
existing equipment
budget
skill
best suited to titration, need dilution for GC
both require solution – equivalent
brown vinegar may obscure endpt detection
equivalent
each could handle this easily
equivalent – pipetting time & replicates extend time for titration
not necessary
no GC
simplest GC still $10k +
technician could handle GC
There is no justification for buying a GC, so the titration is the best choice.
(b) nitrate in soil by HPLC or ion-selective electrode
concentration
physical state
matrix
accuracy
number of analyses
time
multiple analytes
existing equipment
budget
skill
both may require dilution, given larger sample quantities due to nonuniform sample
both require solution – equivalent
suspended solids will require filtration for HPLC
HPLC accuracy not necessary in this instance
ISE a clear winner, long per sample time for HPLC a problem
ISE a clear winner per sample and overall
not necessary
each is available
not necessary
ISE useable by anyone
The ISE is the obvious choice because of its speed where lots of samples are required, and high accuracy
isn’t important.
Chapter 2
2.1
2: less than 100 mg is inaccurate, > 25 g may be difficult to handle/dissolve
3 & 4: accuracy & speed
2.2
(a)
(b)
(c)
(d)
(e)
(f)
11.1 mg/kg (in mg)
222 g/L (g)
3.33 %w/w (mg)
0.444 g/100 mL (mg)
5.55 mg/kg (ug)
0.666 mg/L (ug)
in 1 g or mL
in 100 mL
0.111 mg
0.222 g
33.3 mg
4.44 mg
5.55 ug
0.666 ug
1.11 mg/L
2.22 g/L
333 mg/L
44.4 mg/L
55.5 ug/L
6.66 ug/L
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Dilution (÷)
Conc. (x)
x 5-10
÷2
÷ 5-10
÷ 20
÷2
x5
Project Theory Notes
Answers to Exercises
2.3(a) 1 mL of milk contains 1.5 mg of Ca, so in 100 mL gives 15 mg/L, which is a little low.
2 or 3 mL would give a better Ca concentration, but neither are appropriate volumes.
5 mL is the smallest pipette volume, unless completely unavoidable, which gives a concentration
of 75 mg/L, which is a little too high (eg if recovery checks were required). However, doubling
the final volume to 200 mL avoids this.
(b)
1g of 10 mg/kg contains 10 ug, so gives 100 ug/L in 100 mL – too high.
Either half the mass or double the volume to give 50 ug/L. The volume solution is better – soil
is not homogeneous, so small sample amounts are not desirable.
(c)
1 mL of 40 g/L is 0.04 g, in 100 mL is 0.4 g/L – too low. 10 mL is 4 g/L which is perfect.
(d)
AR grade sodium nitrate can be assumed to be close enough to 100%w/w.
1 g of it contains 729 mg of nitrate (using FW ratio 62/85). In 100 mL, this gives 7290 mg/L, a
little high!
0.1 g of sample diluted to 500 mL is the maximum achievable without serial dilution, and gives
a concentration of 146 mg/L.
This could be then diluted 25 to 100, giving 36 mg/L. If closer to 50 is really necessary, increase
the sample mass to 0.14 g.
2.4(a) 1 g contains 0.1 ug, so in 100 mL this is equal to 1 ug/L.
The simplest way to get to 20-25 ug/L would be 20-25 g of soil, and since this method would
require an extraction and filtration, this large sample mass wouldn't be a problem.
(b)
1 mL contains 0.3 mg, so in 100 mL, you get 3 mg/L => 15 mL would give 45 mg/L.
(c)
1 tablet = 1g which contains 500 mg; in 100 mL gives 5000 mg/L – way too high.
100 mg in 100 mL is 500 mg/L, in 500 mL is 100 mg/L.
The rest must be by dilution – 20x ( 5 to 100) gets 5 mg/L.
(d)
95%w/w means 95 g/100g, so 1 g of steel contains 0.95 g (950 mg) of iron.
In 100 mL, this is 9500 mg/L, a long way from the standards. We want to avoid large serial
dilution.
Start by reducing the sample mass to 0.1 g, so the concentration is now 950 mg/L.
Next, increase the VF to 500 mL, so the concentration is now 190 mg/L.
To get to 5 mg/L requires a dilution factor of 38 - let's say 40 - which can be achieved by 5 mL to
200.
So the method is:
1. Weigh out 0.1 g, dissolve and make up to 500 mL.
2. Pipette 5 mL of this solution into a 200 mL VF and make up to volume.
(e)
1 g of Weetbix contains 2.9 mg of Na, so in 100 mL, this is 29 mg/L.
0.5 g would be equivalent to about 14.5 mg/L, a little high so 0-3.0.4 g would be OK.
(f)
1%w/w = 1 g/100 g or 1000 mg/100 g, giving 10 mg/g.
In 100 mL, this is 100 mg/L. 0.1 g in 100 mL is 10 mg/L. If this sample amount is too small, then
0.5 g in 500 mL gives the same.
(g)
200 mg/100 g means 2 mg/g. Therefore, 2.5 g of sample gives 5 mg.
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Project Theory Notes
2.5
Answers to Exercises
For the sake of space, the full working of amounts (eg how much in 1 mL or g) is not included.
(a)
S
SP
SPS
Procedure
100 mg/L
500 uL of 1000 mg/L
6.1 mg/L
10 mL to 250 mL
Method 1 – mass in original sample
Method 2 – concentration in final solution
S – 10 mL of 100 mg/L is 1 mg.
S – 100 mg/L ÷ 25 (diln factor) = 4 mg/L
SP – 0.5 mL of 1000 mg/L is 0.5 mg.
SP – 0.5 x 1000 ÷ 250 = 2 mg/L
SPS – 6.1 mg/L in 250 mL = 1.525 mg.
SPS – 6.1 mg/L.
%RC = 105%
(b)
S
SP
SPS
Procedure
50 mg/kg
5 mL of 50 mg/L
4.4 mg/L
5 g in 100 mL
Method 1 – mass in original sample
S – 5 g of 50 mg/kg contains 0.25 mg.
Method 2 – concentration in final solution
S – 0.25 mg in 100 mL = 2.5 mg/L
SP – 5 mL of 50 mg/L is 0.25 mg.
SP – 5 x 50 ÷ 100 = 2.5 mg/L
SPS – 4.4 mg/L in 100 mL = 0.44 mg.
SPS – 4.4 mg/L.
%RC = 76%
(c)
S
SP
SPS
Procedure
50 mg/L
2 mL of 1000 mg/L
40 mg/L
50mL to 100 mL
Method 1 – mass in original sample
Method 2 – concentration in final solution
S – 50 ÷ 2 (diln factor) = 25 mg/L
S – 50 mL of 50 mg/L contains 2.5 mg.
SP – 2 x 1000 ÷ 100 = 20 mg/L
SP – 2 mL of 1000 mg/L is 2 mg.
SPS – 40 mg/L.
SPS – 40 mg/L in 100 mL = 4 mg.
%RC = 75%
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Project Theory Notes
Answers to Exercises
2.6(a)
S
SP
SPS
Pr
41.2 mg/100 mL
10 mL of 1000 mg/L
398 mg/L
25 mL to 50 mL
Method 1 - mass in original sample (mg)
Method 2 - mg/L in the analysed solution
S - 25 mL of the sample contains 10.3 mg of SP - 10 mL of 1000 mg/L diluted to 50 mL of
analyte.
200 mg/L.
SPS - the analysed 398 mg/L comes from 50 S - 412 mg/L diluted 25 to 50 becoming 206
mL, so it contains 398 ÷ 20 = 19.9 mg.
mg/L.
SP - 10 mL of 1000 mg/L = 10 mg.
SPS – 398 mg/L
%RC = 96%
(b)
S
SP
SPS
Pr
12.3 %v/v
200 uL
16.5 %v/v
5 mL
Method 1 - volume (mL)
S - 5 mL contains 0.615 mL of ethanol.
Method 2 - %v/v
SP - 0.2 in 5 mL which is 4%
SPS - 5 mL contains 0.825 mL
S & SPS – 12.3 & 16.5%
SP – 0.2 mL
%RC = 105%
(c)
S
SP
SPS
Pr
475 mg/100 g
5 mL of 1000 mg/L
15.9 mg/L
100 mL, 10 to 100 mL, 2.4376 g
Method 1 - mass in original sample (mg)
Method 2 - mg/L in analysed solution
S - 2.4376 g of sample contains 4.75 mg/g x The numbers in the calculation are the same,
2.4376 = 11.58 mg.
just as mg/L, not mg.
SPS - contains 15.9 mg/L x 10 (DF) ÷ 10 (in 100
mL) = 15.9 mg.
SP - 5 mg
%RC = 86%
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Project Theory Notes
Answers to Exercises
2.6(d)
S
SP
SPS
Pr
27.4 mg/L
200 uL of 200 mg/L
145.9 ug
5 mL to 25, 20 mL
Method - mass in original sample (ug)
S - 5 mL contains 0.005 x 27.4 = 137 ug
SP - 0.2 mL x 200 ug/mL = 40 ug.
SPS - 25 mL contains 145.9 x 25/20 = 182.4 ug
%RC = 113%
(e)
S
SP
SPS
Pr
0.115%w/w
5 mL of 500 mg/L
21.6 mg/L
100 mL, 25 to 100 mL, 5.0762 g
Method 1 - mass in original sample (mg)
Method 2 - mg/L in analysed solution
S - 5.0762 g contains 0.115 % x 5.0762 = 5.84
mg.
SPS - 21.6 mg/L x 4 (DF) ÷ 10 (in 100 mL) = 8.64
mg.
SP - 2.5 mg.
S - 5.84 ÷ 0.1 ÷ 4 = 14.6 mg/L.
SP - 5 mL x 500 mg/L to 100 mL then diluted
by 4 giving 6.25 mg/L
SPS – 21.6 mg/L
%RC = 112%
(f)
S
SP
SPS
Pr
10.1%w/w
5 mL of 5 g/L
32.4 mg/L, 0.5682 g
500 mL, 10 to 50 mL
Method 1 - mass in original sample (mg)
Method 2 - mg/L in analysed solution
S – 0.5682 g contains 10.1 % x 568.2 = 57.4 mg.
SPS – 32.4 mg/L x 5 (DF) ÷ 2 (in 500 mL) = 81
mg.
SP – 5 mL of 5000 mg/L = 25 mg.
S - 57.4 ÷ 0.5 ÷ 114.8 mg/L diluted x 5 is 22.96
mg/L.
SP – 5 mL of 5000 mg/L in 500 mL is 50 mg/L,
then diluted by 5 giving 10 mg/L
SPS – 32.4 mg/L
%RC = 94%
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Project Theory Notes
Answers to Exercises
2.6(g)
S
SP
SPS
Pr
2.69%w/w
20 mL of 1000 mg/L K
1.3825 g, 52.3 mg/L
100 mL, 10 mL to 100 mL,
Method 1 - mass in original sample (mg)
Method 2 - mg/L in the analysed solution
S – 2.69%w/w x 1382.5 mg = 37.19 mg.
S – 37.19 mg in 100 mL is 371.9 mg/L. Diluted
by a factor of 10 gives 37.19 mg/L.
SPS – 52.3 mg/L x 10 (dilution) ÷ 10 (/100 mL)
= 52.3 mg.
SP – 200 mg/L in 1st solution, 20 mg/L in
analysed
SP - 20 mg
%RC = 76%
(h)
S
SP
SPS
Pr
385 mg/L
5 mL of 1000 mg/L
138 mg/L
25 mL, 100 mL
Method 1 - mass in analysed solution (mg)
Method 2 - mg/L in the analysed solution
S - 25 mL of 385 mg/L = 9.63 mg.
S - 25 mL of 385 mg/L sample ÷ 100 mL = 96.3
mg/L.
SP - 5 mL of 1000 mg/L ÷ 100 mL = 50 mg/L
spike.
SPS - 100 mL of 138 mg/L = 13.8 mg.
SP - 5 mL of 1000 mg/L = 5 mg.
%RC = 83%
(i)
S
SP
SPS
Pr
395 ug/kg
5 mL of 100 ug/L
62.3 ug/L
2.5346 g, 25 mL
Method 1 - mass in original sample (ug)
Method 2 - ug/L in the analysed solution
S - 395 ug/kg, which is .395 ug/g or 1.00 ug in SP - 0.5 ug/25 mL = 20 ug/L.
the sample mass.
S - 1.00 ug/25 mL = 40 ug/L
SPS - 62.3 ug/L in 25 mL = 1.56 ug.
SP - 5 mL of 100 ug/L = 0.5 ug.
%RC =112 %
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Project Theory Notes
Answers to Exercises
2.6(j)
S
SP
SPS
Pr
147 mg/100 mL
5 mL of 1000 mg/L
6.21 mg/L
5 mL to 100 mL, 5 mL to 100 mL
Method 1 - mass in original sample (mg)
Method 2 - mg/L in the analysed solution
S – 5 mL of 147 mg/100 mL = 7.35
S – 147 mg/100 mL is 1470 mg/L; two 5-100
dilutions gives 3.68 mg/L
SPS – 6.21 x 20 (dilution) ÷ 10 (/100 mL) =
SP – 1000 with two 5-100 dilutions = 2.5
12.42
mg/L
SP – 5 mg
%RC = 101%
(k)
S
SP
SPS
Pr
3.18 %w/w
5 mL of 1000 mg/L
0.2387 g, 44.8 mg/L
250 mL
Method 1 - mass in original sample (mg)
Method 2 - mg/L in the analysed solution
S – 3.18%w/w x 238.7 mg = 7.59 mg
S – 7.59 mg in 250 mL = 30.4 mg/L
SPS – 44.8 mg/L ÷ 4 (/250 mL) = 11.2 mg.
SP – 5 x 1000 ÷ 250 = 20 mg/L
SP - 5 mg
%RC = 72%
(l)
S
SP
SPS
Pr
24.2 %w/w
1.0583 g
9.6571 g, 31.8 g/L
100 mL
Method 1 - mass in original sample (g)
Method 2 - g/L in the analysed solution
S – 24.2 %w/w x 9.6571 g = 2.34 g
S – 2.34 g in 100 mL = 23.4 g/L
SPS – 31.8 g/L ÷ 10 (/100 mL) = 3.18 g
SP – 1.058 g in 100 mL = 10.6 g/L
%RC = 80%
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Project Theory Notes
2.7
Answers to Exercises
Because this makes the method for the recovery check different to that for the sample, eg the
matrix will be more dilute.
2.8(a) (i) 22.7 x 5 x 10 = 1135 mg/L
(ii) 1135 ÷ 100 = 11.35 mg
(iii) 5 mg
(iv) 5 mL of 1000 mg/L
(b)
(i) 0.39 x 4 = 1.56 ug/20 mL = 78/L
(ii) 1.56 ug
(iii) 1 ug
(iv) 10 mL of 100 ug/L (because 1 mL of 1000 isn’t accurate enough).
(c)
There is 5 mg Cl in 10 g sample. Add 2.5 mg; 5 mL of 500 mg/L.
(d)
5 mL contains 0.2 g. 0.1 g pure ethanoic acid could be added directly.
2.9
The problem is occurring during the wet ashing, since the results for a solution spiked after the
ashing are good.
2.10 Shorter time with higher temperature, blend the bread instead of breaking it by hand, don’t predry the basin
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