Exam 1 study guide PDF

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Exam 1 (Sep. 14)

1.  15% of your final grade 2.  18 problems: multiple choices (8, 32 pts), Short answers (5, 31) and workout problems (5, 37)

It’s absolutely critical you know how to solve homework problems with ease and speed. So here they are again.

Homework problems (HW) 1 (p. 22 Harris) 1-15,16,20,22,26,30,34,37 HW 2 (p.61&91) 3-1,2,5(a,c,e,g),13,16(b,d,f),18(b) 4-2,5,9,12,14,19,22,23 HW 3 (p. 114&141) 5-7,22,23,29,30 6-4,6,8,14,20,34-36,39,46-49 (The cutoff is Solubility Product, meaning you won’t see acid/base problems in this test.)

Instead of an old test, which can be misleading, I’m giving you some sample problems of each type I used recently. 1. (Fill the blank) Identify the number of significant figures in the following numbers. For example, 3.680 x 10-9, answer = 4.

2. (Short answer) What is a blank solution?

3. (Multiple choices) The steps in a chemical analysis are ___. (4) ! A) 1. Formulate the question. " !!!!!!!!!2. Select the analytical procedure. " !!!!!!!!!3. Sampling. " !!!!!!!!!4. Prepare the sample. " !!!!!!!!!5. Make replicate measurements of the sample. B) 1. Select the analytical procedure " !!!!!!!!!2. Sampling. " !!!!!!!!!3. Prepare the sample. " !!!!!!!!!4. Make replicate measurements of the sample. " !!!!!!!!!5. Make a clear and complete written report of your findings. C) 1. Formulate the question. " !!!!!!!!!2. Select the analytical procedure. " !!!!!!!!!3. Sampling. " !!!!!!!!!4. Prepare the sample. " !!!!!!!!!5. Make replicate measurements of the sample. " !!!!!!!!!6. Make a clear and complete written report of your findings.

4. (Multi-step calculations) They are very, very similar to HW problems.

The following 2 slides contain equations and tables that will be available to you in the test…

Equations & Tables for Exam 1

If Gcal > Gtable, discard If Gcal < Gtable, keep! !

Key concepts and methods you should know…

Step 3. Chemical analysis (finally!) The need for calibration 2.5 µg/µL

Calibration curves

Chemical concentrations

Preparing solutions Dilution & concentration!

! M1(2),V1(2) : molarity, volume of sample 1(2) ! Example 2.4 Given: a concentrated HCl is 12.1 M Find: V = ? mL if diluted to 1.000 L to make 0.100 M HCl

‘mole’ = Central station of Subway 3050!

wt%, ppm M.W.! g!

m = d!V

V! mol!

Stoichiometry of rxns!

M!

Stoichiometry calculations reactant!whose amount limits the Limiting agent: the reactant reaction (from producing more product)!

Significant figures

π=?

3.1415926535897932…

2 The minimum number of digits needed to write a given value in scientific notation without loss of accuracy.! 3 1

Example 3.2 Compare the number ‘36830’with 3.7 x 104, 3.68 x 104, 3.683 x 104, and 3.6830 x 104

Significant figures: The rules of ‘0’ Zeros are significant if 1) in the middle of a number or 2) at the end of a number on the right-hand side of a decimal point.!

Example 3.3 Find the significant zeros in the following numbers. 106

0.0106

0.106

0.1060

Answer: 106

0.0106

0.106

0.1060

Significant figures: Addition/Subtraction Rule #1: same in, same out. Rule #2: not the same? Then the leastcertain one rules. Rule #3: rounding up/down still applies.

Example 3.4a Write answers for the following calculations with correct number of digits. 1.364 x 10-4 1.364 x 10-4 1.364 x 10-4 + 3.11 x 10-4 + 3.111 x 10-4 4.475 x 10-4

5.838 x 10-4 5.84 x 10-4

Significant figures: Logarithms/Antilogarithms Rule #1: number of digits in mantissa equals to that of the original number. Rule #2: vice versa. log 339 = 2.530 Characteristic!

Mantissa!

Example 3.6 Antilog(-3.42) = 10-3.42 = 3.8019 x 10-4 = 3.8 x 10-4!

Documenting Uncertainties Absolute uncertainty: margin of uncertainty of a measurement (12.35 ± 0.02 mL) !

en !

Absolute uncertainty Magnitude of measurement 0.02 mL! =! = 0.002! 12.35 mL!

Relative uncertainty =!

Percent Relative uncertainty!

=! Relative uncertainty x 100! = 0.2%!

%en!

Propagation of uncertainty from random error: Addition/Subtraction e4 = √e12 + e22 + e32

Example 4.1 Calculate the uncertainty associated with the arithmetic. 1.76 (±0.03) +

1.89 (±0.02)

−

0.59 (±0.02) 3.06 (±?)

e4 = √(0.03)2 + (0.02)2 + (0.02)2 = 0.041

Propagation of uncertainty from random error: Multiplication/Division %e4 = √(%e1)2 + (%e2)2 +(%e3)2

Example 4.2 Calculate the uncertainty associated with the arithmetic. 1.76 (±0.03) × 1.89 (±0.02) 0.59 (±0.02) 1.76 (±1.7%) ×1.89 (±1.1%) 0.59 (±3.4%)

= 4.0% × 5.64 = 0.23 = 5.6 (±0.2) !

= 5.64 (±?) %e4 = √(1.7)2 + (1.1)2 + (3.4)2 = 4.0%

Gaussian distribution

Mean:!

Standard deviation:!

Σx i! i n

x =!

s =!

2 Σ(x i – x) !

√!

i

n–1!

Gaussian distribution

Gaussian curve:!

y =!

1! σ!√!2π!

–(x – µ)2/2σ2

e!

1777–1855!

s! e ≈ 2.718!

x!

Gaussian distribution

z =!

x– µ x– x ≈ ! σ! s!

Confidence intervals An expression stating the true mean (µ) is likely to lie within a certain distance from the measured mean (x).! Confidence interval:!

µ = x ±!

ts √n!

Confidence intervals: What do they mean exactly?

Confidence intervals An expression stating the true mean (µ) is likely to lie within a certain distance from the measured mean (x).! Confidence interval:!

µ = x ±!

ts √n!

t: calculated constant!

Example 5.3 A government chemist has performed a gravimetric analysis for chloride ion on a sample submitted to the laboratory. The following results (%Cl−) were obtained: 27.46, 27.52, 27.48 and 27.10. Given the following data: Degrees of Freedom 98% Confidence Level 1 31.821 2 6.965 3 4.541 4 3.747 5 3.365 What is the 98% confidence interval for this set of data? A. 0.37% ts B. 0.91% µ = x ±! √n! C. 0.43% D. 0.16%

Grubbs test for bad data Example 6.1

G test:!

Mass loss (%) 10.2, 10.8, 11.6, 9.9, 9.4, 7.8,! 10.0, 9.2, 11.3, 9.5, 10.6, 11.6 Is there an outlier in this set of data?

|x? – x| Gcal =! s

If Gcal > Gtable, discard If Gcal < Gtable, keep! ! average = 10.16; s = 1.11 Gcal = 2.13 < Gtable(2.285) Answer: no, keep.

Calibration curves Calibration curves: plots that show the analytical response as a function of analyte concentration.! Standard solutions: solutions prepared with known concentrations of analyte.! Blank solutions: solutions containing everything (reagents & solvents) but added analyte.!

Standard addition The problem: the ‘matrix’ effect! A change in the analytical signal caused by anything in the sample other than analyte.!

Standard addition Assumptions: 1) the addition does NOT perturb the matrix and 2) signal is directly proportional to analyte concentration (i.e., within linear region) ! Concentration of analyte in initial solution! Concentration of (analyte +standard) in final solution!

Standard addition equation!

=!

[X]i [S]f + [X]i

Signal from initial solution Signal from final solution

IX

=! I S+X

Internal standards A known amount of compound, different from analyte, that is added to the unknown.! The problem: variations in instrument response or sample quantity; potential sample loss during preparation! Assumption: signals are directly proportional to analyte concentration! Response factor (F)! AX [X]

=! F !

AS [S]

Calibration methods

AX!

Calibration for X!

Response factor (F)! AX [X]

=! F !

AS [S]

[X]!

Standard addition ! [X]i [S]f + [X]i

AX or AS !

Internal standard! Calibration for X!

Calibration for S!

IX

=! I S+X [X] or [S] !

General rules for calculating/manipulating K aA + bB

cC + dD

[C]c[D]d K =! [A]a[B]b

What does [C] really stand for?! The ratio of the conc. of a species to its concentration at its standard state (1M for solutes, its conc. in pure substances). !

1.  Concentration of solutes in M 2.  Concentration of gases in bar 3.  Concentration = 1 (unity) for pure substances

General rules for calculating/manipulating K 1. Reverse reaction => 1/K 2. If reaction 3 = reaction 1 + reaction 2, then K3 = K1K2 !

Example 6.4 Known: H2O H+ + OH− Kw = 1.0 x 10−14 K1 = 1.8 x 10−5 NH3 (aq) + H2O NH4+ + OH− Ask: NH4+ NH3 (aq) + H+ K2 = ?

Kw =

K2 =!

[H+][OH−]! [NH3][H+]! [NH4

+]!

K1 =!

=!

[NH4+][OH−]! [NH3]!

[NH3]! [NH4

+][OH−]!

!! [H+][OH−] = Kw/K1 !

Solubility product The equilibrium constant for solid-dissolving reactions.! Hg2Cl2 (s)

Hg22+ + 2Cl−

Ksp = [Hg22+][Cl−]2 = 1.2 x 10−18! Analysis: [Hg22+] = Ksp/[Cl−]2 When there’s no chloride in solution, [Cl−] = 2[Hg22+], thus 4[Hg22+]3 = 1.2 x 10−18 [Hg22+] = 6.7 x 10−5 M When, however, [Cl−] = 0.1 M already in solution [Hg22+] = Ksp/[Cl−]2 = 1.2 x 10−16 M

Separation by precipitation Example 6.6 Known: PbI2 (s) Pb2+ (0.01 M) + 2I− Ksp = 7.9 x 10−9 Hg2I2 (s) Hg22+ (0.01 M) + 2I− Ksp = 4.6 x 10−29 Ask: can one use I− to precipitate Hg22+ by 99.990% and only? 99.990% => 0.01% left => [Hg22+] = 1.0 x 10−6 M [Hg22+] = Ksp/[I−]2 therefore, [I−] = 6.8 x 10−12 M Question: would 6.8 x 10−12 M I− be enough to precipitate 0.01 M Pb2+ ? Q = [Pb2+] [I−]2 = 4.6 x 10−25 < Ksp = 7.9 x 10−9 Answer: Yes you can!...