COLORED SOLUTIONS A solution will appear a certain color if it absorbs its complementary color from the color wheel EX2-1 (of 24)

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Presentation transcript:

COLORED SOLUTIONS A solution will appear a certain color if it absorbs its complementary color from the color wheel EX2-1 (of 24)

COLORED SOLUTIONS If a solution appears orange, it is primarily absorbing its complimentary color, blue EX2-2 (of 24)

Fe3+ (aq) + SCN- (aq) ⇆ FeSCN2+ (aq) EQUILIBRIUM CONSTANT DETERMINATION Fe3+ (aq) + SCN- (aq) ⇆ FeSCN2+ (aq) (blood red) 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 EX2-3 (of 24)

Concentration of FeSCN2+ If you measure the absorbance of FeSCN2+ solutions of known concentrations, and plot absorbance vs. concentration , the relation is linear C: 0.25 M 0.50 M 0.75 M 1.00 M A: 0.241 0.478 0.722 0.961 Absorbance Concentration of FeSCN2+ EX2-4 (of 24)

A light bulb emits white light SPECTROPHOTOMETER – A device that measures the amount of light absorbed by a sample A light bulb emits white light A diffraction grating separates the colors of light Light passes through the sample Light passes through a slit to form a narrow beam Another slit allows just one color to pass A detector measures the final amount of light EX2-5 (of 24)

100 photons Incident Light Transmitted Light 10 photons I0 It TRANSMITTANCE (T) – the fraction of the incident light that passes through the sample T = It / I0 T = 10 photons = 0.1 _________________ 100 photons EX2-6 (of 24)

100 photons 10 photons I0 It ABSORBANCE (A) – negative logarithm of the transmittance A = -log (T) A = -log (0.1) = 1 EX2-7 (of 24)

100 photons 1 photon I0 It ABSORBANCE (A) – negative logarithm of the transmittance A = -log (T) A = -log (0.01) = 2 EX2-8 (of 24)

PART A – Preparing the STOCK SOLUTION of FeSCN2+ 10.00 mL 0.200 M Fe(NO3)3 3.00 mL 0.00200 M KSCN 17.00 mL 6 M HNO3 EX2-9 (of 24)

PART A – Preparing the STOCK SOLUTION of FeSCN2+ Calculation 1: Concentration of Fe(NO3)3 in the Stock Solution (Initial) 1. 200 MCVC = MDVD 10 00 MC = VC = 0.200 M 10.00 mL MD = VD = ? M 30.00 mL 00200 3 00 MCVC = MD _______ VD = (0.200 M)(10.00 mL) ________________________ (30.00 mL) 30 00 0667 = 0.0667 M Fe(NO3)3 EX2-10 (of 24)

PART A – Preparing the STOCK SOLUTION of FeSCN2+ Calculation 2: Concentration of KSCN in the Stock Solution (Initial) 2. 200 MCVC = MDVD 10 00 MC = VC = 0.00200 M 3.00 mL MD = VD = ? M 30.00 mL 00200 3 00 MCVC = MD _______ VD = (0.00200 M)(3.00 mL) ___________________________ (30.00 mL) 30 00 0667 000200 = 0.000200 M KSCN EX2-11 (of 24)

PART A – Preparing the STOCK SOLUTION of FeSCN2+ Calculation 3: Concentration of Fe3+ in the Stock Solution (Initial) 3. 200 0.0667 M Fe(NO3)3 x 1 = 0.0667 M Fe3+ 10 00 00200 3 00 30 00 0667 000200 0667 EX2-12 (of 24)

PART A – Preparing the STOCK SOLUTION of FeSCN2+ Calculation 4: Concentration of Fe3+ in the Stock Solution (Initial) 4. 200 0.000200 M KSCN x 1 = 0.000200 M SCN- 10 00 00200 3 00 30 00 0667 000200 0667 000200 EX2-13 (of 24)

PART A – Preparing the STOCK SOLUTION of FeSCN2+ Concentration of FeSCN2+ in the Stock Solution (Equilibrium) Fe3+ (aq) + SCN- (aq) ⇆ FeSCN2+ (aq) Initial M’s Change in M’s Equilibrium M’s 0.0667 0.000200 - x - x + x 0.0667 - x 0.000200 - x x We will assume all of the SCN- is converted to FeSCN2+ at equilibrium EX2-14 (of 24)

PART A – Preparing the STOCK SOLUTION of FeSCN2+ Concentration of FeSCN2+ in the Stock Solution (Equilibrium) Fe3+ (aq) + SCN- (aq) ⇆ FeSCN2+ (aq) Initial M’s Change in M’s Equilibrium M’s 0.0667 0.000200 - 0.000200 -0.000200 + 0.000200 0.0667 – 0.000200 0.000200 – 0.000200 0.000200 We will assume all of the SCN- is converted to FeSCN2+ at equilibrium  the [FeSCN2+] = 0.000200 M EX2-15 (of 24)

PART A – Preparing the STOCK SOLUTION of FeSCN2+ Concentration of FeSCN3+ in the Stock Solution (Equilibrium) 200 10 00 00200 3 00 30 00 0667 000200 0667 000200 000200 EX2-16 (of 24)

PART B – Preparing the STANDARD SOLUTIONS of FeSCN2+ 000200 Must calculate the concentration of FeSCN2+ in each standard solution Solution 1: 0.000200 M FeSCN2+ Solution 2-5: MCVC = MDVD Solutions 0: 0 M FeSCN2+ EX2-17 (of 24)

PART C – Determining the Absorbances of the STANDARD SOLUTIONS Place a colored standard solution in the spectrometer ABSORBANCE SPECTRUM – A graph of the absorbance of a solution at different wavelengths EX2-18 (of 24)

PART C – Determining the Absorbances of the STANDARD SOLUTIONS LAMBDA MAX (λmax) – The wavelength of maximum absorbance When measuring the absorbance of solutions, it is most accurate to measure the absorbance at λmax EX2-19 (of 24)

PART C – Determining the Absorbances of the STANDARD SOLUTIONS Determine the absorbance of each standard solution A-546.0 = mx + b m(slope): 3425 b(y-intercept): - 0.021 C: 0.25 M 0.50 M 0.75 M 1.00 M A: 0.241 0.478 0.722 0.961 y = mx + b A = mc + b A = (3425 M-1)c – 0.021 This is called a CALIBRATION LINE m = Δy ____ Δx = Δ Absorbance _____________________ Δ Concentration = no units ___________ M = M-1 EX2-20 (of 24)

PART C – Determining the Absorbances of the STANDARD SOLUTIONS Determine the absorbance of each standard solution A-546.0 = mx + b m(slope): 3425 b(y-intercept): - 0.021 C: 0.25 M 0.50 M 0.75 M 1.00 M A: 0.241 0.478 0.722 0.961 y = mx + b A = mc + b A = (3425 M-1)c – 0.021 This is called a CALIBRATION LINE m = Δy ____ Δx = Δ Absorbance _____________________ Δ Concentration = no units ___________ M = M-1 EX2-21 (of 24)

PART C – Determining the Absorbances of the STANDARD SOLUTIONS If an unknown solution has an absorbance of 0.351, find its concentration of FeSCN2+ A-546.0 = mx + b m(slope): 3425 b(y-intercept): - 0.021 0.351 = (3425 M-1)c – 0.021 0.372 = (3425 M-1)c 0.372 = c ___________ 3425 M-1 = 0.000109 M EX2-22 (of 24)

ɛ = extinction coefficient 1852 AUGUST BEER Proposed a mathematical explanation for the linear relationship between concentration and absorbance BEER’S LAW : A = ɛ l c A = absorbance ɛ = extinction coefficient (a constant for a given solute at a given 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 EX2-23 (of 24)

A = ɛ l c + b A = mc + b slope = ɛ l Calculate the extinction coefficient for absorbance at a wavelength of 546 nm and using a 1.00 cm cuvet given the calibration line: 0.351 = (3425 M-1)c – 0.021 m = ɛ l m = ɛ ___ l = 3425 M-1 ____________ 1.00 cm = 3425 M-1cm-1 EX2-24 (of 24)