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VOLUMETRIC ANALYSIS

VOLUMETRIC ANALYSIS Volumetric analysis is an analysis in which the amount of the unknown is calculated from a known volume of added solution. TITRATIONS In a titration, the titrant (solution with a known concentration of a reagent) area added to analyte until their reaction is complete.

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VOLUMETRIC ANALYSIS

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  1. VOLUMETRIC ANALYSIS Volumetric analysis is an analysis in which the amount of the unknown is calculated from a known volume of added solution.

  2. TITRATIONS In a titration, the titrant (solution with a known concentration of a reagent) area added to analyte until their reaction is complete. (a) In a titration, the titrant must react with the analyte to completion, with a reproducible stoichiometry, with an adequate reaction rate, and it must be stable over the time required for the analysis. (b) The volume of the titrant and the weight or volume of the sample must be measured to the desired precision and accuracy. (c) Pretreatment of the sample must be carried out to remove interference if the titration reaction is not specific to eliminate matrix effects. (d) There must be a technique to determine the equivalence point. The equivalent point is a point in the titration when the mole of the titrant added is equivalent to the moles of analyte.

  3. In titration methods, the analytical results depend on knowing the amount of one of the reactants used. Direct titration: titrant is added to the analyte until the reaction is complete. Back titration Indirect titration

  4. titration can be divided into for types. • Acid-base or neutralization • Precipitation • Complex formation • Oxidation-reduction (redox)

  5. ACID-BASE TITRATIONS • Titration of a strong acid and strong base • Titration of a weak acid and strong base • Titration of a weak base and strong acid • Titration of a weak acid and weak base

  6. ACID-BASE INDICATORS In an acid-base titration, addition of titrant near the equivalence point causes the solution pH to change drastically. This pH change is detectable with indicators that change color as a function of pH. Indicators are weak acids that change color when they gain or lose their acidic proton(s).

  7. COMMON INDICATORS The point at which the indicator changes color called the end point.

  8. HOW DOES AN INDICATOR WORK? • Weak acids are titrated in the presence of indicators which change under slightly alkaline/base conditions. • Weak bases should be titrated in the presence of indicators which change under slightly acidic conditions. • HIn(aq) + H2O(l) In- (aq) + H3O+(aq) • Acid color A Base color B • The acid and its conjugate base have different colors. • At low pH: [H3O+] is high and so the equilibrium position lies to the left. The equilibrium solution has the color A. • At high pH: [H3O+] is low and so the equilibrium position thus lies to the right and the equilibrium solution has color B.

  9. PROPERTIES OF AQUEOUS ACID-BASE INDICATORS AT 25C

  10. ACID-BASE TITRATIONS CURVES The titration curve for an acid-base titration is a plot of the solution pH, normally on the vertical axis, against the volume of titrant added. TITRATION OF A STRONG ACID AND STRONG BASE We can identify three different regions in this titration experiment: Before the Equivalence Point: The pH is determined by the concentration of unneutralized strong acid. At the Equivalence Point: The pH, 7, is determined by the dissociation of water. After the Equivalence Point: The pH is determined by the concentration of excess strong base that we are adding.

  11. FOUR DISTINCT REGIONS • i) Solution of only strong acid (solution of H3O+) • ii) Excess moles of strong acid + limiting moles of strong base (solution of H3O+) • iii) Equivalence Point with equal moles of strong acid + strong base (solution of H2O) • iv) Excess moles of strong base + limiting moles of strong acid (solution of OH-) iv iii i & ii

  12. Example: Sample: Receiving flask containing 50.00 mL of a 0.100 M HBr(aq) solution. Titrant: Stepwise addition of a 0.200 M KOH(aq) solution from a buret using the following volumes:

  13. a)Addition of 0.00 mL of Strong Base (pH of initial solution) • The solution contains only 0.100 M strong acid. • Strong acids completely ionize in aqueous solution. • HBr(aq) + H2O(l)  H3O+(aq) + Br(aq) • Therefore, [H3O+] = 0.100 M • pH = - log [0.100] • pH = 1.000

  14. b) Addition of 10.00 mL of Strong Base (Excess of strong acid present) First solve for the final concentrations at the end of the strong acid/strong base reaction: HBr(aq) + KOH(aq) H2O(l) + KBr(aq) Initial mmol of H3O+ = 50.00 mL x 0.1 mmol mL-1 = 5. 00 mmol NaOH added = 10.00 mL x 0.20 mmol mL-1 = 2.00 mmol HBr consumed = 2.00 mmol HBr left = 3.00 Total Volume = (50.00 + 10.00) mL = 60 mL Therefore, [H3O+]final = 3.00 mmol / 60mL = 0.0500 M pH = - log 0.0500 pH = 1.30

  15. c) Addition of 25.00 ml of Strong Base (the equivalence point) First solve for the final concentrations at the end of the strong acid/strong base reaction: H3O+(aq) + OH-(aq)  2 H2O(l) Initial mmol of H3O+ = 50.00 mL x 0.1 mmol mL-1 = 5. 00 mmol NaOH added = 25.00 mL x 0.20 mmol mL-1 = 5.00 mmol H3O+ consumed = 5.00 mmol HBr left = 0.00 Total Volume = (50.00 + 25.00) mL = 75.00 mL At equivalent point, [H3O+] = [ OH-] Kw = [H3O+][ OH-] = 1  10-14, pada 25 oC [H3O+] = Kw = 1 x 10-14 = 1.0  10-7 M pH = - log 1.0  10-7 = 7.00

  16. d) Addition of 30.00 mL of Strong Base (excess strong base added) First solve for the final concentrations at the end of the strong acid/strong base reaction: H3O+(aq) + OH-(aq)  2 H2O(l) Initial mmol of H3O+ = 50.00 mL x 0.1 mmol mL-1 = 5. 00 mmol NaOH added = 30.00 mL x 0.20 mmol mL-1 = 6.00 mmol H3O+ consumed = 5.00 (none left) mmol NaOH left = 1.00 Total Volume = (50.00 + 30.00) mL = 80.00 mL Therefore, [OH-]final = 1.00 mmoles / 80.00 mL = 1.25 x 10-2 M pOH = -log 1.25 x 10-2 = 1.90 pH = 14 – 1.90 = 12.10.

  17. - - - - - - - Buret clamp Basic solution in buret (titrant) Acidic solution in Erlenmeyer flask Magnetic stirring bar Magnetic stirrer

  18. TITRATION OF A WEAK ACID WITH A STRONG BASE • The pH of the Starting SolutionAt the start of the titration the solution contains only the weak acid. The pH is calculated from the concentration and its Ka. • Before the Equivalence Point. After titrant been added, the solution consists of buffer. The pH is determined by the Ka, [conjugate base] and [acid]. • At the Equivalence PointThe solution contains only the conjugate base (salt) and the pH is calculated based on the hydrolysis of the salt. • After the Equivalence PointThe excess of strong base titrant represses the basic character that the pH is governed largely by the concentration of the excess titrant.

  19. Example: Let us calculate the pH values along the course of a particular titration, the titration of exactly 20 mL of 0.100 M acetic acid with 0.100 M sodium hydroxide. The stoichiometric reaction, CH3COOH + Na+ + OH-  CH3COO- + Na+ + H2O.

  20. Ka = = 1.76 x 10-5 = • The pH of the Starting SolutionAt the start of the titration the solution contains only the weak acid CH3COOH, for which • CH3COOH + H2O H3O+ + CH3COO- • (0.10 – x) x x • Assume x << 0.10 M, so (0.100 – x)  0.100 M, [H3O+]2 = 1.76 x 10-5 x 0.100 [H3O+] = 1.32 x 10-3 pH = - log 1.32 x 10-3 = 2.88

  21. Before the Equivalence Point (Addition of 10.00 mL NaOH) • Certain portion of the acetic acid has been titrated by the NaOH solution and producing CH3COONa. The solution in the flask become a buffer solution. CH3COOH + NaOH  CH3COONa + H2O • Initial mmol of CH3COOH = 20.00 mL x 0.1 M • = 2. 00 • mmol NaOH added = 10.00 mL x 0.100 M • = 1.00 • mmol CH3COOH consumed = 1.00 • mmol CH3COOH left = 1.00 • mmol CH3COONa produced = 1.00 • Total Volume = (20.00 + 10.00) mL = 30.00 mL

  22. At the Equivalence Point (Addition of 20.00 mL NaOH) • Initial mmol of CH3COOH = 20.00 mL x 0.1 M • = 2. 00 • mmol NaOH added = 20.00 mL x 0.100 M • = 2.00 • mmol CH3COOH consumed = 2.00 • mmol CH3COOH left = 0.00 • mmol NaOH left = 0.00 • At the equivalence point, all the acetic acid solution and NaOH solution react completely. Only a solution of the weak base CH3COONa, sodium acetate is present. • CH3COOH + NaOH  CH3COONa + H2O • mmol of CH3COONa produced = 2. 00 • Total Volume = (20.00 + 20.00) mL = 40.00 mL

  23. Since, only CH3COONa present in the solution, CH3COONa  CH3COO- + Na+ The weak base CH3COO- which hydrolyzes, giving CH3COOH and OH- in a stoichiometric 1:1 ratio, following the reaction equilibrium CH3COO- + H2O CH3COOH + OH-

  24. 4. Beyond/after the Equivalence Point (Addition of 30.00 mL NaOH) Initial mmol of CH3COOH = 20.00 mL x 0.1 M = 2. 00 mmol NaOH added = 30.00 mL x 0.100 M = 3.00 mmol CH3COOH consumed = 2.00 mmol CH3COOH left = 0.00 mmol NaOH left = 1.00 The resulting solution is then a mixture of the strong base NaOH and the weak base CH3COONa. The strong base controls the pH,

  25. COMPARISON OF TITRATION CURVE FOR STRONG ACID-STRONG BASE AND WEAK ACID-STRONG BASE TITRATION CURVE A WEAK ACID-STRONG BASE

  26. Example. A total of 25 mL of ammonium hydroxide, NH4OH, solution which is approximately 0.2000 M is to be titrated with 0.2000 M HCl. (Kb NH4OH = 3.39 x 10–5).

  27. Anda diberi 50 mL larutan HCl, pH larutan tersebut ialah 1.0. Berapa isipadu NaOH 0.25 M diperlukan untuk merubah pH larutan kepada 7.0. Anda diberi 50 mL larutan HCl, pH larutan tersebut adalah 1.0. Berapa isipadu NaOH 0.20 M yang telah ditambah jika pH larutan ialah 6.50. Anda diberi 50 mL larutan HCl, pH larutan tersebut adalah 1.0. Berapa isipadu NaOH 0.25 M yang telah ditambah jika pH larutan adalah 11.15. (Jwp: 20.4 mL)… buktikan

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