Name:
Chem 232
Analytical Chemistry
Test 1-In Class portion
1. (10 points) My water at home is a bit hard. The Ca2+ level is 90 ppm. What is this
concentration in moles/liter?
Ppm = (mass solute/mass solution) x 106
Assume 1 liter = 1 kg
90 = (x/1000g) x106
90/106 = X/1000g
1000(90/106) g = .09g/l
.09g x (1mole/ 40.08g) = 2.25x10-3 M
Faster way 1 ppm = 1 :g/ml = 1 mg/l; 90 ppm = 90 mg/L = .09g/l...
2. (10 points) I am going to make a primary standard solution for a titration. The electronic
balance is broken, so I have to use an old mechanical balance that does not tare. The
weight of my paper is .402g, the weight of my sample and paper together is 1.245g. The
molecular weight of my standard is 204.223 g/mol. I make my solution by placing all of the
standard in a 250 ml volumetric flask. The volumetric flask is marked as 250 ±.12 ml.
What is the molarity of my primary standard solution? What is the uncertainty in this
number?
Wt = 1.245 (±.001) - .402 (±.001) = .843 ±sqrt(.0012 + .0012) (addition/subtraction)
=.843 ±.0014
Molarity = mole/liter = (.843(±.0014) / 204.223)/(.250±.00012)L
(assume uncertainty in MW is not significant)
(Multipilcation/division)
2
 .0014 
 .00012 
.01651 ± 
 +

 .843 
 .250 
2
.01651± .0017(relative)
.01651± .00003(absolute)
1
3. (10 points) For the following data determine the mean, median, and mode
(10,11,10,11,6,12,11,13)
mean = (10+11+10+11+6+12+11+13)/8 = 10.50
Median = middle value
sort in order 6,10,10,11,11,11,12,13
4th and 5 th values are 11 so median is 11
Mode = value that appears the most
= 11 (occurs 3 times)
4. (10 points) I have 2 sets of data. Set 1 has 10 samples with a mean of 100 and a
standard deviation of 5. My second set of data has 8 samples and a standard deviation of
5
What value would the mean of my second set of data have to be, to be considered
statistically different than the first set at a 95% confidence level?
Two equations that apply here are:
t=
x1 − x2
s pooled
s pooled =
n1n2
n1 + n2
s12 (n1 − 1) + s22 (n2 − 1)
n1 + n2 − 2
s pooled =
52 (10 − 1) + 52 (8 − 1)
10 + 8 − 2
=5
With 18 samples total my degrees of freedom is 16, Unfortunately the table I gave you only
had 10 or 20 for degrees of freedom, so you use the 10 and get a t of 2.228 for the 95%
confidence level.
2
100 − X 10 ⋅ 8
5
10 + 8
100 − X
2.228 =
× 2.108
5
2.228 × 5
= 100 − X
2.108
5.285 = 100 − X
2.228 =
X < 94.715 or X > 105.285
5. Describe three different methods that can be used to find the endpoint in a precipitation
reaction involving the Ag+ ion
1. You can use an electrode to follow the electrical potential of the silver in the solution
2. You can use the Fajans method. It is a dye that won’t bind to the crystal lattice when Clis in excess, but will bind to the crystal when Ag+ is in excess to you see a color change.
3. You can use the Volhard method. In this method you add an excess of Ag+, then do a
back titration of the excess Ag+ with SCN-. The endpoint of the second titration is detected
through the use of Fe3+ which forms a red complex with SCN- when you reach pass the
endpoint and have excess SCN- in the solution.
6. (10 points) Calculate the concentrations of [Ag+], [Cl-] and [Na +] in a solution where I mix
50 ml of 1M NaCl with 50 ml of 1M AgNO3. (The K sp of AgCl is 1.8x10-10)
In this solution you are mixing Ag+ and Cl- in equimolar amounts, so you will get a
precipitate. All I want is the Ag+ and Cl- that remains in solution after the precipitation
occurs, and that is described by the K sp equation:
Ksp = 1.8x10-10 = [Ag+][Cl-]
Assume X = [Ag+]= [Cl-];
1.8x10-10 = X 2
X=sqrt(1.8x10-10)
X = 1.34x10-5
3
Name:
Chem 232 Analytical Chemistry
Test 1-Take home portion
Open book - use any resource you want except other people
(but you can ask Dr. Z. if the question is unclear)
SHOW ALL WORK ON SEPARATE PAGES FOR PARTIAL CREDIT
5. (20 points)(Propagation of uncertainty & titration calculation) Consider the titration of
50.(±.1) ml of a mixture of I- and SCN- with 0.0663(±0.0002) M Ag+. The Ksp
of AgI is 8.3x10-17, the K sp of AgSCN is 1.1x10-12
A. Which precipitate first, AgI or AgSCN
B. The first equivalence point in this titration is observed at 13.3 (±0.3) ml, and the second
is observed at 29.3 (±0.2) ml. Find the molarity and the uncertainty in molarity of the
thiocyanate in the first solution
6. (10 points) Below are two sets of data.
A.) Determine the mean, standard deviation, and relative standard deviation and 95%
confidence interval of each data set. The B data set contains an outlier data point - should
this datum be retained or dropped from the data set?
Set A: 45.6, 44.7, 46.3, 45.5, 46.0, 44.9, 47.1
Set B: 45.2, 47.1, 46.7, 47.1, 50.0, 46.2, 45.8
B. Would you consider these sets of data statistically the same or different at the 99%
confidence level? (If you throw out a data point in B set, do the comparison after the datum
is removed.)
7. (10 points) Below is some data used to obtain a line of best fit
x
1
2
3
4
y
1.8
3
5.2
Notice that the Y value for the third point is missing. If the equation for the line of best fit is y
= 1.1x + .7 and the D [ ( n
∑ (x
2
i
) − (∑ xi ) 2 ] is 20, what is the missing y value?
4
5A.
AgI has the lower K sp so it precipitates first
5B. The Ag used to precipitate the SCN- is found by the volume between the 1 st and
second endpoints, thus the volume of Ag+ delivered is:
Volume of Ag+ = 29.3(±0.2)-13.3(±0.3)
=16.0±sqrt[.2 2 + .32]
=16.0±.36(absolute)
Molarity = Mole/volume =
[.016(±.00036) ×.0663(±.0002)]/.05(±.0001)
=.0212 ±sqrt[(.00013/.016)2 + (.0002/.0663)2 + (.0001/.05)2]
=0.0212 ±. 23 (relative)
= .0212±.0005(absolute)
6 ( I used a spreadsheet)
Average
Set 1 45.73
Set 2
46.87
All values
Std dev
.83
relative std dev
1.8%
95% CI
44.96-46.49
1.54
3.3%
45.44-48.30
Checking if the 50 value in the second set can be thrown out using the Q test
Q = (50-47.1)/(50-45.2) = .604
Q table = .51
Q> Q table, may be rejected
Set 2
reject value 46.35
.76
1.6%
45.55-47.15
Are the sets statistically equivalent?
.826 (7 − 1) + .761 (6 − 1)
7 + 6− 2
2
spooled =
= .80
X =
46.35 − 4573
.
.80
2
7⋅6
7+ 6
= 14
.
t table = 3.169, t<t table two groups are not statistically different
5
7
I was hoping you would set up the tables
X
Y
XY
X2
1
1.8
1.8
1
2
3
6
4
3
Y
3Y
9
4
5.2
20.8
16
10+y 28.6+3Y
30
Sum 10
Then, using either the equation for m, solve for y
m =[ 4(sum(xy) - sum(x)sum(y)]D
1.1 = [4(28.6+3y)-10(10+y)]/20
22 = 114.4 + 12y -100-10y
22+100-114.4 = 2y
y=3.8
But the easy way around was to plug the numbers into a spreadsheet, and then start trying
different values of y until you got one that fit. This worked OK, and I accepted this, but
some of your spreadsheets didn’t give enough sig figs for m and b, so you got numbers
that were not =3.8 for Y
6