Last time: Chem 201 lecture 1b. Today: First Experiment: Calcium Exp. Calcium. Brief pointers on lab experiments. Brief pointers on all Lab Expts:

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1 Chem 201 lecture 1b Brief pointers on all Lab Expts: Last time: Orientation Contact information Student conduct Unit conversion Gravimetric factor Brief pointers for the Ca expt. Today: Brief pointers on lab experiments Discuss Ca experiment Mn expt Fe expt Fluoride expt Ni expt CuAA expt ph expt Continue lecture All the lab protocols are posted in the website: Please download these protocols (they will not be supplied by instructor in hard copy form, so you may wish to print them yourself). Determine YOUR sequence of lab experiments. (it depends on your locker #.) Pay careful attention to the instructions! First Experiment: Calcium Exp Calcium This is a complexometric titration. titrimetric analysis Titration rxn: Ca 2+ + EDTA 4 > Ca(EDTA) It s a 1:1 titration! So, M Ca V Ca = M EDTA V EDTA To determine M Ca (in ppm), you need V Ca, M EDTA,V EDTA. Prep. EDTA solution ( 0.01M); But to know M EDTA accurately, standardize it by titrating known Ca 2+ primary standard sol n. Need 0.5 g CaCO 3 dried to constant weight (heated and weighed until mass ±0.0004g for 2 subseq weighings) (outline only) 1

2 Typo in Ca protocol Sample calculation On page 2 of the lab protocol for Ca expt: Under heading: Standard Solution,pure CaCO 3 into a 250 ml Should read:,pure CaCO 3 into a 20 ml Consider the following data: g CaCO 3 standard dissolved in ml solution. Titration of ml of CaCO 3 std requires ml of EDTA to reach endpoint (V e ). What is M EDTA =? M CaCO3 = (0.3945g) (1mol/100.1g) / L = M M EDTA = M Ca V Ca / V EDTA = ( M)(25.00mL)/(19.55mL)= If titration of 50.0 ml of unknown solution requires mls to reach end point, what is ppm CaCO 3 in the unknown? M CaCO3 = M EDTA V EDTA / V CaCO3 = ( M)(34.25mL)/(50.00mL) = M; ppm CaCO 3 = ( mol/l)(100.1g/mol)(10 3 mg/g) =540.5 mg/l = ppm CaCO 3 Manganese (Mn) Beer s Law This uses UVvis spectrophotometry for %Mn in steel No drying of samples necessary. Steel sample Balance platform Weigh samples elevated over platform Steel is dissolved in HNO 3 and the Mn converted to MnO 4 Prepare standard steel together with unk steel and compare. A = εbc A= absorbance, ε b = path length (usu. 1 cm); c = concentration of absorbing species. Note that c is directly proportional to A So [MnO 4 ] is proportional to % Mn and also to A. So, A is proportional to % Mn. Manganese Manganese Use 2 unknowns and 2 standards. (0.5g each close in mass) When purple MnO 4 forms, dilute to 250mLs all 4 solutions Use only 1 of std solutions for serial dilution. I repeat: No dilution of the other solutions. Serial dilution: 50mL 50mL H 2 O How do you prepare the rest of the standards? You are to have: 0.94%, 0.47%, 0.235%, 0.118%,0.059% 0.94% 0.47% 2

3 Manganese Mn Calculations First, determine λ max (wavelength at max. absorption) (i.e. Vary wavelength and measure A) Plot Absorbance vs % Mn with λ = λ max. A unk A %Mn %Mn unk Problem: A 2.0 g of 0.56% Mn standard steel is dissolved in 250 mls in the form of MnO 4 and its absorbance(a)=0.22. For a 1.5 g of unknown steel, dissolved in 250mL, A = What is the %Mn in the unknown steel? A unk A std = "c unkb "c std b = c unk = %Mn unkm unk c std %Mn std m std We can solve for %Mn unk since we know 5 out of 6 variables, m %Mn unk =%Mn std A unk std = (.56%) m unk A std =1.25% Iron Experiment (Fe) Fe expt This is a redox titration experiment Titration reaction: 6Fe 2+ +Cr 2 O 7 >6Fe Cr 3+ etc.. DO NOT DRY SAMPLE! Minimize contact with O 2. Do 3 indicator titrations + 1 titration using potentiometer (looks like ph meter) Plot mv vs mls of Cr 2 O 7 added to get equivalence point Know how to balance the equation! At equiv.pt: M Fe V Fe = 6M Cr V Cr (remember how to do this) Derive the following: %Fe = (6M Cr V Cr )(AW Fe)(100%)/ (g ore (fraction titrated)) The plot will look like: mv ml Cr 2 O 7 V e Fe Expt Fluoride (F) experiment Things to keep in mind: a) don t dry anything to constant weight b) Do titrations on same day if possible c) titration using potentiometer. There s only 1 potentiometer (it s by the window, near oven) Uses ionselective electrode (ISE) specific for F Requires acidic buffered solution (TISAB; = Total Ionic Strength Adjustment Buffer) to minimize the effect of ionic strength variation in solutions being compared. Plot of log ppm vs mv gives straight line 3

4 Nickel Experiment (Ni) Ni procedures and reagents Gravimetric: involves selective precipitation of Ni Ppt n rxn: Ni DMG > Ni(DMG) 2 (red ppt) (ph 9) Ni ore:dry to const wt ( m =±0.0004g); use weighing bottle mass = g cruc.+ppt g cruc. = g Ni(DMG) gNi g Ni(DMG) gNi(DMG) % Ni = 2 x100% g Ni ore Understand the roles of the various reagents and the steps involved in the process for example, nitric acid, tartaric acid, 6M NH 3, ph 89, alcoholic DMG, heating not boiling, Digestion what to do if you don t have enough time to Where s a good place to stop? On day 1, at point in which you added DMG before digestion. Soda Ash Expt. (ph) Soda Ash Expt. (ph) Acidbase titration involving diprotic base, Na 2 CO 3. Dry ASAP: a)1g Na 2 CO 3 and b) Soda ash unknown. Standardize HCl standard using Na 2 CO 3. Unknown ore contains Na 2 CO 3 + inert soluble material Determine %Na 2 CO 3 (or as %Na 2 O) in unknown, 2 methods: a) indicator method; b) phmeter/gran Plot method. Use 2nd eq. pt. for %Na 2 CO 3 determination of unk. ore. Theory: 2 equiv. pts (at 2 diff. ph s, thus 2 diff indicators): (1) CO 3 + H + > HCO 3 V HCl = V e (example: 20mL) (2) HCO 3 + H + > H 2 CO 3 V HCl = V e (example: 20mL) Net: CO H + > H 2 CO 3 V HCl = V 2e (example: 40mL) You derive: Standardization: M HCl =2 (g Na2CO3 /105.99) / V 2e Unk: %Na 2 CO 3 = (1/2)M HCl V 2e (105.99)(100%)/g ore Note relation: Na 2 CO 3 > Na 2 O + CO 2 (g) Sometimes %Na 2 O is calculated instead of %Na 2 CO 3 To determine V 2e Gran Plot Problems often encountered in the 2 titrations using indic. a) Unclear 1st equiv. pt. (phenolphthalein is used) b) Unclear 2nd e.p. (methyl orange usedcolor chg slight?) 3rd tit n needed: ph vs mls of HCl every 1 ml. Use Gran Plot (read it in book!) Gran plot: V HCl x 10 ph (yaxis) vs nl HCl (xaxis) 4

CuAA More concentration units: % AAS uses this %(w/w) = g solution x100% or equivalently, %(w/w) = %(v/v) = ml solute 100 ml solution 100 g solution %(w/v) = 100 ml solution More concentration units:

5 CopperAA CuAA equipment Dry unknown Cu ore for 1 hour (one time only) Prep standard Cu 2+ solutions as described in protocol. Plot Signal (A) vs ppm Cu. It s important for Cu unk absorption to be within the linear range of A vs ppm curve. linear range, absorption is proportional to ppm Cu. A = k[cu], k= some constant. CuAA More concentration units: % AAS uses this %(w/w) = g solution x100% or equivalently, %(w/w) = %(v/v) = ml solute 100 ml solution 100 g solution %(w/v) = 100 ml solution More concentration units: ppm More concentration units: N ppm = g solution x106 ppm = or equivalently, 10 6 g solution For aqueous solutions it s convenient to derive: ppm = 10 6 g solution x = g solution m m = 10 3 g solution 1mL = 1L L solution g 10 3 ml Normality = # mole equivalents solute L solution What s the normality of 0.35 M H 2 SO 4? #N = 0.35 M H 2 SO 4 x (2N/1M) = 0.70 N H 2 SO 4 What s M for 0.25 N HCl? 0.25 M HCl! 5

6 Example of preparing a diluted solution from a stock solution. Concentrations of lab stocks Prepare 500mLs of 0.1 M HNO 3. Stock is 15.8 M HNO 3. Anyone remember the equation to use? M 1 V 1 = M 2 V 2 What to solve for? For example: 1=conc. 2 = diluted V 1 = M 2 V 2 /M 1 Measure 3 mls of stock HNO 3 and add to it enough water to reach a total volume of 500 mls Usually add acid to water but here it s dilute acid so it s OK. Conc of H 2 SO 4 = 18 M = 36 N H 2 SO 4 HCl = 12 M = 12 N HCl HNO 3 = 16M = 16 N NH 4 OH = 15 M = 15 N Note that NH 3 solutions are often labelled NH 4 OH (ammonium hydroxide). Why do you think so? Sample calculation: ppm Titration of Ca 2+ using the hexadentate ligand, EDTA involves a 1:1 titration. If 35.0 ml of Ca 2+ unknown sol n requires 24.0mLs of M EDTA, what is the concentration of Ca 2+ in unknown (as ppm CaCO 3 )? At e.p. #mol Ca 2+ =# mol EDTA; M ca =M EDTA V EDTA /V CaCO3 ppm CaCO 3 = M V EDTA EDTA (gcaco 3 ) (10 3 mg) V CaCO3 mol g ppm CaCO 3 = (0.014mol/L)(24.0mL) (35.0mL) 100.1g 10 3 mg = 960 ppm mol g 6

1.00 Measurements. Chemistry 251. Dr. Fred Omega Garces

1.00 Measurements. Chemistry 251. Dr. Fred Omega Garces 1.00 SI -Units Mass and Weights The Mole Millimole Solutions and Concentration Analytical Molarity Equilibrium Molarity Composition by Parts % Composition ppm, ppb Density and specific gravity Titration