Chapter 26 Gravimetric Analysis Precipitation Titrations and Combustion

Chapter 26 Gravimetric Analysis, Precipitation Titrations, and Combustion Analysis

26 -1 Examples of Gravimetric Analysis Gravimetric analysis Analyte Mass Quantification Precipitation Calculation of moles - For solution mol solute = M • V - For pure substance g solute mol solute = molar mass Ex) Determination of Cl- by precipitation with Ag+ Cl- + Ag+ Ag. Cl(s) (26 -1)

26 -2 Precipitation • Ideal precipitate for gravimetric analysis - Should be very insoluble. Easily filterable (i. e. , large particle). Very Pure. Known and constant composition. • Colloidal suspensions - Dia. =1 ~ 500 nm Suspension in the solution Not easily filterable Tyndall effect

Crystal Growth • Crystallization pathway ①Nucleation ②Particle growth • Super Saturation(SS) - The state a solution contains more solute than should be present at equilibrium - Less Super Saturation → Larger particles Nucleation at more SS → Nucleation rate > Particle growth less SS → tiny particles(or colloid) at less SS → Nucleation rate < Particle growth Rate → tractable particles Particle growth more SS Super saturation • Techniques for minimize super saturation (to promote particle growth) 1) Elevated temperature (to increase solubility or SS) 2) Slow addition of ppt. agent with good stirring (to less SS) 3) Dilute solution (to minimize SS)

Homogeneous Precipitation - The precipitant is generated slowly by a chemical reaction - Forms large particles (∵low SS) Ex) Formation of Fe(III) formate precipitate (26 -2) (the source of OH-) (26 -3) (26 -4)

Digestion: After precipitation, most procedures call for a period of standing in the presence of hot mother liquor. (promotes slow recrystallization of the precipitate. ) : Particle size increases and impurities tend to be expelled from the crystal. Purity • Types of coprecipitation: Impurities are ① Surface adsorption: adsorbed to crystal surface ② Inclusion: placed in crystal lattice instead of analyte ③ Occlusion: physically trapped within pocket - Coprecipitation is a problem because undesired impurities often coprecipitate with the analyte, resulting in excess mass. - Coprecipitation tends to be worst in colloidal precipitates (which have a large surface area).

§ Masking agent: an agent that reacts selectively with a interfering species so that suppress the interfering effect in analysis. - Some impurities (or interfering species) can be treated with a masking agent to prevent them from reacting with the precipitant. Ca 2+ + 2 RH Ca. R 2(s)↓ + 2 H+ Ex) Mixture of Ca 2+ and Mn 2+ : precipitation of only Ca. R 2(s) ? Analyte N-p-Chlorophenyl- Precipitate cinnamohydroxamic acid Mn 2+ Impurity + 6 CN- Mn(CN)64 - Masking agent Stays in solution ⇒Ag+, Mn 2+, Zn 2+, Cd 2+, Hg 2+, Fe 2+, Ga 2+ were masked by excess KCN. Pb 2+, Pd 2+, Sb 2+, Sn 2+, Bi 3+, Zr 4+, Ti 4+, V 5+, Mo 6+ were masked by a mixture of citrate and oxalate. § Postprecipitation: Impurities might collect on the product while it is standing in the mother liquor. (It usually involves a supersaturated impurity that does not readily crystallize. ) § Peptization: The breaking up of the product is called peptization, results in loss of product through the filter. Ex) Washing Ag. Cl(s) - Ag. Cl(s) washing with water Peptize Ag. Cl(s) - Ag. Cl(s) washing with diluted HNO 3(electrolyte) Preserve coagulated Ag. Cl(s). (HNO 3 is volatile so that it will be lost during drying) : Electrolytes to maintain coherence of tiny crystals by screening the surface charges : Volatile electrolytes; HNO 3, HCl, NH 4 NO 3, NH 4 Cl, (NH 4)2 CO 3

• Removal impurities by Reprecipitation - Washing away the mother liquor with pure solvent or electrolyte - Redissolving and reprecipitating the precipitate with solvent (Reprecipitation or Recrystallization)

26 -3 Examples of Gravimetric Calculations Example Relating Mass of Product to Mass of Reactant (26 -6) 0. 3126 g (Sample) 0. 7121 g (ppt) ⇒ wt% of piperazine in sample =? Solution = moles of piperazine

Example Calculating How Much Precipitant to Use - Ni content of steel sample is ~3 wt% (a) We wish to analyze 1. 0 g of steel. What volume of 1. 0 wt% DMG in alcohol solution should be used to give a 50% excess of DMG? (density of the solution is 0. 79 g/m. L). Solution Mass of Ni in 1. 0 g of steel (~3 wt% of Ni)=(0. 03)(1. 0 g)=0. 03 g ⇒ Mole of Ni =(0. 03 g)/58. 69 g/mol)=5. 1× 10 -4 mol Mole of DMG=2×(Mole of Ni)=2(5. 1× 10 -4 mol)=1. 02× 10 -3 mol ⇒ Mass of DMG=(1. 02× 10 -3)(116. 12)=0. 12 g Mass of 50% excess of DMG=(0. 12 g)(1. 5)=0. 18 g From 1. 0 wt% DMG in alcohol solution (density=0. 79 g/m. L) ⇒ 1. 0 wt% DMG=0. 01 g DMG/g solution Required mass of DMG solution =(0. 18 g DMG)/(0. 01 g DMG/g solution )=18 g solution ∴ Volume of DMG solution=mass/density=(18 g)/(0. 79 g/m. L)=23 m. L (b) 0. 1795 g of Ni(DMG)2 ppt from 1. 1634 g steel → wt% of Ni in steel? Solution Mole of Ni(DMG)2 = 0. 1795/288. 91 = 6. 213× 10 -4 mol =mole of Ni Mass of Ni =(6. 213× 10 -4)(58. 69)=0. 03646 g ∴ wt% of Ni

26 -4 Combustion Analysis Analyte + excess O 2(burning) → Mass of gas § Elemental analysis of organic compounds (to determine molar ratio of C, H, N, S) CO 2(g)+H 2 O(g)+N 2(g)+SO 3(g C, H, N, S ) excess O 2 (burning) H 2 O absorption CO 2 absorption

Example Combustion Analysis Calculations C, H, …. . (5. 714 mg) excess O 2 (burning) H 2 O(g)=2. 529 mg CO 2(g)=14. 414 mg Find wt% of C, H in the sample. Solution Mole of C = mole of CO 2 = (14. 414× 10 -3 g)/(44. 010 g/mol) =3. 275× 10 -4 mol Mass of C = (3. 275× 10 -4)(12. 0107)=3. 934 mg ∴ wt% of C Mole of H = 2×(mole of H 2 O) = 2(2. 529× 10 -3 g)/(18. 015 g/mol) =2. 808× 10 -4 mol Mass of H = (2. 808× 10 -4)(1. 0079)= 2. 830× 10 -4 g ∴ wt% of H

26 -5 Precipitation Titration Curves Titrant volume (x axis) vs. Concentration-related variables (y axis) Titration Curves Because concentration varies over many orders of magnitude, y-axis is the p function of analyte (or reagent) or potential of analyte (or reagent). p function: (26 -8) where [X] is concentration of X Sigmoidal Curve (S shape) p. X, (or Potential) (Titrant, m. L)

Titration curve of Precipitation titration Agentiometry: Precipitation titration curves involving silver ion - Standard solution(titrant): Ag. NO 3 - Analyte(X-): halide ions, SCN-, CNO-… - Reaction product: Ag. X(s) ☞ Titration reaction: X- + Ag+ → Ag. X(s) Titration curve: VAg+ vs. p. Ag (or VAg+ vs. p. X) Calculation [Ag+] (or [X-]) : Three distinct regions in titration curve 1) Before equivalence (excess analyte [X-]) 2) At equivalence 3) After equivalence (excess standard solution([Ag+])) Ex) Consider the titration of 25. 00 m. L of 0. 1000 M I- with 0. 05000 M Ag+. Titration reaction: (26 -9) (26 -10) - Since Ksp is so small, each addition of Ag+ reacts completely with I- (K≫ 1) - At equivalence point, all I- has been consumed. (sudden increase in Ag+ concentration) What volume of Ag+ titrant is need to reach the equivalence point (Ve) ? mol I- mol Ag+

Before the Equivalence Point - All titrant [Ag+] is consumed, free [I-] is [I-] that has not been precipitated. - Negligible I- from Ag. I(s) - Concentration of Ag+ is governed by Ksp [Ag+]=Ksp/[I-] (26 -11) Ex. Consider the titration of 25. 00 m. L of 0. 1000 M I- with 10 m. L of 0. 05000 M Ag+ Moles of I- = original moles of I- - moles of Ag+ added Volume is 0. 03500 L (25. 00 m. L + 10. 00 m. L) (26 -12) (26 -13) (26 -14)

At Equivalence Point - Added exactly enough Ag+ to react with all IComplete Ag. I(s) precipitation and only Ag. I(s) exist. [Ag+] independent of the original concentrations [Ag+] dependent on Ksp After Equivalence Point - All Ag+ added before equivalence point has ppt and remains excess Ag+. - [Ag+] is determined by Ag+ added after the equivalence point. Ex. For 2 m. L of Ag+ added past equivalence point (volume of 52 m. L of Ag+ added, total volume=(25+52)m. L=77 m. L) Moles of Ag+ = moles of added Ag+ - original moles of I-

The Shape of Titration Curve Equivalence point is the steepest point of the curve. : Point of maximum slope inflection point second derivative is zero Ve

§ Effect of Ksp on Titration Curve - Lowest solubility gives steepest change at equivalence point - The larger the equilibrium constant (smaller Ksp) for a titration reaction, the more pronounced will be the change in concentration near the equivalence point.

26 -6 Titration of a Mixture - Product with the Smaller Ksp Precipitates First - Two Stage Titration Curve - Assumes significant difference in Ksp (Chapter 13, 14, 15) Fig 26 -10

Titrate mixture of KI and KCl with Ag. NO 3 : Titration curve Ksp(Ag. I) << Ksp(Ag. Cl) (a) I- + Cl(b) IFirst, Ag. I ppt. Iodide end point(b) = p. Ag+ (∵Ag. I ppt. not complete at midpoint) Then, Ag. Cl ppt.

• Effect of Ksp for mixture Curve A(significant difference in Ksp ) : for Ag. I, Ag. Cl : Two equivalence points are evident Curve B(not significant difference in Ksp) : for Ag. Br, Ag. Cl : equivalence point for Cl- is unclear

26 -8 End-Point Detection • Three methods of end points detection in titration with silver nitrate 1) Chemicals: three chemical indicators (this section) 2) Potentiometry: measuring the potential between a Ag electrode and reference electrode (Chapter 14, 15) 3) Amperometry: measuring the current generated a pair of silver microelectrodes (Chapter 16) Apparatus for Potentiometric Titration Methods for end-point detection by indicators Volhard titration: formation of a soluble, colored complex at the end point Fajans titration: adsorption of a colored indicator on the precipitate at the end point Mohr titration: formation of a red silver chromate precipitate at the end point

■ Volhard Titration Analyte: 1) direct determination of SCN 2) indirect determination of halide ion (back titration) Indicator: Fe(III) p. H = Strong acidic condition (①prevent Fe(III)-hydrated oxide precipitation, ②remove interferences of carbonate, oxalate, and arsenate) 1) Direct determination of SCN- Titration reaction: Ag++SCN-=Ag. SCN(s) white Ksp (Ag. SCN)=1. 07 x 10 -12 - Indicator reaction: Fe 3++SCN-=Fe(SCN)2+ Kf (Fe(SCN)2+)=1. 05 x 103 Red, complex ion 2) Indirect determination of halide ion (Back titration) - Titration reaction: Ag+ (excess) + Cl- = Ag. Cl(s) : V 1 of Ag+ white SCN- + Ag+ = Ag. SCN(s) : V 2 of SCNwhite - Indicator reaction: Fe 3+ + 6 SCN- = Fe(SCN)63 Red, complex ion ⇒Total amount of Ag+ is known, so amount consumed by Cl- can be calculated. Subtract excess [Ag+] from total [Ag+] used to ppt. Cl-

■ Fajans Titration Analyte: halide ion Indicator: Fluorescein(adsorption indicator) - Titration reaction: Ag+ + Cl- = Ag. Cl(s) - Indicator reaction: Ag. Cl(s) + Ag+ + In 2 - = Ag. Cl(s)Ag+ln 2 - (red) Fluorescein (HIn 2) Ag. Cl(s) + Ag+ + ln 2 - 2 H+ + Fluoresceinate (ln 2 -) (yellow green) Ag. Cl(s)Ag+In 2 - (red) Before the equivalence point : After the equivalence point : solution in the surface layer of Ag. Cl(s) Cl- excess(Ag. Cl(s) particles are “-” charge) →In 2 - are repelled→yellow green in Ag+ excess(Ag. Cl(s) particles are “+” charge)→ ln 2 - are adsorbed→red color of Ag. Cl(s) Ag+In 2 -

§ Anionic dyes (indicator) - Maximize surface area higher binding stronger color change - small particle size low concentration - must use appropriate p. H to maintain negative charge Changes from greenish yellow to pink Sharper color transition, binds to tightly to Cl- - Fluorescein: p. H 7~10 - Dichlorofluorescein: p. H 4. 4> -Tetrachlorofluorescein(eosin): p. H 2>

■ The Mohr Titration Analyte: Cl-, Br-, CNIndicator: K 2 Cr. O 4 p. H = 7 ~ 10 - Titration reaction : Ag+ + Cl- = Ag. Cl(s) - Indicator reaction : 2 Ag+ white + Cr. O 42 - = Ag 2 Cr. O 4(s) red-brown Ksp (Ag 2 Cr. O 4(s))=12 x 10 -12 • Silver ion concentration at equivalence point • Chromate ion concentration require to initiate formation of Ag 2 Cr. O 4(s)