1

Name:

Chem 232Analytical 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 thisconcentration in moles/liter?

Ppm = (mass solute/mass solution) x 106

Assume 1 liter = 1 kg90 = (x/1000g) x106

90/106 = X/1000g1000(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 electronicbalance is broken, so I have to use an old mechanical balance that does not tare. Theweight of my paper is .402g, the weight of my sample and paper together is 1.245g. Themolecular weight of my standard is 204.223 g/mol. I make my solution by placing all of thestandard 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 thisnumber?

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)

...

..

. . ( )

. . ( )

016510014843

00012250

01651 001701651 00003

2 2

±

+

±±

relativeabsolute

2

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 = 10.50Median = middle value

sort in order 6,10,10,11,11,11,12,134th and 5th 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 astandard deviation of 5. My second set of data has 8 samples and a standard deviation of5

What value would the mean of my second set of data have to be, to be consideredstatistically different than the first set at a 95% confidence level?

Two equations that apply here are:

tx x

sn n

n n

ss n s n

n n

pooled

pooled

=−

+

=− + −

+ −

1 2 1 2

1 2

12

1 22

2

1 2

1 12

( ) ( )

spooled =− + −

+ −=

5 10 1 5 110 8 2

5

2 2( ) (8 )

With 18 samples total my degrees of freedom is 16, Unfortunately the table I gave you onlyhad 10 or 20 for degrees of freedom, so you use the 10 and get a t of 2.228 for the 95%confidence level.

3

2 228100

510 810 8

2 228100

52108

2 228 52108

100

5285 10094 715 105 285

.

. .

..

.. .

=− ⋅

+

=−

×

×= −

= −< >

X

X

X

XX or X

5. Describe three different methods that can be used to find the endpoint in a precipitationreaction involving the Ag+ ion

1. You can use an electrode to follow the electrical potential of the silver in the solution2. You can use the Fajans method. It is a dye that won’t bind to the crystal lattice when Cl-

is 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 aback titration of the excess Ag+ with SCN-. The endpoint of the second titration is detectedthrough the use of Fe3+ which forms a red complex with SCN- when you reach pass theendpoint and have excess SCN- in the solution.

6. (10 points) Calculate the concentrations of [Ag+], [Cl-] and [Na+] in a solution where I mix50 ml of 1M NaCl with 50 ml of 1M AgNO3. (The Ksp of AgCl is 1.8x10-10)

In this solution you are mixing Ag+ and Cl- in equimolar amounts, so you will get aprecipitate. All I want is the Ag+ and Cl- that remains in solution after the precipitationoccurs, and that is described by the Ksp equation:

Ksp = 1.8x10-10 = [Ag+][Cl-]

Assume X = [Ag+]= [Cl-];

1.8x10-10 = X2

X=sqrt(1.8x10-10)X = 1.34x10-5

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Name:

Chem 232 Analytical ChemistryTest 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 of50.(±.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 Ksp 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 secondis observed at 29.3 (±0.2) ml. Find the molarity and the uncertainty in molarity of thethiocyanate 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 - shouldthis datum be retained or dropped from the data set?

Set A: 45.6, 44.7, 46.3, 45.5, 46.0, 44.9, 47.1Set 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 datumis removed.)

7. (10 points) Below is some data used to obtain a line of best fit

x y1 1.82 334 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 [ ( ] is 20, what is the missing y value?n x xi i( ) ( )2 2− ∑∑

5

spooled =− + −

+ −=

. ( ) . ( )

.

826 7 1 761 6 17 6 2

80

2 2

X =− ⋅

+=

46 35 457380

7 67 6

14

. ..

.

5A. AgI has the lower Ksp so it precipitates first

5B. The Ag used to precipitate the SCN- is found by the volume between the 1st andsecond endpoints, thus the volume of Ag+ delivered is:

Volume of Ag+ = 29.3(±0.2)-13.3(±0.3)

=16.0±sqrt[.22 + .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 Std dev relative std dev 95% CI

Set 1 45.73 .83 1.8% 44.96-46.49Set 2

All values 46.87 1.54 3.3% 45.44-48.30

Checking if the 50 value in the second set can be thrown out using the Q testQ = (50-47.1)/(50-45.2) = .604Q table = .51Q> Q table, may be rejected

Set 2reject value 46.35 .76 1.6% 45.55-47.15

Are the sets statistically equivalent?

t table = 3.169, t<t table two groups are not statistically different

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7I was hoping you would set up the tables

X Y XY X2

1 1.8 1.8 12 3 6 43 Y 3Y 94 5.2 20.8 16

Sum 10 10+y 28.6+3Y 30

Then, using either the equation for m, solve for ym =[ 4(sum(xy) - sum(x)sum(y)]D1.1 = [4(28.6+3y)-10(10+y)]/2022 = 114.4 + 12y -100-10y22+100-114.4 = 2yy=3.8

But the easy way around was to plug the numbers into a spreadsheet, and then start tryingdifferent values of y until you got one that fit. This worked OK, and I accepted this, butsome of your spreadsheets didn’t give enough sig figs for m and b, so you got numbersthat were not =3.8 for Y