COLORED SOLUTIONS A solution will appear a certain color if it absorbs its complementary - PowerPoint Presentation

Slide1

EXPERIMENTS 2 and 3 – AN EQUILIBRIUM CONSTANT DETERMINATION

known - x

x

Fe3+ (aq) + SCN- (aq) ⇆

Initial M’sChange in M’sEquilibrium M’s

known

known

0

- x

- x

+ x

known - x

Kc = [FeSCN2+] _______________ [Fe3+][SCN-]

Kc = x _____________________________ (known – x)(known – x)

Must have a way to find x!

FeSCN2+ (aq) is blood red

FeSCN2+ (aq)

The concentration of a colored chemical is proportional to the amount of light it absorbs

EX2-1

(of

21)Slide2

EXPERIMENTS 2 and 3 – AN EQUILIBRIUM CONSTANT DETERMINATION

A darker color means a higher concentration of the colored component

The “darkness” can be determined by measuring the amount of light absorbed by the solution, called its ABSORBANCE

EX2-2 (of

21)

A solution’s absorbance is directly proportional to the concentration of the colored componentSlide3

SPECTROPHOTOMETER – A device that measures the amount of light absorbed by a sample

A

light bulb emits white light

Light

passes through a slit to form a narrow beam

A prism

separates the colors

of light

Another

slit allows just one color to pass

Light

passes through the sample

A detector measures the final amount of light

EX2-3

(of

21)Slide4

If you choose a specific wavelength of light, and measure the

absorbances of FeSCN2+ solutions of known concentrations, and plot absorbance vs. concentration

C:

0.25 M 0.50 M 0.75 M 1.00 MA: 0.241 0.478 0.722 0.961

Concentration

of

FeSCN

2+

Absorbance

EX2-4

(of 21)

, the relation is linear

y = mx + b

A = mC + bSlide5

EX2-5 (of 21)If an equilibrium solution has an absorbance of 0.351 using a wavelength of light of 476.0 nm,

find its concentration of FeSCN2+ 0.351 = (3425 M-1)C – 0.021

0.372 = (3425 M-1)C 0.372 = C ___________ 3425 M-1

= 0.000109 M

A@476.0 = mx + bm(slope): 3425b(y-intercept): - 0.021

m = Δy ____ Δx

= Δ Abs. __________ Δ Conc.

= no units ___________ M = M

-1

A@476 = mC + b

known - x

x

Fe3+ (aq) + SCN- (aq) ⇆known

known0

- x

-

x

+

x

known

-

x

FeSCN

2+

(

aq

)Slide6

PART A – Preparing the STOCK SOLUTION of FeSCN2+

10.00 mL0.200 M Fe(NO3

)3

3.00 mL0.00200 M KSCN

17.00 mL

6 M HNO3

EX2-6

(of 21)STOCK SOLUTION – A solution of known concentration (in this case, of FeSCN2+)? M FeSCN2+Slide7

MCVC

= MDVD

MCVC = MD_______

VD

= (0.200 M)(10.00 mL) ________________________ (30.00 mL)

= 0.0667 M Fe(NO3)3

M

C =VC =

0.200 M10.00 mL

MD

=VD =

? M30.00 mLPART A – Preparing the STOCK SOLUTION of FeSCN2+Calculation 1: Concentration of Fe(NO3)

3 in the Stock Solution (Initial)EX2-7 (of 21)20010 001.002003 0030 00

0667Slide8

MCVC

= MDVD

MCVC = MD_______

VD

= (0.00200 M)(3.00 mL) ___________________________ (30.00 mL)

= 0.000200 M KSCN

M

C =VC =

0.00200 M3.00 mL

MD

=VD =

? M30.00 mLPART A – Preparing the STOCK SOLUTION of FeSCN2+Calculation 2: Concentration of KSCN in the Stock Solution (Initial)

EX2-8 (of 21)20010 002.002003 0030 00

0667000200Slide9

PART A – Preparing the STOCK SOLUTION of FeSCN2+Calculation 3: Concentration of Fe3+ in the Stock Solution (Initial)

EX2-9 (of 21)20010 00

3.002003 0030 000667

0002000667

0.0667 M Fe(NO3)3 = 0.0667 M Fe3+

x 1 mol

Fe3+ __________________ 1 mol Fe(NO3)3Slide10

PART A – Preparing the STOCK SOLUTION of FeSCN2+Calculation 4: Concentration of Fe3+ in the Stock Solution (Initial)

EX2-10 (of 21)4.000200

20010 00002003 0030 00

06670002000667

0.000200 M KSCN= 0.000200 M SCN-

x 1

mol SCN- _______________ 1 mol KCSNSlide11

0.000200 - x

x Fe3+

(aq) + SCN- (aq) ⇆ FeSCN2+ (aq)

Initial M’sChange in M’sEquilibrium M’s

0.0667

0.000200

0

- x

- x

+ x

0.0667 - xPART A – Preparing the

STOCK SOLUTION of FeSCN2+Concentration of FeSCN2+ in the Stock Solution (Equilibrium)Because we have 333 times more Fe3+ than SCN-, we will assume that essentially all of the SCN- is converted to FeSCN2+ at equilibrium

EX2-11 (of 21)Slide12

~0.000200 Fe3+

(aq) + SCN- (aq) ⇆ FeSCN2+ (aq)

Initial M’sChange in M’sEquilibrium M’s

0.0667

0.000200

0

- ~0.000200

- ~0.000200

+ ~0.000200

~0.0665PART A – Preparing the STOCK

SOLUTION of FeSCN2+Concentration of FeSCN2+ in the Stock Solution (Equilibrium) the [FeSCN2+] = [SCN-] = 0.000200 M

EX2-12 (of 21)

~0Slide13

PART A – Preparing the STOCK SOLUTION of FeSCN2+Concentration of FeSCN2+ in the Stock Solution (Equilibrium)

EX2-13 (of 21)00020020010 00

002003 0030 000667000200

0667000200Slide14

PART B – Preparing the

STANDARD SOLUTIONS of FeSCN2+Must calculate the concentration of FeSCN2+ in each standard solution

0 M FeSCN2+

Solution 1:0.000200 M FeSCN2+Solutions 2-5:MCVC = MD

VDSolution 0:

EX2-14 (of 21)0

000200Slide15

PART C – Determining the

Absorbances of the STANDARD SOLUTIONS

EX2-15 (of 21)

Place a colorless solution (called the

BLANK) in the spectrometer and set it to zero absorbanceSlide16

PART C – Determining the

Absorbances of the STANDARD SOLUTIONS

EX2-16 (of 21)

ABSORBANCE SPECTRUM – A graph of the absorbance of a solution at different wavelengths

Place a colored standard solution in the spectrometer

I

f a solution appears orange, it is primarily absorbing its complimentary color, blueSlide17

PART C – Determining the

Absorbances of the STANDARD SOLUTIONSLAMBDA MAX (λmax) – The wavelength of maximum absorbance

When

measuring the absorbance of solutions, it is most accurate to measure the absorbance at λmax

EX2-17 (of 21)Peak of the absorbance graph

λmaxSlide18

This is called a CALIBRATION LINE

C: 0.25 M 0.50 M 0.75 M 1.00 M

A: 0.241 0.478 0.722 0.961

y = mx + bA@476 = (3425 M-1)C – 0.021

A@476.0 = mx + bm(slope): 3425

b(y-intercept): - 0.021R2 = 0.995

EX2-18 (of 21)

PART C – Determining the

Absorbances of the STANDARD SOLUTIONS

Determine the absorbance of each standard solution at

λmaxSlide19

This is called a CALIBRATION LINE

C: 0.25 M 0.50 M 0.75 M 1.00 M

A: 0.241 0.478 0.722 0.961A@476.0 = mx + b

m(slope): 3425b(y-intercept): - 0.021R2 = 0.995

EX2-19 (of 21)

PART C – Determining the

Absorbances of the STANDARD SOLUTIONS

Determine the absorbance of each standard solution at

λmax

y = mx + b

A@476 = (3425 M

-1)C – 0.021Slide20

EX2-20 (of 21)

1852 AUGUST BEERProposed a mathematical explanation for the linear relationship between

concentration and absorbance

A = mC + b

A = absorbance

ɛ = extinction coefficient (a constant for a given solute at a specific λ)l = width

of the cuvet holding the sample (for our cuvets it is 1.00 cm)C = concentration (

in our lab it’s in “M FeSCN2+”)

l

= 1.00 cm

A = ɛ

l CBEER’S LAW :

+ 0Slide21

Calculate the extinction coefficient for absorbance at a wavelength of 476 nm and using a 1.00 cm cuvet given the calibration line:

A@476 = (3425 M-1)C – 0.021

A@476 = ɛ

l C + b

A@

476 = mC + b slope = ɛ l

EX2-21 (of 21)

m = ɛ

___ l

= 3425 M-1 ____________

1.00 cm= 3430 M-1cm-1