Chapter 26 Gravimetric Analysis, Precipitation Titrations, and Combustion Analysis

26-1 Examples of Gravimetric Analysis Calculation of moles Gravimetric analysis - For solution Analyte Mass Quantification mol solute = M•V - For pure substance Precipitation g solute mol solute = molar mass Ex) Determination of Cl- by precipitation with Ag+ Cl- + Ag+  AgCl(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 Particle growth Nucleation Super saturation Rate  at more SS → Nucleation rate > Particle growth → tiny particles(or colloid)  at less SS → Nucleation rate < Particle growth → tractable particles less SS more SS  • 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 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).

Ex) Mixture of Ca2+ and Mn2+ : precipitation of only CaR2(s) ? 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. Ex) Mixture of Ca2+ and Mn2+ : precipitation of only CaR2(s) ? Ca2+ + 2RH  CaR2(s)↓ + 2H+ Analyte N-p-Chlorophenyl- Precipitate cinnamohydroxamic acid Mn2+ + 6CN-  Mn(CN)64- Impurity Masking agent Stays in solution ⇒Ag+, Mn2+, Zn2+, Cd2+, Hg2+, Fe2+, Ga2+ were masked by excess KCN. Pb2+, Pd2+, Sb2+, Sn2+, Bi3+, Zr4+, Ti4+, V5+, Mo6+ 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 AgCl(s) - AgCl(s) washing with water  Peptize AgCl(s) - AgCl(s) washing with diluted HNO3(electrolyte)  Preserve coagulated AgCl(s). (HNO3 is volatile so that it will be lost during drying) : Electrolytes to maintain coherence of tiny crystals by screening the surface charges : Volatile electrolytes; HNO3, HCl, NH4NO3, NH4Cl, (NH4)2CO3

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 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

Calculating How Much Precipitant to Use 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/mL). Solution Mass of Ni in 1.0g of steel (~3 wt% of Ni)=(0.03)(1.0g)=0.03g ⇒ Mole of Ni =(0.03g)/58.69g/mol)=5.1×10-4 mol Mole of DMG=2×(Mole of Ni)=2(5.1×10-4mol)=1.02×10-3 mol ⇒ Mass of DMG=(1.02×10-3)(116.12)=0.12g Mass of 50% excess of DMG=(0.12g)(1.5)=0.18g From 1.0 wt% DMG in alcohol solution (density=0.79 g/mL) ⇒ 1.0 wt% DMG=0.01g DMG/g solution Required mass of DMG solution =(0.18g DMG)/(0.01g DMG/g solution )=18g solution ∴ Volume of DMG solution=mass/density=(18g)/(0.79g/mL)=23 mL (b) 0.1795 g of Ni(DMG)2 ppt from 1.1634g steel → wt% of Ni in steel? Solution Mole of Ni(DMG)2 = 0.1795/288.91 = 6.213×10-4mol =mole of Ni Mass of Ni =(6.213×10-4)(58.69)=0.03646g ∴ wt% of Ni

CO2(g)+H2O(g)+N2(g)+SO2(g)+SO3(g) 26-4 Combustion Analysis Analyte + excess O2(burning) → Mass of gas Elemental analysis of organic compounds (to determine molar ratio of C, H, N, S) C, H, N, S CO2(g)+H2O(g)+N2(g)+SO2(g)+SO3(g) excess O2 (burning) H2O absorption CO2 absorption

Example Combustion Analysis Calculations C, H, ….. (5.714 mg) H2O(g)=2.529 mg CO2(g)=14.414mg excess O2 (burning) Find wt% of C, H in the sample. Solution Mole of C = mole of CO2 = (14.414×10-3g)/(44.010g/mol) =3.275×10-4mol Mass of C = (3.275×10-4)(12.0107)=3.934mg ∴ wt% of C Mole of H = 2×(mole of H2O) = 2(2.529×10-3g)/(18.015g/mol) =2.808×10-4mol Mass of H = (2.808×10-4)(1.0079)= 2.830×10-4g ∴ 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) pX, (or Potential) (Titrant, mL)

Titration curve of Precipitation titration Agentiometry: Precipitation titration curves involving silver ion - Standard solution(titrant): AgNO3 - Analyte(X-): halide ions, SCN-, CN-, CNO-… - Reaction product: AgX(s) ☞ Titration reaction: X- + Ag+ → AgX(s) Titration curve: VAg+ vs. pAg (or VAg+ vs. pX) : Three distinct regions in titration curve 1) Before equivalence (excess analyte [X-]) 2) At equivalence 3) After equivalence (excess standard solution([Ag+])) Calculation [Ag+] (or [X-]) Ex) Consider the titration of 25.00 mL of 0.1000M I- with 0.05000M 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 AgI(s) - Concentration of Ag+ is governed by Ksp [Ag+]=Ksp/[I-] (26-11) Ex. Consider the titration of 25.00 mL of 0.1000M I- with 10 mL of 0.05000M Ag+ Moles of I- = original moles of I- - moles of Ag+ added Volume is 0.03500 L (25.00 mL + 10.00 mL) (26-12) (26-13) (26-14)

After Equivalence Point At Equivalence Point - Added exactly enough Ag+ to react with all I- - Complete AgI(s) precipitation and only AgI(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 mL of Ag+ added past equivalence point (volume of 52 mL of Ag+ added, total volume=(25+52)mL=77mL) 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 AgNO3 : Titration curve Ksp(AgI) << Ksp(AgCl) (a) I- + Cl- (b) I- First, AgI ppt. Then, AgCl ppt. Iodide end point(b) = pAg+ (∵AgI ppt. not complete at midpoint)

Effect of Ksp for mixture Curve A(significant difference in Ksp ) : for AgI, AgCl : Two equivalence points are evident Curve B(not significant difference in Ksp) : for AgBr, AgCl : equivalence point for Cl- is unclear

26-8 End-Point Detection •Three methods of end points detection in titration with silver nitrate 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) pH = 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-=AgSCN(s) Ksp (AgSCN)=1.07 x 10-12 white - Indicator reaction: Fe3++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- = AgCl(s) : V1 of Ag+ white SCN- + Ag+ = AgSCN(s) : V2 of SCN- white - Indicator reaction: Fe3+ + 6SCN- = 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- = AgCl(s) - Indicator reaction: AgCl(s) + Ag+ + In2- = AgCl(s)Ag+ln2- (red) Fluorescein (HIn2) 2H+ + Fluoresceinate (ln2-) (yellow green) AgCl(s) + Ag+ + ln2- AgCl(s)Ag+In2- (red) Before the equivalence point : Cl- excess(AgCl(s) particles are “-” charge) →In2- are repelled→yellow green in solution After the equivalence point : Ag+ excess(AgCl(s) particles are “+” charge)→ ln2- are adsorbed→red color of AgCl(s) Ag+In2- in the surface layer of AgCl(s)

Anionic dyes (indicator) - Maximize surface area  higher binding stronger color change - small particle size  low concentration - must use appropriate pH to maintain negative charge Changes from greenish yellow to pink Sharper color transition, binds to tightly to Cl- Fluorescein: pH 7~10 Dichlorofluorescein: pH 4.4> Tetrachlorofluorescein(eosin): pH 2>

■ The Mohr Titration Analyte: Cl-, Br-, CN- Indicator: K2CrO4 pH = 7 ~ 10 - Titration reaction : Ag+ + Cl- = AgCl(s) white - Indicator reaction : 2Ag+ + CrO42- = Ag2CrO4(s) Ksp (Ag2CrO4(s))=12 x 10-12 red-brown • Silver ion concentration at equivalence point • Chromate ion concentration require to initiate formation of Ag2CrO4(s)