Reactions in Aqueous Solutions

Transcription

1 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.

2 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

3 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 Fe H 2 O 3

4 Example 4.11 In a titration experiment, a student finds that ml of a NaOH solution are needed to neutralize g of KHP. What is the concentration (in molarity) of the NaOH solution?

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8 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 g of KHP?

9 Example 4.11 (2) Solution First we calculate the number of moles of KHP consumed in the titration: moles of KHP = g KHP 1 mol KHP g KHP = mol KHP Because 1 mol KHP 1 mol NaOH, there must be mol of NaOH in 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 = mol NaOH ml soln 1000 ml soln 1 L soln = mol NaOH 1 L soln = MM

10 Example 4.12 How many milliliters (ml) of a M NaOH solution are needed to neutralize 20.0 ml of a M H 2 SO 4 solution? H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4

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13 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 M H 2 SO 4 solution? How many moles of NaOH would this quantity of H 2 SO 4 neutralize?

14 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 = mol H 2SO ml soln 20.0 ml soln = 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 mole, or mole.

15 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 = mol NaOH mol L soln = L or 16.1 ml

16 Example 4.13 A mL volume of M KMnO 4 solution is needed to oxidize 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 Fe H 2 O End 9 am & 10 am class

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19 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 ml of M KMnO 4 solution?

20 Example 4.13 (2) Solution The number of moles of KMnO 4 in ml of the solution is moles of KMnO 4 = mol KMnO mL soln ml = 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 = mol KMnO 4 5 mol FeSO 4 1 mol KMnO 4 = mol FeSO 4

21 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 = mol FeSO ml soln 1000 ml soln 1 L soln = MM