Fe3+ (aq) + SCN- (aq) ⇆ FeSCN2+ (aq)

(blood red)

EQUILIBRIUM CONSTANT DETERMINATION

COLORED SOLUTIONS

A solution will appear a certain color if it absorbs its complementary color from the color wheel

If a solution appears red, it is primarily absorbing its complimentary color, green

COLORED SOLUTIONS

White light is a combination

of all colors

Sample absorbs green, but transmits

all other colors

Your eye sees the remaining combination

of colors as red

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

LAMBDA MAX (λmax) – The wavelength of maximum absorbance

For best accuracy, when measuring the absorbance of solutions, it is best to measure the absorbance at λmax

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 diffraction grating 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

Transmitted Light

The fraction of light that gets through is the TRANSMITTANCE

I0 It

TRANSMITTANCE (T) – the fraction of the incident light that passes through the sample

T = It / I0

100 photons

0 photons

T = 0 photons = 0 ________________

100 photons

Incident Light

10 photons

T = 10 photons = 0.10 ________________

100 photons

ABSORBANCE (A) – negative logarithm of the transmittance

A = -log (T)

The fraction of light that gets through is the TRANSMITTANCE

I0

100 photons

10 photons

A = -log (0.1) = 1

The fraction of light that doesn’t get through (is blocked or

absorbed) is the ABSORBANCEIt

A = -log (0.01) = 2

A = -log (1) = 0

The darker the color, the higher concentration of the colored component, the higher the absorbance of the solution

A = absorbance

ɛ = extinction coefficient (a constant for a given solute at a given wavelength)

b = 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+)

b = 1.00 cm

A = ɛbc

BEER’S LAW – The mathematical relationship between concentration and absorbance

In our lab ɛ and b are constants, so A and c are the variables

A cA = ɛbc

y = mx + b

+ 0

This means a graph of A vs. c will produce a straight line

A cA = ɛbc

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

y = mx + b

+ 0

You will produce several solutions of known concentration:

This is called a CALIBRATION LINE because it shows the relationship between the measured absorbance and the concentration of FeSCN2+

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

0 0.2 0.4 0.6 0.8 1 1.20

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.9612 x − 0.000199999999999978

Beer's Law Graph

M FeSCN2+

Abso

rban

ce

m = Δy ____

Δx

A = ɛbc

A = (0.9612 M-1)c

= Δ Absorbance _____________________

Δ Concentration

= no units ___________

M

= M-1

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

If an unknown solution has an absorbance of 0.351, find its concentration

0.351 = (0.9612 M-1)c

0.351 = c_____________

0.9612 M-1

0.365 M = c

A = ɛbc

A = (0.9612 M-1)c0 0.2 0.4 0.6 0.8 1 1.2

0

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.9612 x − 0.000199999999999978

Beer's Law Graph

M FeSCN2+

Abso

rban

ce

Calculate the extinction coefficient of this substance, with units

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

m = ɛb

m = ɛ ___

b

A = ɛbc

A = (0.9612 M-1)c

= 0.9612 M-1

_____________

1.00 cm

= 0.961 M-1cm-1

0 0.2 0.4 0.6 0.8 1 1.20

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.9612 x − 0.000199999999999978

Beer's Law Graph

M FeSCN2+

Abso

rban

ce

PART A – Preparing the STOCK SOLUTION

10.00 mL0.200 M Fe(NO3)3

3.00 mL0.00200 M KSCN

17.00 mL6 M HNO3

MCVC = MDVD

MCVC = MD

_______

VD

= (0.200 M)(10.00 mL) ________________________

(30.00 mL)

= 0.0667 M Fe(NO3)3

MC =

VC =

0.200 M

3.00 mL

MD =

VD =

? M

30.00 mL

PART A – Preparing the STOCK SOLUTION

Concentration of Fe(NO3)3 in the Stock Solution:

PART A – Preparing the STOCK SOLUTION

Concentration of Fe3+ in the Stock Solution:

x 1 mol Fe3+

___________________

1 mol Fe(NO3)3

0.0667 mol Fe(NO3)3

__________________________

L

= 0.0667 M Fe3+

0.000200 - x x

Fe3+ (aq) + SCN- (aq) ⇆ FeSCN2+ (aq)

Initial M’s

Change in M’s

Equilibrium M’s

0.0667 0.000200 0

- x - x + x

0.0667 - x

PART A – Preparing the STOCK SOLUTION

Concentration of FeSCN2+ in the Stock Solution:

0.000200 – 0.000199 0.000199

Fe3+ (aq) + SCN- (aq) ⇆ FeSCN2+ (aq)

Initial M’s

Change in M’s

Equilibrium M’s

0.0667 0.000200 0

- 0.000199 -0.000199 + 0.000199

0.0667 – 0.000199

PART A – Preparing the STOCK SOLUTION

Concentration of FeSCN2+ in the Stock Solution:

We will assume all of the SCN- is converted to FeSCN2+ at equilibrium

the [FeSCN2+] = 0.000200 M

PART B – Preparing the STANDARD SOLUTIONS

Must calculate the concentration of FeSCN2+ in each standard solution

MCVC = MDVD

Solution 0: 0 M FeSCN2+

Solution 1: 0.00200 M FeSCN2+

Solution 2:

EXPERIMENT 3 – DETERMINATION OF THE Keq

MCVC = MD

_______

VD

= (0.00200 M)(5.00 mL) __________________________

(10.00 mL)

= 0.00100 M Fe(NO3)3 = 0.00100 M Fe3+

MCVC = MD

_______

VD

= (0.00200 M)(2.00 mL) __________________________

(10.00 mL)

= 0.000400 M KSCN = 0.000400 M SCN-

EXPERIMENT 3 – DETERMINATION OF THE Keq

0.000400 - x x

Fe3+ (aq) + SCN- (aq) ⇆ FeSCN2+ (aq)

Initial M’s

Change in M’s

Equilibrium M’s

0.00100 0.000400 0

- x - x + x

0.00100 - x

Kc = [FeSCN2+] _______________

[Fe3+][SCN-]

Kc = x _________________________________

(0.00100 – x)(0.000400 – x)

0 0.2 0.4 0.6 0.8 1 1.20

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.9612 x − 0.000199999999999978

Beer's Law Graph

M FeSCN2+

Abso

rban

ceEXPERIMENT 3 – DETERMINATION OF THE Keq

Measured absorbance of solution: 0.246

A = (3425 M-1)c - 0.021

0.246 = (3425 M-1)c - 0.021

0.267 = (3425 M-1)c

0.267 = c__________

3425 M-1

0.00007796 M = c

EXPERIMENT 3 – DETERMINATION OF THE Keq

0.000400 - x x

Fe3+ (aq) + SCN- (aq) ⇆ FeSCN2+ (aq)

Initial M’s

Change in M’s

Equilibrium M’s

0.00100 0.000400 0

- x - x + x

0.00100 - x

x = 0.00007796 M

= 0.000922 M

= 0.0003220 M

= 0.000077956 M

[Fe3+]eq =

[SCN-]eq =

[FeSCN-]eq =

0.00100 – x

0.000400 – x

x

Kc = [FeSCN2+] _______________

[Fe3+][SCN-]

Kc = (0.000077956 M) _________________________________

(0.000922 M)(0.0003220 M)

= 260 M-1