Cumulative Review Packet 1 PDF

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CUMULATIVE REVIEW PACKET #1

SPECTROPHOTMETRIC ANALYSIS – BEER’S LAW Used to determine both the concentration (MOLARITY) of a solution that has a COLOR!! This is accomplished by passing light of a specific wavelength through a colored solution and measuring the amount of light that the solution absorbed as the light passes through the solution!

IMPORTANT!

This process only works with solutions that have a color!!! Sodium chloride is a colorless solution. So SPECTROPHOTMETRIC ANALYSIS could not be used to determine the molarity/concentration of a NaCl(aq) solution. Remember solutions which show a color have a transition metal present… Cr2O7-2 - Orange

CrO4-2  Yellow

Cu2+  Blue

MnO4-1  Purple

SO HOW DO I DETERMINE THE CONCENTRATION OF A SOLUTION USING THIS METHOD?

1.

An instrument called a Spec 20 (shown later in the document) measures

the

ABSORPTION of light of a particular color/wavelength as the light passes through your sample whose concentration is not known. 2. The ABSORBANCE of light of as measured by the SPEC 20 is DIRECTLY related to the concentration/molarity of your solution. The HIGHER the ABSORBANCE reading the more CONCENTRATED the sol’n. By measuring ABSORBANCE of light by sol’n of differing concentrations, you can make a graph of ABSORBANCE vs CONCENTRATON.

GRAPH 1

NOTE: A straight line is produced. This means that ABSORBANCE and CONCENTRATION are DIRECTLY related!

You can now determine the MOLARITY/CONCENTRATION of your sol’n as long as you know its ABSORBANCE!  BEER’S LAW relates ABSORBANCE of light by the sample to the samples concentraton/molarity. Mathematically represented below.

A – absorbance as measured by a SPEC 20! € - a constant which is given – many time the letter ‘a’ is used instead of epsilon. b – path length of light as it passes through the cuvette – usually 1 cm. C – the concentration of your solution – MOLARITY! The equation above would say that the greater the concentration of a solution, the greater the amount of light absorbed by the solution. The greater the path length of light passing through the solution, the greater the absorbance. Since CONCENTRATION of a solution and ABSORBANCE of a solution are DIRECTLY related, the following RATIOS of ABSORBANCE to CONCENTRATION of two solution with differing concentrations also apply. ABSORBANCE (sol' n1) MOLARITY (sol' n1)

=

ABSORBANCE (sol ' n 2) MOLARITY (sol ' n 2)

Let’s say in an experiment, you are asked to determine the concentration/molarity of a CoCl2 sol’n – which is pink….here is the process.

1.

You would place a few milliliters of your sol’n in a CUVETTE (fancy term for a test tube). VERY IMPORTANT that you NOT hold the cuvette with your hands/fingers as doing so would leave smudges on the cuvette.

Why? You want to measure how much light is ABSORBED by your sol’n. If there are greasy, grimy ( )fingerprints on the cuvette, it would prevent light from passing through the cuvette and making the absorbance reading TOO HIGH!

2.

You would then place your cuvette in the sample holder of a SPEC 20.

3.

The monochromator selects one particular energy wavelength or color of light from the source that will cause maximum absorbance of the sample.

4.

How do I know which wavelength of light will cause maximum absorbance? Interpreting an ABSORBANCE vs WAVELENGTH graph that must be given.

You are looking for a wavelength of light that causes MAXIMUM absorbance!! By looking at the graph above, seems to be approximately 560 (nanometers). A nanometer is a unit of length for wavelength. 1 nm = 10-9 m. The detector measures the amount of light that the sample absorbs.

Since some light is absorbed as it passes through the sample, ‘I’, which is light transmitted is < ‘Io” the intensity of the light before passing through the sample.

5.

By repeating this process with solutions of differing concentrations, you

can construct an ABSORBANCE vs CONCENTRATION graph similar to GRAPH 1 shown above ON PAGE 1! SAMPLE MULTIPLE CHOICE QUESTION. 1. To determine the number of moles of Cu in the sample of the mixture, the students measured the absorbance of known concentrations of Cu(NO3)2(aq) using a spectrophotometer. A cuvette filled with some of the solution produced from the sample of the mixture was also tested. The data recorded by one student are shown in the table above. On the basis of the data provided, which of the following is a possible error that the student made? [Cu2+] 0.025 M 0.050 M 0.100 M 0.150 M UNKNOWN

ABSORBANCE 0.059 0.235 0.117 0.468 0.330

(A) The Cu(NO3)2(aq) from the sample of the mixture was not diluted properly. (B) The spectrophotometer was calibrated with tap water instead of distilled water. (C) The student labeled the cuvettes incorrectly, reversing the labels on two of the solutions of known concentration. (D) The spectrophotometer was originally set to an inappropriate wavelength, causing the absorbance to vary unpredictably. Answer is ‘C’. Why? As concentration increases the ABSORBANCE should also increase. The 0.117 ABSORBANCE should be with the 0.050 M and NOT the 0.100 M.

FRQ EXAMPLE WITH ANSWERS 1. One approach to determine the concentration of C2H5OH(aq) in a solution is based on the reaction represented below. 3 C2H5OH(aq) + 2 Cr2O72-(aq)  Orange

4 Cr3+(aq) + 3 CH3COOH(aq) + 11 H2O(l)

Blue-green

A student has an initial Cr2O72- (aq) concentration of 1.0 x 10-3 M and an initial C2H5OH(aq) concentration of 0.0500 M. The solution contains enough strong acid to keep the pH essentially constant throughout the reaction. The student places a sample of the solution in a cuvette that has a path length of 0.50 cm and places it into a spectrophotometer set to measure absorbance at 440 nm. Dichromate ion, Cr2O72-(aq) is the only species in the reaction mixture that absorbs light at this wavelength. The absorbance of Cr2O72-(aq) in the solution is monitored as the reaction proceeds; the table below shows the absorbance as a function of time for the first trial. Time (min) 0.00 1.50 3.00

Absorbance at 440 nm 0.782 0.553 0.389

4.50 6.00

0.278 0.194

(a) Calculate the value of [Cr2O72-] at 1.50 min. USE THE EQUATION.

ABSORBANCE (1) MOLARITY (1)

0.782 1 x 10−3 M

=

=

ABSORBANCE (2) MOLARITY (2)

0.553 MOLARITY (2)

Cross multiply to solve for MOLARITY(2)

MOLARITY 2 = 7.1 x 10-4 M (b) The student runs a second trial but this time uses a cuvette that has a path length of 1.00 cm. Describe how the experimental set up should be adjusted to keep the initial absorbance at 0.782. Just your answer with respect to the factors that influence the absorbance of a sample in a spectrophotometer. REMEMBER BEER’S LAW STATES THE A = (a) x (b) x (C) The initial path length is 0.50 cm. If the path length (b) is doubled, and the molar absorptivity constant (a) remains the same, the [Cr2O7-2 ]o must be ½ of its value to maintain the same value for ‘A’ – the absorbance! (c) In a DIFFERENT experiment to spectrophotometrically determine the mass percent of cobalt in an ore containing cobalt and some inert materials, solutions with known [Co+2] are prepared and the absorbance of each of the solutions is measured at the wavelength of optimum absorbance. The data are used to create a calibration plot, shown below.

A 0.630 g sample of the ore is completely dissolved in concentrated HNO3. The mixture is then diluted with distilled water to a final volume of 50.00 mL. Assume that all the cobalt in the ore sample is converted to Co+2.

(i) What is the [Co+2] in the solution if the absorbance of a sample of the solution is 0.74?

Reading from the graph, an absorbance of 0.74 corresponds to a concentration of 0.0130 M. Answers from 0.0125 M to 0.0135 M are accepted.

(ii) Calculate the number of moles of Co+2 in the 50.00 ml solution. NEVER forget that MOLARITY x VOLUME(L) = moles. You just found the molarity of Co+2 to be 0.0130 M. The volume is 50.00 mL so…. Moles of Co+2 = 0.0130M x 0.050L =

6.50 x 10-4 moles Co+2

(iii) CALCULATE THE MASS PERCENT OF CO+2 IN THE 0.630g SAMPLE OF THE ORE. The ore sample had an initial mass of 0.630 g. To find the mass percent that IS cobalt Mass % = Mass of Co in sample is

Mass of Co∈sample Mass of sample

x 100

6.50 x 10-4 mol Co x 58.93 g/mol = 0.0383 g Co.

Mass % Co =

0.0383 0.630

x 100 = 6.00% (approximately)

ANOTHER GOOD FRQ TO LOOK AT! 

VIDEO LINKS TO LOOK AT! 

https://www.youtube.com/watch?v=qbCZbP6_j48

https://www.youtube.com/watch?v=QPHo5lFWgT0

HOW TO PROPERLY MAKE A SOLUTION FOR LAB!  Solutions (aq) can be made in one of two ways…

#1.

THE SOLUTE IS A SOLID TO BE DISSOLVED IN A SOLUTE – A LIQUID ALWAYS BEING WATER. NEVER USE TAP WATER!!!!

Here are the materials needed for accomplishing this. 1. Volumetric flask of correct size.

2. Electronic balance like we have in lab.

3. Plastic weighing boat or weighing paper. 4. Distilled/Deionize (D.I.) water which could be in a plastic water bottle.

EXAMPLE: THE SOLUTE IS A SOLID. For lab, a student needs to make 250. mL of 0.100 M NaOH solution. Describe the process to accomplish this… 

First determine the mass in grams of the solute needed. To do this, we determine the number of MOLES of solute needed. REMEMBER!!! MOLARITY x VOLUME (L) = moles so in this case. Moles of NaOH = 0.100 M x 0.250 L = 0.0250 moles Now convert moles to mass. 40.00 g NaOH = 1.00 g NaOH needed. 1 mol NaOH Using the plastic weighing tray and electronic balance mass 1.00 g of NaOH. Transfer the 1.00 g of NaOH to the 250. mL Volumetric flask. Add sufficient DISTILLED water to dissolve the NaOH. Swirl flask. Add enough D.I. water to reach the mark on the neck of the flask. Swirl. MASS = 0.0250 moles NaOH x

   

Very uncomplicated – similar to how you would make some tea!!! Measure out what you need – add it to water and STIR!

Just MAKE SURE you mention proper equipment to be used. A beaker is NEVER used as it is nothing more than a container!

#2.

THE SOLUTE IS A SOLUTION TO BE DILUTED WITH DISTILLED WATER. – NEVER USE TAP WATER!!!! USED DISTILLED WATER.

1. Volumetric flask of correct size. 2. Pipet or Buret (like the titration buret)

3. Plastic water bottle with Deionized water – D.I. (shown above). 4. NO electronic balance needed as we are working with solutions!

5. Eye dropper pipet…

EXAMPLE: For lab, a student needs to make 250. mL of 0.100 M NaOH solution. In stock – previously made - is 0.750 M NaOH. Describe the procedure for diluting the 0.750 M NaOH solution to a 250. mL of concentration of 0.100 M using a volumetric flask, pipet or buret and D.I. water. 

Again, the first thing that needs to be done is to determine the volume of the 0.750 M NaOH solution that will be diluted. Equation to accomplish this is…

M(stock) x V(stock) = M(dilute) x V(dilute) We want to solve for the Volume of the stock solution so….

V(stock)

=

M (dilute) x V ( dilute ) M (stock )

=

0.100 x 250. 0.750

= 33.3 mL of 0.750 M stock needed .



Since we know the amount of stock solution needed we measure that volume using either a pipet or buret. So next step is….



Using a pipet or buret measure 33.3 mL of 0.750 M NaOH and add to a 250 mL volumetric flask.



Add enough D.I. water to get close to the mark on the flask near its top.



Now using an eye dropper, add enough D.I. water to reach this mark.



Stopper and mix.



This is what is meant as the FILL MARK on the NECK of the volumetric flask is shown in the diagram below.

FRQ EXAMPLE FROM AP TEST A student is assigned the task of making a 0.200 M CuSO4 solution for a spectrophotometric lab . (a) The student is provided with a stock solution of 1.00 M CuSO4, a 50. mL volumetric flask, a dropper pipet, buret and distilled H2O. (i) Calculate the volume, in mL, of 1.00 M CuSO4 that the student should use for preparing 50. mL of 0.200 M CuSO4. M(stock) x V(stock) = 1.00 M

M(dilute) x V(dilute)

x V(stock) = 0.200 M

x 50.0 mL

Vstock = 10.0 mL

(ii) Briefly list the steps of an appropriate and safe procedure for preparing the 50. mL of 0.200 M CuSO4. Only materials selected from those provided to the student (listed above) may be used.  Using a buret measure 10.0 mL of 1.00 M CuSO4 into a 50.0 mL volumetric flask.  Using a water bottle of D.I. water, add enough water to bring the solution volume near the mark on the neck of the 50.0 mL volumetric flask.  Using an eye dropper, add enough water to reach the mark on the neck of the flask .

OXIDATION – REDUCTION REACTIONS AKA REDOX REACTIONS  REDOX REACTIONS describe all chemical reactions in which there is a net change in the charge of an element from the reactant side to the product side. GENERAL PRINCIPLES….     



We know that the charge on a sodium ion – or any ion in column 1 is a +1. We will now call this CHARGE the OXIDATION NUMBER for the sodium ion. As an ATOM, sodium has NO CHARGE which is true for ANY atom. Consider the reaction below.  2 Na(s) + Cl2(g)  2 NaCl(s) On the REACTANT side, we have a sodium ATOM. It has NO CHARGE as an atom so its OXIDATION NUMBER is 0!  On the PRODUCT side, the sodium atom has become a sodium ION in the compound sodium chloride!! Now the OXIDATION NUMBER for the sodium ion is +1. This is an example of a REDOX reaction in which an element changed its charge/oxidation number from what it was on the REACTANT side to a new value on the PRODUCT side

TYPES OF REACTIONS THAT ARE REDOX. 1. SYNTHESIS which is the formation of a compound from its elements 2. All COMBUSTION reactions. (REMEMBER – O2 MUST BE A REACTANT) 3. Reactions that generate ELECTRICITY

IMPORTANT!!!  HOW TO ASSIGN OXIDATION NUMBERS TO ATOMS IN VARIOUS PARTICLES

GENERAL RULES (O.N. means OXIDATION NUMBER) 1. For an atom in its elemental form itS oxidation number is ZERO. Cl atom – O.N. = 0

Mg atom – O.N. = 0

ANY ATOM – O.N. = 0 

2. For the diatomic molecules – (HOFBrINCl) – the oxidation number of the atom in the molecule is ZERO! H2 – Oxidation number of ‘H’ is 0 I2 – Oxidation number of ‘I’ is 0!

3. For a monatomic ion the oxidation number is the charge on the ion. Magnesium ion – Column 2 – Mg+2 – oxidation number is +2 Iron(III) ion – Fe+3 – oxidation number is +3. Rules 4 and 5 are a little TRICKIER! But use this as a guide…. NOTE!!!!!!!!!!!!!! In MOST cases, ‘H’ in a compound has an oxidation number of +1. In MOST cases, ‘O’ in a compound has an oxidation number of -2. (Exceptions do exist)

4. IN A POLYATOMIC ION, THE SUM OF THE OXIDATION NUMBERS ON THE INDIVIDUAL ATOMS IN THE ION EQUALS TO THE CHARGE OF THE POLYATOMIC ION! PHOSPHATE ION. PO4-3 How do I assign the oxidation numbers for ‘P’ and ‘O’? Oxygen will have an O.N. of -2 – but there are 4 of them. So the TOTAL the four oxygens would be -8! RULE 4 says that ‘P’ total + ‘O’ total must equal a -3! So…………

‘P’ total + (-8) = -3 ONE ‘P’ is a +5!

PERMANGANATE ION. MnO4-1 How do I assign the oxidation numbers for ‘Mn’ and ‘O’? Oxygen will have an O.N. of -2 – but there are 4 of them. So the TOTAL the four oxygens would be -8! RULE 4 says that ‘Mn’ total + ‘O’ total must equal a -1 So…………

‘Mn’ total + (-8) = -1 ONE ‘Mn’ is a +1

DICHROMATE ION. Cr2O7-2 How do I assign the oxidation numbers for ‘Cr’ and ‘O’? Oxygen will have an O.N. of -2 – but there are 7 of them. So the TOTAL the four oxygens would be -14 (gasp!) RULE 4 says that ‘Cr’ total + ‘O’ total must equal a -2 So………… ‘Cr’ total + (-14) = -2 ‘Cr’ TOTAL IS +12 BUT THERE ARE TWO OF THEM SO ONE ‘Cr’ WOULD HAVE AN OXIDATION NUMBER OF +6!

5.

In a molecule, the sum of the oxidation numbers of the atoms that constitute the molecule must equal 0!

SULFURIC ACID - H2SO4 From statement at top of page – O.N. for ‘H” is +1 but there are 2 of them so total for ‘H’ is +2 There are 4 oxygens – each at -2 for a total of -8. So how do we find the oxidation number for ‘S’? Total for ‘H’ + total for ‘S’ + total for ‘O’ must = 0 2 + total for ‘S’ + (-8) =0

ONE ‘S’ is +6

ETHANOL – C2H5OH Let’s see if we can shorten this up. There is 1 ‘O’ for a total of -2.

There are 6 ‘H’ for a total of +6. So what about the two carbons?

Total ‘C’ + Total ‘H’ + Total ‘O’ = 0 Total ‘C’ + 6 + -2 =0 Total ‘C’ must be -4 but since there are 2 ‘C’ the O.N. for one ‘C’ is -2.

6.

The ONE exception for oxygen being a -2!!

HYDROGEN PEROXIDE – H2O2 Total ‘H’ + Total ‘O’ = 0 +2 + Total ‘O’ = 0 Total ‘O’ = -2 but since there are two ‘O’, one ‘O’ must = -1. So the ONE exception you need to know is that in H2O2 – hydrogen peroxide – the ‘O’ has an oxidation number of -1. Determine the oxidation number (O.N.) of each element in these compounds: a) CaO (s) b) KNO3 (s) c) NaHSO4 (aq) d) CaCO3 (s) Solution to Example

a) CaO

Ca = +2 O = -2

+1 ? -6

b)

KNO3 +1

c)

+1 ? -8

NaHSO4 +2

d)

N = 0-(+1)-3(-2) = +5

S = 0-(+1)-(+1)-4(-2) = +5

? -6

CaCO3

C = 0-(+2)-3(-2) = +4

COMMON AP QUESTION Which of the following reactions are REDOX? To answer this question, you must assign OXIDATION NUMBERS to each element on the reactant side and product side. If any element has its oxidation number change – then the reaction is REDOX! EXAMPLE For the following examples….. The number ABOVE the element represents the total for that element. The number BELOW the element represents the O.N. for ONE atom of the element +2

+4 -6

+2

-2

+4 -4

(This line shows total O.N. for the element)

CaCO3 + HEAT  CaO + CO2 +2

+4 -2

+2

-2

+4 -2

(This line shows O.N. on ONE atom of the element)

This is NOT a REDOX reaction as Ca has an O.N. of +2 on reactant and product side. C has an O.N. of +4 on reactant and product side. O has an O.N. of -2 on reactant and product side.

0

+1 +5 -6

+2

3 Cu + 8 HNO3  3 Cu 0

+1 +5 -2

+5 -6

+2

+2

-1 3

+ 6 NO

+5 -2

+2

-2

+2 -2

+ 4 H2O + 2 NO +1

-2

+2 -2

This IS a REDOX reaction as Cu has an O.N. of 0 on REACTANT and +2 on PRODUCT N has an O.N. of +5 on REACTANT and +2 on PRODUCT NOTE!! COEFFICIENTS PLAY NO ROLE IN ASSIGNING OXIDATION NUMBERS!  OK – SO LET’S FINISH UP WHAT IS NEEDED FOR YOUR AP TEST ON THIS TOPIC!!!  (Top row of number represents TOTAL O.N. on the element. Bottom represents O.N. one atom only)

0

+1 +5 -6

+2

3 Cu + 8 HNO3  3 Cu 0

+1 +5 -2

+2

+5 -6

+2

-1 3

+ 6 NO

+5 -2

+2

-2