Chemistry 315: EDTA Titration Calculations

• Exam for Quantitative Analysis, Chemistry 315 was held by D.C. Muddiman on March 7, 2024 and was worth 200 points

Titration Experiment with EDTA

• 50.00 mL of 0.0100 M Ca2+ at pH 10 is titrated with 0.0100 M EDTA solution.
• Equilibrium constant (K) is 5.0 × 10^10, and αY4- is 0.35 at pH 10.
• Equilibrium expression: CdY2- ⇌ Y4- + CT.
• Titration reaction: Ca2+ + Y4- → CdY2-

Concentration of Y4- Form of EDTA

• The concentration of Y4- is calculated as [Y4-] = α4CT = 0.35 × 0.0100 M = 3.50 × 10^-3 M

Initial pCa Calculation

• The initial pCa, before adding any EDTA, is calculated as -log[Ca2+] = -log[0.0100 M] = 2.00

pCa After Titration with 10.00 mL EDTA

• [Ca2+] ≈ (50.00 mL × 0.0100 M - 10.00 mL × 0.0100 M) / (50.00 + 10.00) mL + CT
• ст is negligible due to the high formation constant (~10^10).
• [Ca2+] = (0.5 mmoles Ca2+ – 0.1 mmoles EDTA) / 60 mL = 0.0067 M
• pCa = -log(0.0067 M) = 2.17

Volume of Titrant at Equivalence Point

• The volume required to reach the equivalence point is calculated as (0.0100 M Ca2+)(50.00 mL) / 0.0100 M EDTA = V2
• V2 = 50.00 mL

pCa at Equivalence Point

• CCaY2- = (50.00 × 0.0100) mmol / (50.00 + 50.00) mL
• [Ca2+] = C1, and [CaY2-] = CCaY2- - [Ca2+] ≈ CCaY2-
• KCay = [CaY2-] / ([Ca2+]CT) = CCaY2- / [Ca2+]²
• [Ca2+] = √(CCaY2- / KCay) = √(5.35 × 10^-7)
• pCa = -log[Ca2+] = 6.27

pCa After Titrating with 60.00 mL

• CCay2- = (50.00 × 0.0100) mmol / (50.00 + 60.00) mL
• CEDTA = (60.00 × 0.0100) mmoles – (50.00 × 0.0100) mmoles / 110.00 mL
• [Ca2+] = √(CCay2-] / (KCaY CEDTA)) = √(0.0045 / (1.75 × 10^10 × 0.000909)) = 2.83 × 10^-10
• pCa = 9.55

Molar Absorptivity and Beer's Law

• Molar absorptivity for aqueous solutions of phenol at 211 nm is 6.17x10^3 L mol^-1 cm^-1.
• The highest recommended molar concentration for phenol analysis by absorption spectrophotometry in a 1.00-cm cell is calculated using Beer's Law.
• Given A = εbC, A = 1.00 = (6.17 x 10^3 L mol^-1 cm^-1)(1.00 cm)C
• C = 1.00 / ((6.17 x 10^3 L mol^-1 cm^-1)(1.00 cm)) = 1.62 × 10^-4 M

Fluorescence and Excitation Wavelengths

• An analyte with a fluorescent peak at 530 nm can be caused to fluoresce by excitation wavelengths of 350 nm and 370 nm.

Factors Related to Fluorescence and Absorption Experiments

• A solution with 2.5% transmittance at a given wavelength is strongly absorbing.
• The maximum signal recorded in an absorption experiment corresponds to the blank.
• The limit of detection in a fluorescence experiment decreases with a higher intensity light source.
• Decreasing sample temperature increases the fluorescence emission signal.
• For a given analyte, phosphorescence typically occurs at the longest wavelength.
• A positive deviation from Beer's law occurs when less light reaches the detector than predicted.

Fluorescence Calibration Curve and Factors for Increasing its Slope

• Factors which increase the slope of a fluorescence calibration curve:
• Exciting fluorescence with a more intense light source.
• Selecting a wider wavelength range for detecting emitted light as long as it includes wavelengths where the analyte emits.
• Using a longer path length.
• Observation: As sample concentration increases, self-absorption increases, causing fluorescence intensity to decrease due to photon absorption by other molecules in the ground state.

Atomic Absorption vs. Flame Emission Spectroscopy

• Atomic absorption is sometimes more sensitive than flame emission spectroscopy because, at flame temperature, ground-state atoms are much more abundant than excited atoms.
• Hollow cathode discharge tubes in atomic absorption have greater radiant power output than ordinary flames.

Molar Absorptivity and Dilution Calculation

• Compound with molecular weight 100 g/mol, molar absorptivity 1.00 × 10^5 L mol^-1 cm^-1.
• To achieve an absorbance of 0.500 after a 200-fold dilution, the required mass: C = A / εb = 0.5 / (1.00 × 10^5 L mol^-1 cm^-1 * 1.00 cm) = 5 × 10^-6 M
• Initial concentration: (5 × 10^-6 M)(200) = 0.001 M
• Mass required: 0.001 M * 100 g/mole = 0.1 grams

Line Spectra Emission

• Line spectra are emitted by excited atoms and monatomic ions.

Atomic Absorption Spectroscopy

• In atomic absorption spectroscopy, the cathode in the hollow cathode lamp is constructed from the element to be investigated.

Extra Credit Questions

• The only number spelled with letters in descending order is ONE.
• It is TRUE that -40 degrees Celsius is equal to -40 degrees Fahrenheit.
• The number with the most synonyms is ZERO.