1. Version 2012 Updated on 042512 Copyright © All rights reserved Dong-Sun Lee, Prof., Ph.D. Chemistry, Seoul Women’s University Chapter 11 Volumetric Titrimetry

2. Principles of volumetric analysis Volumetry : the volume of standard reagent needed to react with analyte is measured Titrimetry : increments of the titrant are added to the analyte until their reaction is complete. Titrant vs Analyte N V = N’V’ nMV = n’M’V’ N= known V= measure N’= unknown V’= known

3. Titration : A procedure for determining the amount of some unknown substance (the analyte ) by quantitative reaction with a measured volume of a solution of precisely known concentration (the titrant ). Titrant : The substance that quantitatively reacts with the analyte in a titration. The titrant is usually a standard solution added carefully to the analyte until the reaction is complete. The amount of analyte is calculated from the volume of titrant required for complete reaction. Indicator : A substance that undergoes an sharp, easily observable physical change when conditions in its solutions change. See, for example, acid-base indicator and redox indicator.

4. Equivalence point : point in a titration at which equivalent amounts of titrant is the exact amount necessary for stoichiometric reaction with the analyte. N= known V= measure N’= unknown V’= known

5. End point : the point in a titration when a physical change occurs that is associated with the condition of chemical equivalence. The experimental estimate of the equivalence point in a titration titration error : e T = V ep – V eq V ep = actual volume at end point, V eq = theoretical volume of equivalence point Blank titration Direct titration : titrant is added to the analyte until the reaction is complete. Back titration : Adding a known excess of reagent to the analyte, then, a second reagent is used to titrate the excess of the first reagent. AT1AT1 Excess T 1 T 2 Ex. PO 4 3– + 3Ag +  Ag 3 PO 4 (s) Ag + + SCN –  AgSCN (s)

6. Evolution of the buret Buret, buret stand, clamp, white porcelain base, wide-mouth Erlenmeyer flask

7. Normally, the buret is filled titrant solution to within 1 or 2 mL of the zero position at the top. The initial volume of the buret is read to the nearest 0.01 mL.

8. Standard solution : A solution of precisely known concentration. Primary standard : An ultra-pure (99.9% purity) compound that serves as the reference material for a titrimetric method of analysis. Secondary standard : A compound whose purity has been established by chemical analysis and that serves as the reference material for a titrimetric method of analysis Standardization : a process in which the concentration of a solution is determined by using the solution to titrate a known amount of another reagent. 

9. V’ ? N’ ? Beaker Steps in a titrimetry  Preparation of primary standard solution Calculation Normality N V Weighing 99.9% pure primary standard reagent xx.xxxx g Dissolve in a volumetric flask Titration unknown sample solution Preparation of unknown sample solution N’V’ Find V’ ml Find N’ normality Titration other sample solution 2ndary standard Buret Transfer pipet Erlenmeyer Flask V ml indicator

10. Calculation of molarity or normality for standard solution 1 mol / 1000 mL = 1 M 1 equivalent / 1000 mL = 1 N x mol / y mL = z M Ex: Na 2 CO 3 (105.99 g / 1 mol) / 1000 mL = 1 M = 2 N x g / 250 mL = 0.01 M = 0.02 N x = 0.264975 g ≈ 0.2650 g

11. Calculation of molarity or normality for standard solution Ex: MnO 4 – + 8 H + + 5 e = Mn 2+ + 4 H 2 O Permanganate manganous 2 KMnO 4 + 5 Na 2 (COO) 2 + 8 H 2 SO 4 = 2 MnSO 4 + K 2 SO 4 + 5 Na 2 SO 4 + 10 CO 2 + 8 H 2 O 2 mol ×158.03 g ≡ 5 mol × 134.01 g ≡ 10 equivalents / 1000 mL = 10 N 3.161 g/L ≡ 6.7005 g/L ≡ 0.1 N 0.1 N 1 mL ≡ 0.0067 g if KMnO 4 solution titration Ve = 42.31 mL ≡ Na 2 (COO) 2 0.2121 g [KMnO4] = {(0.2121 / 0.1341) mmol × ( 2/5) } / 43.31 mL = 0.01462 M

12. Ex Volumetric determination of calcium in urine Ca 2+ Titration reaction : Ca 2+ + (COO) 2 2– + H 2 O  Ca (COO) 2 H 2 O (solid) Ca (COO) 2 + H 2 SO 4  CaSO 4 + H 2 (COO) 2 5 H 2 (COO) 2 + 2 KMnO 4 +3H 2 SO 4  K 2 SO 4 +2MnSO 4 +10CO 2 + 8H 2 O purple colorless Standardization : 5 Na 2 (COO) 2  2 KMnO 4 5 × 134.00 g / L  2 ×158.004 g / L  10 N 0.3562 g / 250 mL = x N x = 0.021266 N KMnO 4 : 0.021266 N×10.00 mL = y N×48.36 mL y = 0.004397 N Titration of urine sample : urine 5.00 mL (COO) 2 2–  Ca 2+ 0.004397 N × 16.17 mL = z N ×5.00 mL z = 0.01421 N = 0.00711 M Oxalate Pot. permanganate 

13. Ex Volumetric determination of chloride in biological fluids Cl – ( in serum, cerebrospinal fluid, or urine) Titrate with mercuric ion e.p.: violet blue complex of Hg 2+ with Ind.( diphenylcarbazone) Titration reaction : 2Cl – + Hg 2+  HgCl 2 (aq) Standardiztion : 2 NaCl  2Cl –  Hg 2+ 2× 58.443 g/L  2×35.453 g/L  200.59 g/L = 2N 147.6 mg /25mL = x N  x = 0.10102 N NaCl vs Hg(NO 3 ) 2 0.10102 N× 25mL = y N ×28.06 mL  y = 0.090 N Titration of urine sample : Urine 2 mL  Hg(NO 3 ) 2 solution 22.83 mL z N ×2 ml = 0.090N×22.83 ml  z =1.02735 N Calculation : Cl – 35.453 mg/mL = 1N X mg/mL = 1.02735 N  X = 36.423 mg/mL

14. Ex. Kjeldahl nitrogen determination Digestion : N(in protein)  NH 4+ Distillation of NH 3 : NH 4+ + OH –  NH 3 (g) +H 2 O Collection of NH 3 in excess HCl : NH 3 + H +  NH 4+ Titration of unreacted HCl : H + + OH –  H 2 O Ex. 0.500 mL aliquot of protein 0.02140 M HCl 10.00 mL 0.0198 M NaOH 3.26 mL Protein concentration(mg/mL)? Solution: 0.0198 M×3.26 mL = 0.02140 M×V mL V= 3.016 mL 10.00 mL – 3.016 mL = 6.984 mL 0.02140 M×6.984 mL= x M ×0.5 mL x = 0.2989 M N = 14.0067 g/L = 1 M y mg/0.5 mL = 0.2989 M y = 2.093 mg If N : protein = 16.2 w/w%  protein =2.093 ×(100/16.2)=12.92 mg  protein = 12.92 mg/0.500 mL=25.84mg/mL 

15. (a)Kjeldahl digestion flask with long neck to minimize loss by spattering. (b)Six-port manifold for multiple samples provides for exhaust of fumes. Kjeldahl distillation unit employes electric immersion heater in flask at left to carry out distillation in 5 min. Beaker at right collects liberated ammonia in standard HCl.

17. Finding the End Point with a pH Electrode End point : Inflection point maximum slope End point : greatest dpH/dV End point : d 2 pH/dV 2 = 0 (a) Experimental points in the titration of 1.430mg of xylenol orange, a hexaprotic acid, dissolved in 1.000mL of aqueous 0.10 M NaNO 3. The titrant was 0.065 92M NaOH. (b) The first derivative, △ pH/ △ V, of the titration curve. ( C) The second derivative, △ ( △ pH/ △ V)/ △ V, which is the derivative of the curve in panel (b). Derivatives for the first end point are calculated in Table 12.3. End points are taken as maxima in the derivative curve and zero crossings of the second derivative. 

18. Calculation of the titration curve for 50.00 mL of 0.02000 M KOH treatrd with 0.1000 M HBr A1B1C1 First derivative D1E1 Second derivative F1 mL HBr added (Va) pHmL HBr added (va)mL HBr added (Va) 0.0012.300.50- 0.061.000.00 1.0012.241.50- 0.062.000.00 2.0012.182.50- 0.063.00- 0.02 3.0012.123.50- 0.084.00- 0.01 4.0012.044.50- 0.095.00- 0.01 5.0011.955.50- 0.106.00- 0.03 6.0011.856.50- 0.137.00- 0.06 7.0011.727.50- 0.198.00- 0.12 8.0011.538.50- 0.318.88- 0.39 9.0011.229.25- 0.609.48- 2.56 9.5010.929.70- 1.759.82- 38.21 9.9010.229.95- 11.119.97- 4217.78 9.999.2210.00- 222.0010.000.00 10.007.0010.01- 222.0010.034217.78 10.014.7810.06- 11.1110.1838.21 10.103.7810.30- 1.7510.532.60 10.503.0810.75- 0.5811.130.37 11.002.7911.50- 0.3012.000.13 12.002.4912.50- 0.1713.000.05 13.002.3213.50- 0.12 14.002.20 =SLOPE(B3:B4,C3:C4 ) =SLOPE(D3:D4, C3:C4)

21. Experiments. Standardization of 0.1000 M NaOH

22. Primary standard KHP 204.22 g/1000 mL= 1.0000 M 0.51055 g/25 mL = x M x = 0.1000 M End point = 5.00 mL 0.1000 N×5.00 mL= x N×5.00 mL X = 0.1000 N = 0.1000 M Titration of 5.00 mL of 0.1 M KHP standard with NaOH Experiments. Standardization of 0.1000 M NaOH

23. Summary Volumetric analysis, titration, titrant, analyte, indicator equivalence point, end point blank titration, direct titration, back titration standardiztion, primary standard, calculation in volumetric analysis Kjeldahl nitrogen determination Titration curve