Reactions in Aqueous Solutions Chapter 4 Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Titrations In a titration, a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point the point at which the reaction is complete Indicator substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 2
Titrations (1) Titrations can be used in the analysis of: Acid-base reactions H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 Redox reactions 5Fe 2+ + MnO 4 + 8H + Mn 2+ + 5Fe 3+ + 4H 2 O 3
Example 4.11 In a titration experiment, a student finds that 23.48 ml of a NaOH solution are needed to neutralize 0.5468 g of KHP. What is the concentration (in molarity) of the NaOH solution?
5
6
7
Example 4.11 (1) Strategy We want to determine the molarity of the NaOH solution. What is the definition of molarity? molarity of NaOH = mol NaOH L soln need to find want to calculate given The volume of NaOH solution is given in the problem. Therefore, we need to find the number of moles of NaOH to solve for molarity. From the preceding equation for the reaction between KHP and NaOH shown in the text we see that 1 mole of KHP neutralizes 1 mole of NaOH. How many moles of KHP are contained in 0.5468 g of KHP?
Example 4.11 (2) Solution First we calculate the number of moles of KHP consumed in the titration: moles of KHP = 0.5468 g KHP 1 mol KHP 204.2 g KHP = 2.678 10 3 mol KHP Because 1 mol KHP 1 mol NaOH, there must be 2.678 10 3 mol of NaOH in 23.48 ml of NaOH solution. Finally, we calculate the number of moles of NaOH in 1 L of the solution or the molarity as follows: molarity of NaOH soln = 2.678 10 3 mol NaOH 23.48 ml soln 1000 ml soln 1 L soln = 0.1141 mol NaOH 1 L soln = 0.1141 MM
Example 4.12 How many milliliters (ml) of a 0.610 M NaOH solution are needed to neutralize 20.0 ml of a 0.245 M H 2 SO 4 solution? H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4
11
12
Example 4.12 (1) Strategy We want to calculate the volume of the NaOH solution. From the definition of molarity [see Equation (4.1)], we write need to mol NaOH find L soln = want to calculate molarity given From the equation for the neutralization reaction just shown, we see that 1 mole of H 2 SO 4 neutralizes 2 moles of NaOH. How many moles of H 2 SO 4 are contained in 20.0 ml of a 0.245 M H 2 SO 4 solution? How many moles of NaOH would this quantity of H 2 SO 4 neutralize?
Example 4.12 (2) Solution First we calculate the number of moles of H 2 SO 4 in a 20.0 ml solution: moles of H 2 SO 4 = 0.245 mol H 2SO 4 1000 ml soln 20.0 ml soln = 4.90 10 3 mol H 2 SO 4 From the stoichiometry we see that 1 mol H 2 SO 4 2 mol NaOH. Therefore, the number of moles of NaOH reacted must be 2 4.90 10 3 mole, or 9.80 10 3 mole.
Example 4.12 (3) From the definition of molarity [see Equation (4.1)], we have liters of soln = moles of solute molarity or volume of NaOH = 9.80 10 3 mol NaOH 0.610 mol L soln = 0.0161 L or 16.1 ml
Example 4.13 A 16.42-mL volume of 0.1327 M KMnO 4 solution is needed to oxidize 25.00 ml of a FeSO 4 solution in an acidic medium. What is the concentration of the FeSO 4 solution in molarity? The net ionic equation is 5Fe 2+ + MnO 4 + 8H + Mn 2+ + 5Fe 3+ + 4H 2 O End 9 am & 10 am class
17
18
Example 4.13 (1) Strategy We want to calculate the molarity of the FeSO 4 solution. From the definition of molarity need to find want to calculate molarity of FeSO 4 = mol FeSO 4 L soln given The volume of the FeSO 4 solution is given in the problem. Therefore, we need to find the number of moles of FeSO 4 to solve for the molarity. From the net ionic equation, what is the stoichiometric equivalence between Fe 2+ and MnO 4? How many moles of KMnO 4 are contained in 16.42 ml of 0.1327 M KMnO 4 solution?
Example 4.13 (2) Solution The number of moles of KMnO 4 in 16.42 ml of the solution is moles of KMnO 4 = 0.1327 mol KMnO 4 1000mL soln 16.42 ml = 2.179 10 3 mol KMnO 4 From the net ionic equation we see that 5 mol Fe 2+ 1 mol MnO 4 Therefore, the number of moles of FeSO 4 oxidized is moles of FeSO 4 = 2.179 10 3 mol KMnO 4 5 mol FeSO 4 1 mol KMnO 4 = 1.090 10 2 mol FeSO 4
Example 4.13 (3) The concentration of the FeSO 4 solution in moles of FeSO 4 per liter of solution is molarity of FeSO 4 = mol FeSO 4 L soln = 1.090 10 2 mol FeSO 4 25.00 ml soln 1000 ml soln 1 L soln = 0.4360 MM