Chemistry 360
First Examination
October 29, 1997

Directions: There are 5 questions, each worth equal points. Some are more difficult than others. Read over all questions before you start to answer. Answering the harder questions last will insure that you obtain the maximum score. You must show your work and pertinent formula's to get full credit. The final number answer is only about ten percent of the total score.

Useful formulas:

Useful data:

Values of Q for datum rejection

Student's t Values

1. Measurements and Experimental Error

1. What are the main steps in a chemical analysis?

1. Understand Problem
2. Formulate Hypothesis
3. Sample
4. Separation
5. Measurement
6. Data Analysis
7. Test Hypothesis
8. Report

B) For each patterned target below, indicate whether the result is; accurate, precise, both, or neither.

C) Define the following concentration units;

1. molarity

moles of solute per liter of solution

2. weight percent

(weight of substance/total weight)x100

3. formal concentration

formula weights of solute per liter of solution

4. parts-per-million

(amount of substance/total amount)x10⁶

2. Statistics

A) Determine whether or not the 22.56 value should be rejected from the following data set;

20.49, 20.95, 20.43, 20.39, 21.23, 22.56

Show your work.

• Using the Q-test, calculate Q = Gap/Range and compare the result to the table value.
• If Q > Q_TABLE, then the datum may be rejected.
• Gap = |22.56 – 21.23| = 1.33, Range = |22.56 – 20.39| = 2.17, Q = 1.33/2.17 = 0.613
• The Table value for 6 data is Q_TABLE = 0.56
• Since Q>Q_TABLE, the datum, 22.56, is rejected from the set.

B) Based on the above results, calculate the 95% confidence interval and report the statistically probable mean value at 95% confidence. Show your work using appropriate formulas.

The resulting data set consists of N = 5 points; {20.49, 20.95, 20.43, 20.39, 21.23}

Student’s t for 5 – 1 degree of freedom at 95% confidence level is t = 2.776

C) The intrinsic standard deviation of the number of analyte particles in a sample, s _N , is related to the total number of particles sampled, N, and the probability of obtaining an analyte particle, p. What is the analyte particle standard deviation of a sample containing 10⁶ particles if the probability of obtaining an analyte particle is 1%?

From the problem N=10⁶ and p=0.01 If the particle is not Analyte, then it is something else. Since the total probability of the particle being either analyte or something else is 1, q=1–p. Using the number count standard deviation formula

3. Chemical Equilibrium

Lead iodate, Pb(IO₃)₂, (F.W. 557.0) has K_sp=2.5´ 10^-13. How many grams lead iodate will dissolve in...

A) 1.00 L pure water ?

B) 1.00 L of a 0.050 M (F) solution of soluble LiIO₃?

4. Gravimetric Analysis

A) State four desirable properties of a gravimetric precipitate.

1. Complete reaction. Formed from low Ksp reaction.
2. Large precipitate particles for easy filtration.
3. Definite and known final chemical composition.
4. High molecular weight for increased measurement precision.

B) A 0.773 g sample containing Ni (F.W. 58.693) was dissolved in acid and subsequently treated with dimethylglyoxime (F.W. 116.12) precipitating all Ni as bis(dimethylglyoximate) nickel(II) complex (F.W. 288.91). Find the weight % of Ni in the sample if the precipitate weighed 2.990 g.

First, calculate the weight of Ni in the sample

Second, calculate the wt % in the sample,

5. Volumetric Analysis

A) Why are standards used in volumetric analysis?

Standardization allows accurate reporting of analyte amounts. Practically speaking, it is easier to standardize than it is to prepare solutions and maintain solutions of high accuracy.

B) What is the difference between end point and equivalence point in a titration?

The equivalence point occurs when the volume of reagent added is sufficient to react with all of the analyte. The end point occurs when a change in a physical property is observed. The end point is often observed by detecting a color change in an indicator. The indicator is chosen to change color near the equivalence point.

C) Sketch the titration curve obtained by plotting pBr versus mL AgNO₃ titrant for the titration of 40 mL of 0.1 F NaBr with 0.1 F AgNO₃ to form AgBr(s) (K_sp=5.0´ 10^-13). First, find pBr is at the start and at the equivalence point, and the volume titrant needed to reach the equivalence point.

starting pBr:

pBr = -log₁₀[Br^-]

pBr = -log₁₀(0.1) = 1

pBr at equivalence point:

pBr = pK_sp/2 = 6.15

titrant volume at equivalence point:

40 mL x 0.1 M Br^- = 4.0 mmole Br^-

Since Ag⁺ and Br^- react in a 1:1 ratio

4.0 mmole Br^- = 4.0 mmole Ag⁺

4.0 mmole Ag⁺/0.1 M = 40 mL

October 03, 2003