*11Reactions in Aqueous Solutions II: Calculations

*Chapter GoalsAqueous Acid-Base Reactions Calculations Involving Molarity Titrations Calculations for Acid-Base Titrations Oxidation-Reduction Reactions Balancing Redox Equations Adding in H+, OH- , or H2O to Balance Oxygen or Hydrogen Calculations for Redox Titrations

*Concentrations of SolutionsSecond common unit of concentration:Molarity is defined as the number of moles of solute per literof solution. Molarity =M x L = moles solute and M x mL = mmol solute

Calculations Involving MolarityThe reaction ratio is the relative number of moles of reactants and products shown in the balanced equation. H2SO4 + 2NaOH Na2SO4 + 2H2O 1mol 2mol 1mol 2molReaction ratio 1mol H2SO4 2mol NaOH 2mol NaOH 1mol H2SO4*or

Example 11-1: Acid-Base ReactionsIf 100.0 mL of 1.00 M NaOH and 100.0 mL of 0.500 M H2SO4 solutions are mixed, what will the concentration of the resulting solution be?What is the balanced reaction?It is very important that we always use a balanced chemical reaction when doing stoichiometric calculations. 2NaOH+ H2SO4 Na2SO4 + 2H2O M=mole/liter mole= M x liter

NaOH =1.00M x 0.1L =0.1 mole H2SO4 =0.500M x 0.1L =0.05 mole

*Calculations Involving Molarity

2NaOH+ H2SO4 Na2SO4 + 2H2O

What is the total volume of solution?100.0 mL + 100.0 mL = 200.0 mLWhat is the sodium sulfate amount, in mol?0.05 mol What is the molarity of the solution?M = 0.05 mol/0.2 L = 0.250 M Na2SO4

*Calculations Involving Molarity

Reaction Ratio2 mol1 mol1 mol2 molBefore Reaction0.1 mol0.05 molAfter Reaction0 mol0 mol0.05 mol0.1 mol

*Calculations Involving MolarityExample 11-2: Acid-Base ReactionsIf 130.0 mL of 1.00 M KOH and 100.0 mL of 0.500 M H2SO4 solutions are mixed, what will be the concentration of KOH and K2SO4 in the resulting solution? What is the balanced reaction?2 KOH+ H2SO4 K2SO4 + 2H2O M=mole/liter mole= M x liter

KOH =1.00M x 0.130L =0.130 mole H2SO4 =0.500M x 0.1L =0.05 mole

2KOH+ H2SO4 K2SO4 + 2H2O

What is the total volume of solution?130.0 mL + 100.0 mL = 230.0 mLWhat are the potassium hydroxide and potassium sulfate amounts?0.03 mol & 0.05 mol What is the molarity of the solution?M = 0.03 mol/0.230 L = 0.130 M KOHM = 0.05 mol/0.230 L = 0.217 M K2SO4

*

Reaction Ratio2 mol1mol1 mol2 molBefore Reaction0.130 mol0.05 molAfter Reaction0.030 mol0 mol0.05 mol0.1 mol

*Example 11-3: Acid-Base ReactionsWhat volume of 0.750 M NaOH solution would be required to completely neutralize 100 mL of 0.250 M H3PO4?

3 NaOH+ H3PO4 Na3PO4 + 3H2O M=mol/liter mol= M x liter liter = mol/M? mol H3PO4 =0.250M x 0.1L =0.025 molecomplete neutralize need 0.025x3=0.075 mol NaOH?L NaOH = 0.075 mol / 0.750M =0.10 L

*TitrationsAcid-base Titration TerminologyTitration A method of determining the concentration of one solution by reacting it with a solution of known concentration. Primary standard A chemical compound which can be used to accurately determine the concentration of another solution. Examples include KHP and sodium carbonate .Standard solution A solution whose concentration has been determined using a primary standard.Standardization The process in which the concentration of a solution is determined by accurately measuring the volume of the solution required to react with a known amount of a primary standard. (standarddization)

*TitrationsAcid-base Titration TerminologyIndicator A substance that exists in different forms with different colors depending on the concentration of the H+ in solution. Examples are phenolphthalein and bromothymol blue.Equivalence point The point at which stoichiometrically equivalent amounts of the acid and base have reacted. End point The point at which the indicator changes color and the titration is stopped.bromothymol blue: Colorless in acidic solution, reddish violet in basic solution

*TitrationsAcid-base Titration Terminology

*Calculations for Acid-Base TitrationsThe properties of an ideal primary standard include the following:It must not react with or absorb the components of the atmosphere, such as water vapor, oxygen, and carbon dioxide. It must react according to one invariable reactionIt must have a high percentage of purity ()It should have a high formula weight to minimize the effect of error in weighing ()It must be soluble in the solvent of interest It should be nontoxic ()It should be readily available (inexpensive )It should be environmentally friendly

*Calculations for Acid-Base TitrationsPotassium hydrogen phthalateis a very good primary standard.It is often given the acronym, KHP.KHP has a molar mass of 204.2 g/mol.

(potassium hydrogen phthalateKHP) () () ()KHP (monoprotic acid)11 (1)HOOCC6H4COOK(aq) + NaOH(aq) C6H4(COO-)2(aq)+ K+ (aq) + Na+(aq) + H2O(l)*

*Example 11-4: Calculate the molarity of a NaOH solution if 27.3 mL of it reacts with 0.4084 g of KHP.NaOH + KHP NaKP +H2O?mole NaOH = 0.4084 g KHP xx= 0.002 mol NaOH?M NaOH == 0.0733 M NaOH

*Example 11-5: Calculate the molarity of a sulfuric acid solution if 23.2 mL of it reacts with 0.212 g of Na2CO3.Na2CO3 + H2SO4 Na2SO4 + CO2 + H2O?mole H2SO4 = 0.212 g Na2CO3 xx= 0.002 mol H2SO4?M H2SO4 == 0.0862 M H2SO4

*Example 11-6: An impure sample of potassium hydrogen phthalate, KHP, had a mass of 0.884 g. It was dissolved in water and titrated with 31.5 mL of 0.100 M NaOH solution. Calculate the percent purity of the KHP sample. Molar mass of KHP is 204.2 g/mol.NaOH + KHP NaKP + H2O?mole NaOH = 31.5x10-3L solution x= 0.00315 mol NaOH?g KHP = 0.00315mol NaOH xx= 0.643g KHP% KHP =x 100% == 72.7%x 100%

*Example 11-5: Standardization of an Acid SolutionCalculate the molarity of a solution of H2SO4 if 40.0ml of the solution neutralizes 0.364g of Na2CO3? Na2CO3 + H2SO4 Na2SO4 + CO2 + H2O ?mole H2SO4= 0.364 g Na2CO3 xx= 0.00343 mol H2SO4?M H2SO4= = 0.0858 M H2SO4Exercise 26

*Example 11-6: Standardization of a Base SolutionA 20.00-ml sample of a solution of NaOH reacts with 0.3641g of KHP. Calculate the molarity of the NaOH solution. ?mole NaOH = 0.3641 g KHP xx= 0.001783 mol NaOH?M NaOH= = 0.08915 M NaOHExercise 28NaOH + KHP NaKP +H2O

*Example 11-7: TitrationThe titration of 36.7ml of a hydrochloric acid solution requires 43.2 ml of standard 0.236M sodium hydroxide solution for complete neutralization. What is the molarity of the HCl solution? ?mmole HCl = 43.2ml NaOH x= 10.2 mmol HCl?M HCl= = 0.278 M HClHCl + NaOH NaCl +H2Ox

Example 11-8: TitrationThe titration of 36.7ml of a sulfuric acid solution requires 43.2 ml of standard 0.236M sodium hydroxide solution for complete neutralization. What is the molarity of the H2SO4 solution? ?mmole H2SO4 = 43.2ml NaOH x= 5.1 mmol H2SO4?M H2SO4 = = 0.139 M H2SO4H2SO4 +2 NaOH Na2SO4 +2 H2OxExercise 38*

*Example 11-9: Dtermination of Percent AcidOxalic acid, (COOH)2, is used to remove rust stains and some ink stains from fabrics. A 0.1743-g sample of impure oxalic acid required 39.82 ml of 0.08915 M NaOH solution for complete neutralization. No acidic impurities were present. Calculate the percentage purity of the (COOH)2. ?mole (COOH)2 = 0.03982L NaOH x= 0.001775 mol (COOH)2?g (COOH)2 = 0.001775 mole x = 0.1598g (COOH)2(COOH)2 +2 NaOH Na2(COO)2 + 2H2OxExercise 38% purity == 91.68% x 100%

*Oxidation-Reduction ReactionsWe have previously gone over the basic concepts of oxidation & reduction in Chapter 4.Rules for assigning oxidation numbers were also introduced in Chapter 4.Refresh your memory as necessary.We shall learn to balance redox reactions using the half-reaction method.

Oxidation-Reduction Reactions: An IntroductionOxidation is an increase in the oxidation number. Corresponds to the loss of electrons ().Reduction is a decrease in the oxidation number. Good mnemonic reduction reduces the oxidation number.Corresponds to the gain of electrons ()*

*Balancing Redox ReactionsHalf reaction method rules:Write the unbalanced reaction.Break the reaction into 2 half reactions:One oxidation half-reaction andOne reduction half-reactionEach reaction must have complete formulas for molecules and ions.Mass balance each half reaction by adding appropriate stoichiometric coefficients. To balance H and O we can add:H+ or H2O in acidic solutions.OH- or H2O in basic solutions.

*Balancing Redox ReactionsCharge balance the half reactions by adding appropriate numbers of electrons.Electrons will be products in the oxidation half-reaction.Electrons will be reactants in the reduction half-reaction.Multiply each half reaction by a number to make the number of electrons in the oxidation half-reaction equal to the number of electrons reduction half-reaction.Add the two half reactions.Eliminate any common terms and reduce coefficients to smallest whole numbers.

*Example 11-12 Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation.Starting ReactionSn2+ +Br2 Sn4+ + Br-Mass balance the half-reactionSn2+ Sn4+ Charge balance the half reactionSn2+ Sn4+ + 2e-Electrons are products thus thisis the oxidation half-reaction Mass balance the half-reactionBr2 2Br- Charge balance the half reactionBr2 + 2e- 2Br- This is the reduction half-reaction

Add the two half-reactionSn2+ Sn4+ + 2e-Br2 + 2e- 2Br-Sn2+ +Br2 Sn4+ + 2Br-

*Example 11-13 Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction.Starting ReactionCr2O72- + Fe2+ Cr3+ + Fe3+Mass balance the half-reactionFe2+ Fe3+ Charge balance the half reactionFe2+ Fe3+ + e-This is the oxidation half-reaction Mass balance the half-reactionCr2O72- 2Cr3+Cr2O72- 2Cr3+ + 7H2O14H+ + Cr2O72- 2Cr3+ + 7H2OCharge balance the half reaction14H+ + Cr2O72-+ 6e- 2Cr3+ + 7H2OThis is the reduction half-reaction Add the two half-reaction 6(Fe2+ Fe3+ + e-)14H+ + Cr2O72- + 6e- 2Cr3+ + 7H2O6Fe2+ 14H+ + Cr2O72- 6Fe3+ + 2Cr3+ + 7H2O+6

*Example 11-14 In basic solution hydrogen peroxide oxidizes chromite ions, Cr(OH)4-, to chromate ions, CrO42-. The hydrogen peroxide is reduced to hydroxide ions. Write and balance the net ionic equation for this reaction.Basic solutionCr(OH)4- + H2O2 CrO42-Mass balance the half-reactionCr(OH)4- + OH- CrO42-Cr(OH)4- + OH- CrO42- + H2O Cr(OH)4- + 4OH- CrO42- + 4H2O Charge balance the half reactionCr(OH)4- + 4OH- CrO42- + 4H2O +3e-oxidation half-reaction Mass balance the half-reactionH2O2 2OH-Charge balance the half reactionH2O2 + 2e- 2OH- reduction half-reaction Add the two half-reaction2 (Cr(OH)4- + 4OH- CrO42- + 4H2O +3e-)3 (H2O2 + 2e- 2OH-)

2Cr(OH)4- +8OH- + 3H2O2 2CrO42- + 8H2O+ 6OH-

2Cr(OH)4- +2OH- + 3H2O2 2CrO42- + 8H2O+3+6

*Example 11-15 When chlorine is bubbled into basic solution, it forms hypochlorite ions and chloride ions. Write and balance the net ionic equation.This is a disproportionation redox reaction. The same species, in this case Cl2, is both reduced and oxidized.Basic solutionCl2 ClO- +Cl-Mass balance the half-reactionCl2 + 4OH- 2ClO- +2H2OCharge balance the half reactionCl2 + 4OH- 2ClO- +2H2O+2e-oxidation half-reaction Mass balance the half-reactionCl2 2Cl-Charge balance the half reactionCl2 + 2e- 2Cl- reduction half-reaction Disproportionation Reaction is a redox reaction in which the same element is oxidized and reduced.+0+1 Add the two half-reaction Cl2 + 4OH- 2ClO- + 2H2O +2e- Cl2 + 2e- 2Cl-

2Cl2 + 4OH- 2ClO- + 2Cl-+ 2H2O Cl2 + 2OH- ClO- + Cl-+ H2O

*Concentrations of SolutionsSecond common unit of concentration:Molarity is defined as the number of moles of solute per literof solution.Molarity =M x L = moles solute and M x mL = mmol solute

*Example 11-16 What volume of 0.200 M KMnO4 is required to oxidize 35.0 mL of 0.150 M HCl? The balanced reaction is:2KMnO4 + 16HCl 2KCl + 2MnCl2+ 5Cl2+ 8H2O?mmole HCl = 35ml solution x 0.150M = 5.25 mmol HCl?mmol KMnO4 = 5.25 mmol HCl x= 0.656 mmol KMnO4? ml KMnO4 = 0.656 mmol KMnO4 x = 3.28ml KMnO4

*Example 11-17 A volume of 40.0 mL of iron (II) sulfate is oxidized to iron (III) by 20.0 mL of 0.100 M potassium dichromate solution. What is the concentration of the iron (II) sulfate solution? From Example 11-19 the balanced equation is:6Fe2+ 14H+ + Cr2O72- 6Fe3+ + 2Cr3+ + 7H2O?mmole Cr2O72- = 20.0ml solution x 0.100M = 2.00 mmol Cr2O72-?mmol Fe2+ = 2.00 mmol Cr2O72- x= 12.0 mmol Fe2+? M Fe2+ = 12.0 mmol KMnO4 x = 0.30 M Fe2+

*Example 11-10: Balancing Redox EquationsA useful analytical procedure involves the oxidation of iodide to free iodine. The free iodine is then titrated with a standard solution of sodium thiosulfate, Na2S2O3. Iodine oxidizes S2O32- ions to tetrathionate ions, S4O62- and is reduced to I- ions. Write the balanced net ionic equation for this reaction. Exercise 38Starting ReactionI2 + S2O32- I-+ S4O62-balance the ox. half-reactionS2O32- S4O62-2 S2O32- S4O62-2 S2O32- S4O62-+ 2e- balance the red. half-reactionI2 I- I2 2I- I2 + 2e- 2 I- Add the two half-reaction2 S2O32- S4O62-+ 2e- I2 + 2e- 2 I- I2 + 2S2O32- 2I-+ S4O62-

*Example 11-11: Net Ionic EquationsPermanganate ions oxidize ion(II) to iron(III) in sulfuric acid solution. Permanganate ions are reduced to manganese(II) ions. Write the balanced net ionic equation for this reaction. Exercise 38Starting ReactionFe2+ + MnO4- Fe3++ Mn2+balance the ox. half-reactionFe2+ Fe3+ Fe2+ Fe3++ e-

balance the red. half-reactionMnO4- Mn2+MnO4- + 8H+ Mn2++ 4H2OMnO4- + 8H++ 5e- Mn2++ 4H2O Add the two half-reaction 5 (Fe2+ Fe3++ e- ) MnO4- + 8H++ 5e- Mn2++ 4H2O5Fe2+ +MnO4- + 8H+ 5Fe3+ +Mn2++ 4H2O

*Example 11-13: Balancing Redox equationsIn basic solution, hypochlorite ions, ClO-, oxidize chromite ions, CrO2-, to chromate ions, CrO42-, and are reduced to chloride ions. Write the balanced net ionic equation for this reaction. Exercise 54Starting ReactionClO- + CrO2- Cl-+ CrO2-balance the ox. half-reactionCrO2- CrO2-CrO2- + 4OH- CrO42- + 2H2O CrO2- + 4OH- CrO42- + 2H2O + 3e-

balance the red. half-reactionClO- Cl-ClO- + H2O Cl-+ 2OH-ClO- + H2O+ 2e- Cl-+ 2OH-

Add the two half-reaction 2(CrO2- + 4OH- CrO42- + 2H2O + 3e-) 3(ClO- + H2O+ 2e- Cl-+ 2OH-) 2CrO2- + 8OH- + 3ClO- + 3H2O 2CrO42- + 4H2O + 3Cl-+ 6OH-

2CrO2- + 2OH- + 3ClO- 2CrO42- + H2O + 3Cl-

*Example 11-14: Redox TitrationWhat volume of 0.0200 M KMnO4 solution is required to oxidize 40.0ml of 0.100M FeSO4 in sulfuric acid solution? Exercise 54MnO4- + 8H+ + 5Fe2+ 5Fe3+ + Mn2+ + 4H2O?mmol Fe2+ = 40.0ml solution x 0.100M = 4.00 mmol Fe2+?mmol MnO4- = 4.00 mmol Fe2+ x= 0.8 mmol MnO4- ? ml KMnO4 = 0.8 mmol KMnO4 x = 40.0 ml KMnO4

*Example 11-15: Redox TitrationA 20.00 ml sample of Na2SO3 was titrated with 36.30 ml of 0.5130 M K2Cr2O7 solution in the presence of H2SO4 solution. Calculate the molarity of the Na2SO3 solution. Exercise 663SO32- + Cr2O72- + 8H+ 3SO42- + 2Cr3+ + 4H2O?mmol Cr2O72- = 36.30ml solution x 0.5130M = 1.862 mmol Cr2O72-?mmol SO32- = 1.862 mmol Cr2O72- x= 5.586 mmol SO32- ? M Na2SO3 = 5.586 mmol Na2SO3 / 20ml Na2SO3

=0.2793 M Na2SO3

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