Presentation on theme: "Titrations. 13.1 Titrations A. Titrations – is an experimental procedure in which a standard solution is used to determine the concentration of an unknown."— Presentation transcript:
13.1 Titrations A. Titrations – is an experimental procedure in which a standard solution is used to determine the concentration of an unknown solution. 1. Standard Solution – is one of known concentration. The titration involves the gradual addition of one solution to another until the solute in the first solution has completely reacted with the solute in the second solution. a) This point is called the equivalence point (same as stoichiometry point). b) The equivalence point is detected using an indicator. c) The point at which the indicator changes color is called the end point of the titration. 2. The most common titrations involve the reaction of an acid solution with a basic solution. 3. The reaction of the acid and base is termed a neutralization.
B. The three types of titration reactions. 1. An acid with a base to give a soluble salt and water 2. A soluble salt with a second soluble salt to give a precipitate.. 3. An oxidizing material with a reducing material. C. The reaction between a strong acid (e.g., HCl or HNO 3 ) and a strong base (e.g., NaOH) gives salts (e.g., NaCl or NaNO 3 ) Since these salts are products of strong acids and strong bases, the resulting solution is neutral. D. For a strong acid / strong base titration, the pH at the equivalence point is 7; but only a small amounts of reagent causes a major pH change. E. The titration curve for the neutralization reaction is shown in Figure 25.1a. 1. Curves can be produced using a pH meter connected to chart recorder. 2. The indicator selected should change color in pH range from about 4 to 10. 3. Phenolphthalein is usually used since it is easy to detect visually a slight pink color from a colorless liquid.
F. Reaction between a strong acid (HCl, HNO 3, or H 2 SO 4 ) and a weak base (NH 4 ) also produces salts (NH 4 Cl, NH 4 NO 3, or (NH 4 ) 2 SO 4 ). 1. These salts hydrolize to form slightly acidic solutions. 2. The titration curve for this reaction is shown Figure 24.1b. 3. Methyl orange can be used as a indicator because of the low pH region in which it changes color. G. The reaction between a weak acid (CH 3 COOH) and a strong base (NaOH) gives a salt (NaCH 3 COO). H. The concentration of the acid and basic solution will change the position of the curve only slightly (especially at the start and completion of the titration), in relation to pH. I. Sample Problem #1 If 20cm 3 of a 0.300M solution of NaOH is required to neutralize completely 30cm 3 of sulfuric acid solution, what is the molarity of the H 2 SO 4 solution? Solve solution on next page
Step A: First write a balanced equation. 2NaOH + H 2 SO 4 Na 2 SO 4 + 2H 2 O Step B: Since the concentration of the base is given, determine the moles of NaOH. Number of moles = 20 cm 3 0.300 mol NaOH 1 dm 3 = 1000 cm 3 = 0.00600 mol NaOH Step C: From the coefficient of the balanced equation, 2 moles of base are required for reaction with 1 mole of acid. Number of moles = 0.00600 mol NaOH 1 mol H 2 SO 4 = 2 mol NaOH = 0.00300 mol H 2 SO 4
Step D: Determine the molarity of the acid. Molarity = 0.00300 mol H 2 SO4 1000 cm 3 = 30 cm 3 1dm 3 = 0.100 mol H 2 SO 4 / dm 3 J. Sample Problem #2 What volume of 0.500M HNO 3 is required to neutralize 25.0 cm 3 of a 0.200M solution? Solving Process: Step A: First write a balanced equation HNO 3 + NaOH NaNO 3 + H 2 O
Step B: Since the concentration of the base is given, determine the moles of NaOH. mol NaOH = 25cm3 0.200 mol NaOH 1dm 3 = 0.00500 1.00 dm 3 solution 1000cm 3 mol NaOH Step C: From the coefficients of the balanced equation, 1 mole of acid will react completely with 1 mole of base. 0.00500 mol NaOH = 0.00500 mol HNO3 Step D: Therefore volume = 0.00500 mol HNO 3 1.00 dm 3 solution 1000 cm 3 0.500 mol HNO 3 1dm 3 = 10.0 cm 3 solution
13.2 Titrations with Normal Solutions A. Another quantitative expression for the concentration of solutions is normality. A 1N solution contains one equivalent mass of solute per dm 3 of solution. Normality = number of equivalents of solute dm 3 of solution 1. An equivalent is that quantity of a substance that provides 1-mole of charge. Consider these three substances. NaNO 3 Na + NO 3 Na 2 SO 4 2Na + SO 4 Na 3 PO 4 3Na + PO 4 2. NaNO 3 contains 1 equivalent per mole, Na 2 SO 4 contains 2 equivalent mole. The equivalent masses of these substances. Examples on Next Page
NaNO 3 = 85.0g 1 mol = 85.0 g / eq 1mol 1 eq Na 2 SO 4 = 142.1g 1 mol = 71.05 g / eq 1mol 2 eq 3. Titration problems can also be solved using the concentration expression normally used and substituting into a mathematical equation. 4. Consider a reaction between 1.00N HCl solution and a 1.00N NaOH solution. a) It is necessary to add equal volumes of the acid and base to have a complete reaction. b) An equal number of H 3 O + ions and OH - ions react to produce water. c) The other product, NaCl, is a salt of a strong acid and a strong base, which produces a neutral solution.
5. The equivalent mass of a substance is that mass of material that will produce 1-mole of charge. One equivalent mass of hydrogen ions, will react with one equivalent mass of hydroxide ion. a) Therefore, a equal number of equivalents of acid and base will react completely to form salt. b) normality acid x volume acid will equal normality base x volume base N a x V a = N b x V b B. If any three values are known, the fourth value can be calculated. 1. For the factors to be equivalent, both volume terms must be in the same unit. C. Acids and bases with one reacting hydrogen ion or hydroxide ion per formula unit contain 1 equivalent per mole D. Acids and bases with two reacting hydrogen ions or hydroxide ions contain 2 equivalents per mole.
one reacting ion HCl 2.00M HCl = 2.00N HCl (1eq = 1mol) NaOH 0.300M NaOH= 0.300N NaOH two reacting ions H2SO4 0.750M H2SO4 = 1.50N H2SO4 (2eq = 1mol) Example Problem #1 If 20.0cm 3 of a 3.00N solution of NaOH is required to neutralize 30.0 cm 3 of a sulfuric acid solution. What is the normality of the H 2 SO 4 ? Solving Process: equivalents of acid = equivalents of base or N a x V a = N b x V b change cm 3 to dm 3 and then substitute: Na = 3.00N 0.0200 dm 3 = 2.00N 0.0300 dm 3
F. Sample Problem #2 How many grams of KOH are required to neutralize completely 2.00.0 cm 3 of a 4.00N solution of HNO 3 ? Solving Process: For neutralization to be complete, the number of equivalents of acid must be equal to the number of equivalents of base. First, calculate the number of equivalents of acid. Then convert this answer to grams of KOH. Equivalents HNO 3 = 4.00 eq HNO 3 0.200 cm 3 = 0.800 eq 1 dm 3 equivalents acid = equivalents base mass KOH = 0.800 eq KOH 1mol KOH 56.1g KOH = 44g 1 eq KOH 1 mol KOH