Environmental Analysis

name______Environmental Analysis Analytical Chemistry Exam I

This exam is open textbook (and calculator) only. No notes, homework or workshop answers are to be used. Remember to show all work for full credit, and to express all answers with the correct number of significant figures. Please circle your final answer. If you have any questions, please do not hesitate to come up and ask

-5 1. The solubility product for lead chloride, PbCl2, is Ksp = 1.7 x 10 . Equilibrium constants for the following formation reactions are listed

2+ - + Pb (aq) + Cl (aq) <= => PbCl (aq) K1 = 38.9

+ - PbCl (aq) + Cl (aq) <= => PbCl2(aq) K2 = 1.62

Combine the chemical equations for the three reactions above to determine the value of the equilibrium constant for the reaction

PbCl2(s) <= => PbCl2(aq)

First we write the chemical equation for the solubility reaction and then combine this reaction with the above two reactions to get the net desired reaction in the following manner.

2+ - PbCl2(s) <= => Pb (aq) + 2 Cl (aq) Ksp

2+ - + Pb (aq) + Cl (aq) <= => PbCl (aq) K1

+ - PbCl (aq) + Cl (aq) <= => PbCl2 (aq) K2

PbCl2(s) <= => PbCl2(aq)

Since we did not have to reverse any of these reactions the combine equilibrium constants by multiplication (pps. 100-101)

K = Ksp x K1 x K2

K = 1.7 x 10-5 x 38.9 x 1.62 = 1.1 x 10-3

2. Calculate the solubility of Pb(IO3)2 (FW = 557.0) , lead iodate, in distilled water if the Ksp is 2.5 x 10-13. Express the answer in molarity and in grams/liter.

The equilibrium reaction is 2+ - Pb(IO3)2(s) <= => Pb (aq) + 2 IO3 (aq)

The equilibrium constant is 2+ - 2 -13 Ksp = [Pb ][IO3 ] = 2.5 x 10

1 Two iodate ions are produced for every lead cation that goes into solution, so the concentration table looks like 2+ - Pb(IO3)2(s) <= => Pb (aq) + 2 IO3 (aq) Before solid 0 0 After solid x 2x Substituting into the Ksp we get

x (2x)2 = 2.5 x 10-13 4x3 = 2.5 x 10-13 ______3 -14 -5 x = √ 6.25 x 10 = 3.97 x 10

2+ - The equilibrium concentrations of Pb and IO3 , are therefore

2+ -5 - -5 [Pb ] = x = 4.0 x 10 M and [IO3 ] = 2x = 7.9 x 10 M

2+ Since one mole of Pb(IO3)2 contains one mole of Pb , the solubility of lead iodate is the same as the concentration of Pb2+; thus, the solubility of lead iodate is 4.0 x 10-5 M.

-5 3.97 x 10 moles/L x 557.0 g/mol = 0.022 g Pb(IO3)2 /L

3. a) Calculate the solubility (expressed in M and g/L) of Pb(IO3)2, lead iodate, in a solution that already contains 0.10 M Pb(NO3)2, lead nitrate. b) Calculate the ionic strength of the final solution in part a). a) This is a problem that involves the common ion effect.

We begin by setting up a concentration table:

2+ - Pb(IO3)2(s) <= => Pb (aq) + 2 IO3 (aq)

Before solid 0.10 0 After solid + x + 2x

Equilibrium concentration solid 0.10 + x 2x

The Ksp equation is now Ksp = (0.10 + x)(2x)2 = 2.5 x 10-13

This is a difficult equation to solve, but from Le Chatelier’s principle, we expect that the large initial concentration of lead ions will significantly decrease the solubility of lead iodate. Thus from problem two above, we expect [Pb2+] to be less than 4.0 x 10-5. Thus it will be much less than 0.10 and we can approximate the Ksp equation with

Ksp = 0.10*4x2 = 2.5 x 10-13 0.40x2 = 2.5 x 10-13 x2 = 6.25 x 10-13 ______-13 -7 x = √ 6.25 x 10 = 7.91 x 10 M

2 We see that our assumption was correct, x is much smaller than 0.10 M.

2+ The solubility equals x, the [Pb ]. As expected the solubility of Pb(IO3)2 decreases in the presence of a solution that already contains one of its ions. We know this as the common ion effect, but it is also explained by Le Chatelier’s Principle.

-7 7.91 x 10 mol/L x 557.0 g/mol = 0.00044 g Pb(IO3)2 /L or 0.44 mg/L b) The ionic strength depends on the amount of ions in the solution. From part a) we see that the concentration of iodate and lead ions from dissolved Pb(IO3)2(s) is small compared to the ions from the Pb(NO3)2 already in solution. So we can ignore the iodate term and the lead dissolved from the salt. Since we have 0.10 M Pb(NO3)2, a soluble salt (all nitrates are soluble), the means 2+ - as ions in solution [Pb ] = 0.10 M and [NO3 ] = 0.20 M. Thus

2+ 2 - 2 μ = ½ { [Pb ] (2) + [NO3 ] (-1) } = ½ { (0.10)*4 + (0.20)*1 } = ½ { 0.40 + 0.20 } = 0.30 M

4. The following computation was performed on a calculator giving the indicated result.

[14.3(±0.2) – 11.6(±0.2)] x 0.050(±0.001) = 1.725129385(± ?) x 10-6 [820(±10) + 1030(±5)] x 42.3(±0.4) a) Calculate the difference (subtraction) in the numerator and determine both the difference and its uncertainty. For addition and subtraction use absolute uncertainties (page 52). . . 14.3 – 11.6 = 2.7; uncertainty = √ (±0.2)2 + (±0.2)2 = ± 0.283 so the difference is 2.7 ± 0.283) b) Calculate the sum (addition) in the denominator and determine both the sum and its uncertainty. . . 820 + 1030 = 1850; uncertainty = √ ( ± 10)2 + ( ± 5)2 = ± 11.18 so the sum is 1850 (± 11.2) c) Rewrite the original equation. It should now have only products and quotients. Calculate the absolute uncertainty of the answer. Express the answer to the appropriate number of significant figures. For multiplication and division, use percent relative uncertainty.

2.7 (± 0.283) x 0.050(±0.001) = 1.725129385(± ?) x 10-6 1850 (± 11.2) x 42.3(±0.4)

We calculate the relative uncertainties

0.283/2.7 x 100% = 10.48% 0.001/0.050 x 100% = 2.00% 11.2/1850 x 100% = 0.61% 0.4/42.3 x 100% = 0.95%

3 . . %uncertainty = √ (10.48)2 + (2.00)2 + (0.61)2 + (0.95)2 = 10.73% to get back to absolute uncertainty we multiply

1.725129385 x 10-6 x 0.1073 = ± 0.185 x 10-6

The real rule of significant figures (p. 54) says that the first uncertain figure of the answer is the last significant figure. So the answer is

1.7 ( ± 0.2) x 10-6

5. Two independent analytical testing labs [Shaky Hands Testing and High Accuracy, Inc.] were competing for a large contract with various municipal water plants to determine the concentration of nitrate in drinking water samples. Both labs were given a National Institute of Standards and Technology (NIST) standard to analyze and results of 5 measurements by each lab are shown below.

Shaky Hands Testing: 3.23, 3.21, 3.29, 3.30, 3.24 ppm

High Accuracy, Inc.: 3.50, 3.57, 3.38, 3.47, 3.41 ppm

(a) Calculate the mean, standard deviation, and percent relative standard deviation for each laboratory. [You do not need to show all your work, but may want to identify the appropriate equations for partial credit in case you enter values incorrectly on your calculator.] (b) Calculate a confidence interval for each laboratory at the 95% confidence level. If the NIST standard was known to have a concentration of 3.40 ppm, which laboratory wold you hire? Why? (c) Which laboratory gave results that : were more accurate? ______were more precise? ______had more random error? ______had more systematic error? ______

(a) Shaky Hands Testing: x(bar) = 3.254; s = 0.039; %RSD = 0.039/3.254 x 100% = 1.2 % High Accuracy, Inc.: x(bar) = 3.466; s = 0.075; %RSD = 0.075/3.466 x 100% = 2.2 %

(b) 95% confidence t value for 4 degrees of freedom (5-1) is 2.776 from page 67. Shaky Hands  = x(bar)  ts/n = 3.254  2.776* 0.039/5 = 3.254  0.048 at 95% confidence High Accuracy  = x(bar)  ts/n = 3.466  2.776* 0.075/5 = 3.466  0.093 at 95% confidence

Hire High Accuracy, Inc. Their 95% confidence interval 3.373 to 3.559 includes the NIST value of 3.40. This is not true for Shaky Hands Testing 3.206 to 3.302. Shaky Hands had a precise analysis, but you should not be confident in the numbers they report.

(c) were more accurate? _High Accuracy____ were more precise? ___ Shaky Hands ____

had more random error? _High Accuracy had more systematic error? _ Shaky Hands__