In-Class Section ______
Take-Home Section ______
Combined Score ______ Name ________________________

Quantitative Analysis
Exam 3
In-Class Section
May 1, 2003

(3 pts.) 1. The anion of ethylenediaminetetraacetic acid is a:

(3 pts.) 2. What indicator was used in the limestone EDTA titration?

Erichrome black T

(3 pts.) 3. Oxidation occurs at which electrode?

oxidation always occurs at the anode

(6 pts.) 4. Balance the following redox reaction:

5Fe⁺² + 8H⁺ + MnO₄^- à 5Fe⁺³ + Mn⁺² + 4H₂O

(4 pts.) 5. What is the purpose of chromatography?

Chromatography is used for the separation of mixtures.

(8 pts.) 6. Describe the four basic components of all optical spectrometers.

The four components common to all spectrophotometers are:
• source of radiaiton - this is the part of the instrument that produces the radiation that is directed at the sample
• monochromator - this takes the continuum of the radiation and selects a single wavelength of radiation or a narrow band of wavelengths to be analyzed by the detector
• detector - the device which monitors variations in the intensity of the radiation and converts the signal into an electrical signal
• signal processor and readout device - takes the electrical signal and processes it into a magnitude and format for the readout device which displays the information for the operator

(3 pts.) 7. The source in a atomic absorption spectrometer is abbreviated HCL. What do these three letters stand for?

Hollow Cathode Lamp

(4 pts.) 8. Why is it necessary to immediately titrate the iron(II) with potassium permanganate?

Iron(II) ions in solution are subject to oxidation to iron(III) by oxygen in the air. Therefore, it is necessary to titrate the sample immediately after the reduction step to prevent this source of error.

(6 pts.) 9. Describe the parameters of Beer’s Law.

A = abc

A => absorbance
a => absorptivity, a function of the substance being analyzed and the wavelength being used
b => path length of the light shinning through the sample
c => concentration of the solution being analyzed, units must agree with the units of absorptivity

Name _________________________________

Quantitative Analysis
Exam 3
Take-Home Section
May 1, 2003

(6 pts.) 10. A solution containing 0.4025 g of CoCl₂*xH₂O was exhaustively electrolyzed to deposit 0.0994 g of metallic cobalt on a platinum cathode by the reaction Co⁺² + 2e^- à Co_(s). Calculate the number of moles of water per mole of cobalt in the reagent.

# moles Co = (0.0994g Co)(1 mol/58.93 g Co) = 1.69x10^-3mol Co

# g Cl = (0.00169mol Co)(2 mol Cl/1 mol Co)(35.453 g Cl/1 mol Cl) = 0.120g Cl

# g Co + Cl = 0.0994g Co + 0.120g Cl = 0.219g

#g water = 0.4025g complex - 0.219g Co + Cl = 0.183g water

# mol water = (0.183g water)(1 mol water/18.0g water) = 0.0102mol water

#mol water/mol Co = (0.0102mol water/0.00169mol Co) = 6.04 ~ 6 mol of water

thus the correct formula would be CoCl₂*6H₂O

(6 pts.) 11. The first excited state of Cu is reached by the absorption of 327 nm light. What is the energy difference (kJ/mol) between the ground and excited states.

E = (hC/l) = (6.63E-34 J*s/photon/1.00E+03 J/kJ)(3.00E+08m/s/3.27E-07m)(6.02E+23 photons/mol)

= 3.66E+02 kJ/mol

(10 pts.) 12. A 0.267 g quantity of a compound with a molar mass of 337.59 was dissolved in 100.0 mL of ethanol. Then 2.000 mL was withdrawn and diluted to 100.0 mL. The spectrum of this solution exhibited a maximum absorbance of 0.728 at 428 nm in a 2.000 cm cell. Find the molar absorptivity of the compound.

#mol = (0.267 g)(1mol /337.59g/mol) = 0.000791mol

A = 0.728

b = 2.00 cm

c = (0.000791mol /0.1L)(2.00mL /100.0) = 0.000158mol/L

a = A/bc = 0.728 / (2.00cm)(0.000158mol/L)

= 2.30x10³L/mol*cm

(5 pts.) 13. Find the conditional equilibirum constant for CaEDTA^-2 at pH 10.00.

K_MY = 4.9x10¹⁰

a = 0.36

K'_MY = K_MY*a = 4.9x10 ¹⁰*0.36 =

1.8x10¹⁰

(6 pts.) 14. Find the concentration of Ca⁺² in 0.050M Na₂[CaEDTA] at pH 10.00.

([CaY^-2])/([Ca⁺²]C_T) = K'_CaY

let x = [Ca⁺²] = C_T and 0.0500 M>> x

(0.0500 M)/(x2) = 1.8 x 10¹⁰M^-1

x = ((0.0500 M)/(1.8 x 10¹⁰))^1/2 = 1.7 x 10^-6 M = [Ca⁺²]

15. Calculate pFe⁺² at each of the following points in the titration 25.00 mL of 0.02026 M EDTA by 0.03855 M Fe⁺² at pH 6.00.


K_MY = 2.0893E+14 ;a = 2.30E-05 ; K'_MY = 4.8e9

(5 pts.) (a.) 12.00 mL;V_EDTAM_EDTA > V_FeM_Fe pre-equivalence point

C_T= (V_EDTAM_EDTA - V_FeM_Fe)/(V_Fe + V_EDTA)

C_T = ((25.00 mL)(0.02026 M) - (12.0 mL)(0.03855 M))/((25.00 + 12.00)mL) = 1.19 x 10^-3M

[FeY^-2] = (V_EDTAM_EDTA)/(V_Fe + V_EDTA)

[FeY^-2] = ((25.00 mL)(0.02026 M))/((25.00 + 12.00)mL) = 0.01368 M

[FeY^-2] = (([FeY^-2])/(C_T*K'_FeY))

[FeY^-2] = (0.01328 M)/(1.19e-3*4.8e9)

[Fe⁺²] = 2.33E-09 M

pFe = - log([Fe⁺²]) = 8.63

(5 pts.) (b.) V_e

V_e =(V_EDTAM_EDTA)/V_Fe

V_e = ((25.00mL)(0.02026M))_EDTA/(0.03855M)_Fe = 13.14 mL

M_FeY = (25.00mL)(0.02026)_EDTA/(25.00 + 13.14)mL = 0.01328M

let x = M_Fe = C_T

(0.01328 M)/(x2) = 4.8 x 10⁹M^-1

x = ((0.01328 M)/(4.8 x 10⁹))^1/2 = 1.66 x 10^-6 M = [Fe⁺²]

pFe = 5.78

(5 pts.) (c.) 14.00 mL V_FeM_Fe > V_EDTAM_EDTA, post-equivalence point

[Fe⁺²] = (V_FeM_Fe - V_EDTAM_EDTA)/(V_Fe + V_EDTA)

[Fe⁺²] = 8.51e-4M

pFe = 3.07

(10 pts.) 16. Identify the oxidizing and reducing agents among the reactants below and write a balanced chemical equation for:

S₂O₄^-2 + TeO₃^-2 + OH^-1 ó SO₃^-1 + Te + H₂O

TeO₃^-2 is the oxiding agent as it is reduced while oxidizing the S₂O₄^-2

S₂O₄^-2 is the reducing agent as it is oxidized while oxidizing the TeO₃^-2

The equation had a typo. It should have read:

S₂O₄^-2 + TeO₃^-2 + OH^-1 ó SO₃^-2 + Te + H₂O

balanced it would have been

2S₂O₄^-2 + TeO₃^-2 + 2OH^-1 ó 4SO₃^-2 + Te + H₂O

everyone got full credit on the balancing part.

(2 pts.) 17. What is the purpose of a salt bridge in an electrochemical cell.

The salt bridge completes the circuit in an electrochemical cell by providing a pathway for ions to flow from one half cell to the other.