Unit B Chapter 4 Electrolytes

Unit B Chapter 4 Electrolytes
Precipitation (aka Double Replacement or Metathesis) Reactions Acids & Bases Neutralization Reactions Oxidation Reduction Concentration of Solutions Solution Stoichiometry

1.0 0.85 0.50 0.33 none of the above No Calculator
When 30. ml of 0.50 M K2SO4 are added to 70. mL of 0.50 M of KMnO4, the resulting concentration of potassium is No Calculator 1.0 0.85 0.50 0.33 none of the above 2

When 30. ml of 0. 50 M K2SO4 are added to 70. mL of 0
When 30. ml of 0.50 M K2SO4 are added to 70. mL of 0.50 M of KMnO4, the resulting concentration of potassium is 1.0 0.85 0.50 0.33 none of the above No Calculator

KMnO4 SnCl2 MgCl2 Sn4+ Sn2+ K+ MnO2 Mg2+ Cl−
When a basic solution of KMnO4 is added to an SnCl2 solution, a brown precipitate of MnO2 forms and Sn4+ remains in solution. When the same basic solution of KMnO4 is added to an MgCl2 solution, no reaction occurs. Which of the substances involved in these reactions serves as the best reducing agent (i.e. the substance that is best at causing reduction in some other substance) ? KMnO4 SnCl2 MgCl2 Sn4+ Sn2+ K+ MnO2 Mg2+ Cl−

KMnO4 SnCl2 MgCl2 Sn4+ K+ MnO2 Mg2+ Cl−
When a basic solution of KMnO4 is added to an SnCl2 solution, a brown precipitate of MnO2 forms and Sn4+ remains in solution. When the same basic solution of KMnO4 is added to an MgCl2 solution, no reaction occurs. Which of the substances involved in these reactions serves as the best reducing agent (e.g. the substance that is best at causing reduction in some other substance) ? The Mn in the KMnO4 was reduced, and by SnCl2, not MgCl2. KMnO4 SnCl2 MgCl2 Sn4+ K+ MnO2 Mg2+ Cl−

Which solution represented below is a nonelectrolyte?
AX AY AZ

Which solution represented below is a nonelectrolyte?
AX there are no ions in the solution Which representation would coduct electricity the best? AY AZ

MgCl → Mg(aq) + Cl(aq) MgCl → Mg+ + Cl− MgCl2 → Mg2+ + (Cl−)2
Write the reaction that best represents the dissolving of magnesium chloride in water MgCl → Mg(aq) + Cl(aq) MgCl → Mg+ + Cl− MgCl2 → Mg (Cl−)2 MgCl2 → Mg Cl−2 MgCl2 → Mg + Cl2 MgCl2 → Mg Cl−

Write the reaction that best represents the dissolving of magnesium chloride in water
MgCl → Mg(aq) + Cl(aq) MgCl → Mg+ + Cl− MgCl2 → Mg (Cl−)2 MgCl2 → Mg Cl−2 MgCl2 → Mg + Cl2 MgCl2 → Mg Cl− Be sure you know the correct charges, and show the charge on the dissociated ions. Remember that multiple amounts of the same ion separate from each other, 2Cl− not Cl2−

HClO4 → H+ + ClO4− HClO4 → H+ + 4ClO− HClO4 → H+ + Cl− + 4O2−
Write the reaction that best represents the dissolving of perchloric acid in water HClO4 → H+ + ClO4− HClO4 → H ClO− HClO4 → H+ + Cl− + 4O2−

HClO2 → H+ + ClO2− HClO2 → H+ + 2ClO− HClO2 → H+ + Cl− + 2O2−
Write the reaction that best represents the dissolving of chlorous acid in water HClO2 → H+ + ClO2− HClO2 → H ClO− HClO2 → H+ + Cl− + 2O2− HClO2  H+ + ClO2− HClO2  H ClO−

Write the reaction that best represents the dissolving of chlorous acid in water
HClO2 → H+ + ClO2− HClO2 → H ClO− HClO2 → H+ + Cl− + 2O2− Polyatomic ions do NOT break up in aqueous solutions. The polyatomic ion, chlorite ion does not break apart in solution. HClO2  H+ + ClO2− This is a weak acid so you should use the double arrows to indicate only some of the acid ionizes. HClO2  H ClO−

Rank the following in order of increasing conductivity in water: (equal moles of each substance put in the water) C6H12O6, KBr, HF, Zn(OH)2 Least conductive → more conductive C6H12O6 < KBr < HF < Zn(OH)2 Zn(OH)2 < C6H12O6 < HF < KBr C6H12O6 < Zn(OH)2 < HF < KBr C6H12O6 < HF < Zn(OH)2 < KBr C6H12O6 < Zn(OH)2 < KBr < HF KBr < Zn(OH)2 < HF < C6H12O6

Rank the following in order of increasing conductivity in water: (equal moles of each substance put in the water) C6H12O6, KBr, HF, Zn(OH)2 Remember that conductivity is related to the amount of ions in solution C6H12O6 < KBr < HF < Zn(OH)2 Zn(OH)2 < C6H12O6 < HF < KBr C6H12O6 < Zn(OH)2 < HF < KBr Molecular sugar is least conductive, next is zinc hydroxide, because although it is a base, it is very insoluble, hydrofluoric acid is weak, and most is the soluble salt. Time to memorize the 7 strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4 C6H12O6 < Zn(OH)2 < KBr < HF KBr < Zn(OH)2 < HF < C6H12O6

C3H7OH < HClO < KCl < AlCl3
Rank the following in order of increasing conductivity in water: (equal moles of each substance put in the water) C3H7OH, KCl, HClO, AlCl3 C3H7OH < HClO < KCl < AlCl3 AlCl3 < C3H7OH < KCl <HClO HClO < C3H7OH < KCl < AlCl3 C3H7OH < HClO < KCl = AlCl3 C3H7OH < KCl < AlCl3 < HClO

Rank the following in order of increasing conductivity in water: (equal moles of each substance put in the water) C3H7OH, KCl, HClO, AlCl3 C3H7OH < HClO < KCl < AlCl3 C3H7OH is an alcohol, it is molecular and therefore least conductive. HClO, perchlorous acid is a weak acid. KCl, AlCl3 are salts and you might think are equally conductive, but because AlCl3 dissociates into 4 ions (Al+3 and 3 Cl−) it produces more ions per equal concentration and thus is a better conductive solution. AlCl3 < C3H7OH < KCl <HClO HClO < C3H7OH < KCl < AlCl3 C3H7OH < HClO < KCl = AlCl3 C3H7OH < KCl < AlCl3 < HClO

In the following reaction, which ions are spectator ions: AgNO3 + NaCl → NaNO3 + AgCl ?
Ag+ and NO3− Na+ and Cl− Cl− and NO3− Na+ and NO3− Ag+ and Cl− Do the problem without your solubility table.

In the following reaction, which ions are spectator ions: AgNO3 + NaCl → NaNO3 + AgCl ?
Ag+ and NO3− Na+ and Cl− Cl− and NO3− Na+ and NO3− Since silver chloride is the precipitate, the nitrate and alkali salts are the spectator ions. Ag+ and Cl− Do the problem without your solubility table.

Identify the precipitate when sodium hydroxide is combined with aluminum nitrate.
NaNO3 AlOH3 NaOH AlOH Al(OH)3 Do the problem without your solubility table.

Alkali nitrates will always be soluble.
Identify the precipitate when sodium hydroxide is combined with aluminum nitrate. NaNO3 AlOH3 NaOH AlOH Al(OH)3 Alkali nitrates will always be soluble. By process of elimination, the precipitate must be the aluminum hydroxide. Do the problem without your solubility table.

Which of the compounds listed below are insoluble:
NiCO3 Ba(NO3)2 C6H12O6 Na3PO3 ZnS K2CrO4 Al2(SO4)3 SnF2 Do the problem without your solubility table.

Which of the compounds listed below are insoluble:
NiCO3 Ba(NO3)2 C6H12O6 Na3PO3 ZnS K2CrO4 Al2(SO4)3 SnF2 Remember that nitrates and alkali salts are always soluble. Sugar is also of course soluble. Sulfates are usually soluble Carbonates and sulfides are usually NOT soluble Do the problem without your solubility table.

H2SO4 + KOH → K2SO4 + H2O H2SO4 + 2KOH → K2SO4 + 2H2O 2H+ + OH− → 2H2O
Select the reaction below that best represents the net ionic equation for sulfuric acid with potassium hydroxide. H2SO4 + KOH → K2SO4 + H2O H2SO4 + 2KOH → K2SO4 + 2H2O 2H+ + OH− → 2H2O 2H+ + 2OH− → 2H2O H+ + OH− → H2O

H2SO4 + KOH → K2SO4 + H2O H2SO4 + 2KOH → K2SO4 + 2H2O 2H+ + OH− → 2H2O
Select the reaction below that best represents the net ionic equation for sulfuric acid with potassium hydroxide. H2SO4 + KOH → K2SO4 + H2O H2SO4 + 2KOH → K2SO4 + 2H2O 2H+ + OH− → 2H2O 2H+ + 2OH− → 2H2O this would be ok too, but it is probably best to reduce it. H+ + OH− → H2O

None of them are since there is no oxygen in the reaction.
In the reaction below, which element is oxidized? NiCl2 + Al → AlCl3 + Ni Ni Cl O Al None of them are since there is no oxygen in the reaction. Perhaps you would balance this equation and turn it in to a net ionic equation.

None of them are since there is no oxygen in the reaction.
In the reaction below, which element is oxidized? NiCl2 + Al → AlCl3 + Ni Ni Cl O Al None of them are since there is no oxygen in the reaction. Aluminum loses three electrons per atom as it changes from atom form to Al3+ ion form. Perhaps you would balance this equation and turn it in to a net ionic equation. 3NiCl2 + 2Al → 2AlCl3 + 3Ni 3Ni2+ + 2Al3+ → 2Al3+ + 3Ni2+

On your white board, write out the net ionic equation for this reaction. NiCl2 + Al → AlCl3 + Ni

On your white board, write out the net ionic equation for this reaction. NiCl2 + Al → AlCl3 + Ni
Ni Al → Al3+ + Ni

For the reaction below, select the oxidizing agent (the substance that causes oxidation of some other substance): Fe2O3(s) + 3H2(g) → Fe(s) + H2O(L) O H Fe H2O Fe2O3

Thus the substance that the iron is part of is the oxidizing agent.
For the reaction below, select the oxidizing agent: Fe2O3(s) + 3H2(g) → Fe(s) + H2O(L) O H Fe H2O Fe2O3 The oxidation number of iron changes from 3+ as a reactant to 0 as a product, gaining electrons, and is therefore reduced. Thus the substance that the iron is part of is the oxidizing agent.

Which reaction below is not an oxidation reduction reaction?
3Co2+ + 2Al → 3Co + 2Al3+ 2Na + 2H2O → 2NaOH + H2 2HCl + Zn → ZnCl2 + H2 2HNO3 + Mg(OH)2 → 2H2O + Mg(NO3)2 CH4 + 2O2 → CO2 + 2H2O

Which reaction below is not an oxidation reduction reaction?
3Co2+ + 2Al → 3Co + 2Al3+ 2Na + 2H2O → 2NaOH + H2 2HCl + Zn → ZnCl2 + H2 2HNO3 + Mg(OH)2 → 2H2O + Mg(NO3)2 CH4 + 2O2 → CO2 + 2H2O Double replacement reactions are never oxidation reduction reactions since none of the atoms change oxidation numbers.

Reading the label of a cobalt supplement, the bottle tells us that each there are 35 mg of elemental cobalt delivered in 150 mg of cobalt(II) acetate in a 457 mg tablet. Is the cobalt(II) acetate in the pill a hydrate (if so, with how many water?) or anhydrate?

Two ways to think about this….
The iron pill bottle tells us that each there are 68 mg of elemental iron delivered in 338 mg of iron(II) sulfate. Is the iron sulfate in the pill a hydrate (if so,with how many water?) or is it an anhydrate? Two ways to think about this…. Yes it is a hydrate, 7 water Set up a ratio to find the mole of the iron in 68 mg 68/55.85 = mole Fe in FeSO4 = mole of FeSO4 Find the MM of the FeSO4 compound in the tablet 338 g / mol = 277 g/mol Determine the mass, then moles of this compound that must be water 277 g g = 125 g 125 / 18 = ~7 H2O Yes it is a hydrate, 7 water Set up a ratio to find the mass of of the iron compound in the tablet 68g/338g = 55.85g/x then solve for x = 277.6g Subtract the mass caused by the anhydrate, FeSO4 277.6g g = 126g water Determine the moles of water 126 / 18 = 7 H2O

Reactions in Aqueous Solutions
Unit B Chapter 4 slide view

soluble ionized dissociated aqueous concentrated dilute
133 g of aluminum chloride in 1 L of water would be considered (select all that apply) soluble ionized dissociated aqueous concentrated dilute weak electrolyte strong electrolyte acidic

133 g of AlCl3 in 1 L of water would be considered
AlCl3 → Al Cl− soluble ionized dissociated aqueous concentrated dilute weak electrolyte strong electrolyte acidic (for reasons you don’t know yet…) generally <0.1 M is considered dilute slide show

Electrolytes Compounds that dissolve and dissociate into ions
Strong or weak = amount ionized Concentrated or dilute = amount dissolved Electrolytes are Soluble ionic compounds Acids or bases Molecular compounds tend to be non-electrolytes

Dissolving & Dissociating
When ionic compounds dissolve they dissociate Observe the orientation of the water molecules as the solvate around the dissociated ions When molecular compounds dissolve, they do not dissociate Slide View

No Calculator 0.05 M 0.10 M 0.15 M 0.20 M 0.50 M 2.0 M 0.40 M 0.75 M
Calculate the concentration of chloride ion when 5.8 g of sodium chloride and 4.8 g of magnesium chloride is dissolved to produce 400 ml of solution. No Calculator 0.05 M 0.10 M 0.15 M 0.20 M 0.50 M 2.0 M 0.40 M 0.75 M

Calculate the concentration of chloride ion when 5
Calculate the concentration of chloride ion when 5.8 g of sodium chloride and 4.8 g of magnesium chloride is dissolved to produce 400 ml of solution. 0.05 M 0.10 M 0.15 M 0.20 M 0.50 M 2.0 M No Calculator

Write a balanced equation to represent silver nitrate dissolving in water.
On notebook paper….

Write a balanced equation to represent solid silver nitrate dissolving in water.
AgNO3(s) → Ag+ + NO3−

Write a balanced equation to represent magnesium chloride dissolving in water.

Write a balanced equation to represent magnesium chloride dissolving in water.
MgCl2(s) → Mg Cl−

Write a balanced overall (molecular) equation to represent the reaction that occurs when a silver nitrate solution reacts with magnesium chloride solution. Then turn your overall Rx into a balanced net ionic equation.

Write a balanced overall (molecular) equation to represent the reaction that occurs when a silver nitrate solution reacts with magnesium chloride solution. 2AgNO3(aq) + MgCl2(aq) → 2AgCl(s) + Mg(NO3)2(aq) Which really means: 2Ag+ + 2NO3− + Mg2+ + 2Cl− → 2AgCl(s) + 2NO3− + Mg2+ which we rewrite as a net ionic Ag+ + Cl− → AgCl(s)

Solution Stoichiometry
If you’re not part of the solution, you’re part of the precipitate

100 ml of M lead(II) nitrate solution is combined with 100 ml of molar sodium chloride solution. Here’s a series of questions.... Write out the balanced overall (aka molecular) and net ionic equations for this reaction On your notebook paper.

100 ml of 0. 050 M lead(II) nitrate is combined with 100 ml of 0
100 ml of M lead(II) nitrate is combined with 100 ml of molar sodium chloride. Here’s a series of questions.... Write out the overall and net ionic equations for this reaction Pb(NO3)2(aq) + 2NaCl(aq) → 2NaNO3(aq) + PbCl2(ppt) Pb2+ + 2Cl− → PbCl2

100. 0 ml of 0. 050 M lead(II) nitrate solution is combined with 100
100.0 ml of M lead(II) nitrate solution is combined with ml of molar sodium chloride solution. Calculate the mass of precipitate formed (MM PbCl2 = 278 g/mol) Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 Pb+2 + 2Cl− → PbCl2

Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2
100.0 ml of M lead(II) nitrate is combined with ml of molar sodium chloride. Calculate the mass of precipitate formed. (MM PbCl2 = 278 g/mol) Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 (0.05 M)(0.1L) = mol Pb(NO3)2 (0.08 M)(0.1L) = mol NaCl use the “limiting trick” to determine that NaCl limits

100 ml of M lead(II) nitrate is combined with 100 ml of molar sodium chloride. What are the concentrations of the sodium ions and nitrate ions in solution after the reaction is complete? Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2

100 ml of M lead(II) nitrate is combined with 100 ml of molar sodium chloride. What is the concentration of the sodium and nitrate ions in solution after the reaction is complete? Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 All of the starting sodium and nitrate ions are still present, since they are spectator ions, swimming in the 200 ml total volume. (0.05 M)(100 ml) = 5 mmol Pb(NO3)2 (0.08 M)(100 ml) = 8 mmol NaCl

100 ml of M lead(II) nitrate is combined with 100 ml of molar sodium chloride. What is the concentration of the lead ions and chloride ions in solution after the reaction is complete? Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2

100 ml of M lead(II) nitrate is combined with 100 ml of molar sodium chloride. What is the concentration of the lead and chloride ions in solution after the reaction is complete? Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 Since the NaCl limited, we can assume all the Cl− ion are in the precipitate not the solution, thus Cl− = 0 M Some of the lead ions are in the precipitate, some are still in solution. (0.05 M)(100 ml) = 5 mmol Pb2+ to start 5mmol Pb2+ start − 4 mmol Pb2+ reacted = 1 mmol Pb2+ remain in soln

100. ml of 0. 050 M (mol/L) cobalt(III) nitrate is combined with 200
100. ml of M (mol/L) cobalt(III) nitrate is combined with 200. ml of molar potassium chromate. Better write a balanced equation. What is the mass of the precipitate? What is the concentration of each of the ions left in the solution? molar mass of cobalt chromate = g/mol

A 1.0 L sample of an aqueous solution contains mol of KCl and mol of AlCl3. What is the minimum number of moles of AgNO3 that must be added to the solution in order to precipitate all of the Cl− as AgCl(s)? No calculator 59

A 1.0 L sample of an aqueous solution contains mol of KCl and mol of AlCl3. What is the minimum number of moles of AgNO3 that must be added to the solution in order to precipitate all of the Cl− as AgCl(s)? No calculator 0.09 mol AgNO3 must be added 60

A 1.0 L sample of an aqueous solution contains mol of KCl and mol of AlCl3. What is the minimum number of moles of Pb(NO3)2 that must be added to the solution in order to precipitate all of the Cl− as PbCl2(s)? No calculator 61

A 1.0 L sample of an aqueous solution contains mol of KCl and mol of AlCl3. What is the minimum number of moles of Pb(NO3)2 that must be added to the solution in order to precipitate all of the Cl− as PbCl2(s)? No calculator 0.045 mole Pb(NO3)2 must be added 62

When 10. milliliter of 0. 4-molar sodium sulfate is added to 20
When 10. milliliter of 0.4-molar sodium sulfate is added to 20. milliliters of 0.10-molar aluminum sulfate the number of moles of Pb2+ that must be added to precipitate out all of the SO42− would be No calculator 0.020 mol mol 0.010 mol 10. mol mol 0.10 mol 0.005 mol 1.0 mol 63

When 10. milliliter of 0. 4-molar sodium sulfate is added to 20
When 10. milliliter of 0.4-molar sodium sulfate is added to 20. milliliters of 0.10-molar aluminum sulfate the number of moles of Pb2+ that must be added to precipitate out all of the SO42− would be No calculator 0.020 mol mol 0.010 mol 10. mol mol 0.10 mol 0.005 mol 1.0 mol 64

A 10.0-milliliter sample of molar K3PO4 solution is added to 20.0 milliliters of molar Ba(NO3)2 solution. Barium phosphate precipitates. The concentration of barium ion, Ba2+, in solution after reaction is No calculator 65

A 10.0-milliliter sample of molar K3PO4 solution is added to 20.0 milliliters of molar Ba(NO3)2 solution. Barium phosphate precipitates. The concentration of barium ion, Ba2+, in solution after reaction is No calculator 2 mmol PO43−, 9 mmol Ba2+, 3 mmol Ba2+ used 6 mmol Ba2+ left over/30 ml = 0.2 M Ba2+ after ppt 66

Acid Review Weak? Strong? Making Solutions

Strong vs Weak Strong electrolytes dissociate completely
HNO3 → H NO3− Strong acids are strong electrolytes Weak electrolytes only partly dissociate. Some molecules remain “whole” while some dissociate. HF ⇋ H F− Weak acids are weak electrolytes

Strong vs Weak HF ⇋ H+ + F−
The double arrow indicates that both the forward and reverse reactions are occurring at the same rate, so the reaction seems to “stop”. It does not mean that the reaction occurs only as far as the half way point. This balance is called chemical equilibrium.

Acids Arrhenius defined acids as substances that produce H+ ions in an aqueous solution. HBr → H Br− HBr + H2O → H3O+ + Br− Bronstead & Lowrey defined acids as proton donors. Monoprotic acids produce 1 H+ Di- or Tri- protic acids produce 2 or 3 H+ H2SO4 can produce 2 H+ per acid molecule. We can think of the dissociation occuring in steps: H2SO4 → H+ + HSO4− HSO4− ⇋ H+ + SO42− In this case, only the first ionization is complete, thus a solution of sulfuric acid will actualy contain a mixture of H+, HSO4−, and SO42−

Bases Substances that react with H+ ions
Arrhenius defined bases as substances that produce OH- ions NaOH → Na OH− Ba(OH)2 → Ba OH− Bronstead & Lowrey defined bases as proton acceptors because some common weak bases appear not to contain OH- ions. Ammonia is a common weak base. It reacts with water, accepts an H+, and produces OH− NH3 + H2O ⇋ NH OH− When you see the double arrow, it tells us the reaction “comes to equilibrium” and it is a weak base.

Memorize the Strong Acids & Strong Bases
Strong acids (All the rest we will encounter in AP are weak) HCl, HBr, HI HClO3, HClO4 HNO3 H2SO4 Strong bases - metal hydroxides that dissolve (most metal hydroxides do not dissolve very well) Alkali hydroxides LiOH, NaOH, KOH, RbOH, CsOH Heavy alkaline earth hydroxides Ca(OH)2, Sr(OH)2, Ba(OH)2

Acids React with Bases Neutralization reaction HNO3 + KOH → H2O + KNO3
aka Double Replacement Rx HNO3 + KOH → H2O + KNO3 Acid base water + salt Net Ionic: H+ + OH- → H2O

How do you make up 500 ml of a 2 M solution of sodium hydroxide?

How do you make 500 ml of a 2 M solution of sodium hydroxide?
Use the formula M = moles/Liter to determine that you need 1 mole in the 500 ml of solution. NaOH, MM = 40g/mol Mass 40 g of NaOH into a volumetric flask. Add ⅔ of the water and swirl to dissolve Add the remaining water to bring the solution to 500 ml total.

How to make 500 ml of a 2 M solution of hydrochloric acid starting with concentrated HCl, 12 M?
slide show

How to make 500 ml of a 2 M solution of hydrochloric acid starting with concentrated HCl, 12 M?
This is a dilution problem, and during a dilution the number of moles in the flask does not change while adding water to dilute the more concentrated solution. Use the formula M×V = M×V (2M)(500 ml) = (12M)(V) V = 83.3 ml Add water to the flask, ⅔ full. Measure out the 83.3ml of concentrated HCl into a graduated cylinder and add to the water in the volumetric flask. Add the remaining water to bring the solution to 500ml total.

On paper write an overall and a net ionic equation to describe the reaction between a 500 ml of 2M solution of NaOH and 500 ml of a 2M solution of HCl.

HCl + NaOH → H2O + NaCl H+ + OH− → H2O
On your white boards write an overall and a net ionic equation to describe the reaction between a 500 ml of 2M solution of NaOH and 500 ml of a 2M solution of HCl. Calculate the moles of water that would form during this reaction. HCl + NaOH → H2O + NaCl H+ + OH− → H2O

H+ + OH− → H2O (2M)(.5L) = 1mol H+, combined with
On your white boards write an overall and a net ionic equation to describe the reaction between a 500 ml of 2M solution of NaOH and 500 ml of a 2M solution of HCl. Calculate the moles of water that would form during this reaction. What volume of water is this? H+ + OH− → H2O (2M)(.5L) = 1mol H+, combined with (2M)(.5L) = 1 mol OH− = 1 mole H2O

(2M)(.5L) = 1mol H+, combined with (2M)(.5L) = 1 mol OH- = 1 mol H2O
On your white boards write an overall and a net ionic equation to describe the reaction between a 500 ml of 2M solution of NaOH and 500 ml of a 2M solution of HCl. Calculate the moles of water that would form during this reaction. What volume of water is this? H+ + OH− → H2O (2M)(.5L) = 1mol H+, combined with (2M)(.5L) = 1 mol OH- = 1 mol H2O Density of water slide show

In class, 500 ml of 2M solution of NaOH and 500 ml of a 2M solution of HCl were poured together into a ml flask. At first, you may have been surprised to see ~18 ml more than 1000 ml of solution, however, after the calculations shown below, you could see why this must occur. H+ + OH− → H2O (2M)(.5L) = 1mol H+, combined with (2M)(.5L) = 1 mol OH- = 1 mol H2O of water will form 1mol H2O = (18g)(1ml/1g)= 18 ml + 500 ml mark 1000 ml In the demonstration done in class, the extra water above the line was measured and turned out to be ~18 ml = slide show

Write a balanced equation for the reaction that represents the combustion of magnesium in air.
This is a redox reaction in which magnesium is oxidized. This is a redox reaction in which oxygen is oxidized. This is a redox reaction in which magnesium is reduced. This is not a redox reaction.

Write a balanced equation for the reaction that represents the combustion of magnesium in air. 2Mg + O2 → 2MgO or Mg + ½O2 → MgO This is a redox reaction in which magnesium is oxidized. This is a redox reaction in which oxygen is oxidized. This is a redox reaction in which magnesium is reduced. This is a redox reaction in which oxygen is reduced. This is not a redox reaction. Combustion reactions are always redox reactions. In redox reactions, particles must be both oxidized and reduced. slide view

Oxidation - Reduction Oxidation (LEO) [OIL] Reduction (GER) [RIG]
When an element loses electrons Reduction (GER) [RIG] When an element gains electrons The name oxidation is used because this type of reaction was first studied by reacting metals with oxygen. Rusting: Fe + O2 → Fe2O3 It was called reduction because when elements are isolated from minerals, we say they are “reduced” to their elemental form. AlCl3 → Al(s) + Cl2(g)

Select the true statement(s) about the reaction below: Mg(OH)2 + HCl → H2O + MgCl2
This is a redox reaction in which chlorine is reduced. This is a redox reaction in which magnesium is reduced. This is a redox reaction in which hydrogen is reduced. This is a redox reaction in which oxygen is reduced. This is a redox reaction in which chlorine is oxidized. This is a redox reaction in which magnesium is oxidized. This is a redox reaction in which hydrogen is oxidized. This is a redox reaction in which oxygen is oxidized. This is not a redox reaction.

Select the true statement(s) about the reaction below: Mg(OH)2 + HCl → H2O + MgCl2
This is a redox reaction in which chlorine is reduced. This is a redox reaction in which magnesium is reduced. This is a redox reaction in which hydrogen is reduced. This is a redox reaction in which oxygen is reduced. This is a redox reaction in which chlorine is oxidized. This is a redox reaction in which magnesium is oxidized. This is a redox reaction in which hydrogen is oxidized. This is a redox reaction in which oxygen is oxidized. This is not a redox reaction. Acid base neutralization is never redox.

Insoluble (Dilute) Strong Base
Mg(OH)2 slide view

Insoluble (Dilute) Strong Bases
Mg(OH)2 is an insoluble ionic compound in H2O In an acid solution, Mg(OH)2 does dissolve Demonstrate The acid ions, H3O+ (or H+) attack the OH− ions that are in the solid crystal structure creating water molecules H+ + OH− → H2O H3O+ + OH− → 2H2O

Adding acid to “Milk of Magnesia” a suspension of Mg(OH)2
Write an overall equation for the reaction between a suspension of magnesium hydroxide and hydrochloric acid. Mg(OH)2 + HCl → H2O + MgCl2 Convert to a net ionic equation. Mg(OH)2 + H+ → H2O + Mg2+ Mg(OH)2 must be written as a compound because it is not soluble in water (to any appreciable degree, and it is simply a suspension in the water.) HCl is soluble, thus the Cl− ions are spectator ions.

If 40.0 ml of a 0.30 M solution of barium nitrate is combined with 25.0 ml of 0.35 M sodium phosphate solution, calculate the mass of the precipitate that should form. If 1.32 g of precipitate did form in the laboratory, calculate the % yield. Determine the concentration (molarity) of any ions still floating in the solution.

type in the number transfered
What is the total number of electrons transferred when magnesium is burned in oxygen? type in the number transfered

For which balanced equation? 2Mg + O2 → 2MgO (4 e-1)
What is the total number of electrons transferred when magnesium is burned in oxygen? For which balanced equation? 2Mg + O2 → 2MgO (4 e-1) OR Mg + ½O2 → MgO (2 e-1) Thus you might say: 2 e-1 per mole of MgO formed.

Oxidation Numbers more practice

What is the oxidation number of sulfur in SF2?
+2 -2 +4 -4

What is the oxidation number of sulfur in SF2?
+2 -2 +4 -4 Since fluorine is -1 × 2 = −2 the sulfur must be +2 to balance.

What is the oxidation state of chlorine in KClO4 ?
−2 −4 +4 +7 +8

What is the oxidation state of chlorine in KClO4 ?
−2 −4 +4 +7 +8 O is −2 × 4 = −8, the K is +1 thus the Cl must be the remaining +7.

What is the oxidation state of nitrogen in N2O5 ?
+5 +2 -10 +10 -5

What is the oxidation state of nitrogen in N2O5 ?
+5 +2 -10 +10 −5 Since oxygen is −2 × 5 = -10, the nitrogens must total +10, but 2 nitrogens cause the +10, so each one individually is +5. slide show

Balancing Aqueous Redox Equations
Chapter 4 and 20 (section 1 & 2)

Redox Reactions NOT double replacement Always combustion
precipitation acid base Always combustion Always single replacement Sometimes synthesis Sometimes decomposition Complex redox reactions in solution

Redox Reactions NO3− + Al → Al3+ + NO
Determine which element is oxidized Determine which element is reduced How many electrons are transferred?

Redox Reactions SO32− + MnO4− → SO42− + Mn2+
Determine which element is oxidized Determine which element is reduced how many electrons are transferred?

Identifying the players
6H+ + 2MnO4− + 5H2O2 → 2Mn O2 + 8H2O Identify the element that is oxidized. Identify the element that is reduced. +1 +7 +2 −2 −1

Disproportionation Reactions
A solution of hypochlorous acid and hydrochloric acid is formed when chlorine gas is dissolved in cold water. Write this reaction on your white board and determine the oxidation numbers for all elements in this reaction. The reaction is: Cl2 + H2O → HOCl + H+ + Cl−

Disproportionation Reactions
Determine the oxidation numbers for all elements in this reaction. Cl2 + H2O → HOCl + H+ + Cl− , → +1, -2, When an element is both oxidized and reduced in the same reaction, it is known as a disproportionation reaction. The chlorine disproportionates.

AuCl4− + Cu → Au + Cl− + Cu2+ In an acidic solution
Identify which elements are oxidized and reduced Balance the number of redox atoms by inspection Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) balance any other elements (other than H and O) balance oxygen by adding water as necessary balance hydrogen by adding H+ as necessary recheck by confirming that ion charge is balanced

AuCl4− + Cu → Au + Cl− + Cu2+ Identify which elements are oxidized and reduced +3, → Au reduced 3e- gained, Cu oxidized 2e- lost Balance the number of redox atoms by inspection − they are already Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 2AuCl4− + 3Cu → 2Au + Cl− + 3Cu2+ (a total of 6e− transferred) balance any other elements (other than H and O) 2AuCl4− + 3Cu → 2Au + 8Cl− + 3Cu2+ balance oxygen by adding water as necessary − not necessary balance hydrogen by adding H+ as necessary − not necessary recheck by confirming that charge is balanced − 2− = 2−

2AuCl4− + 3Cu → 2Au + 8Cl− + 3Cu2+ It will be important later, for us to identify the number of electrons transferred. in this reaction, there are 6 electrons lost and the same 6 electrons are gained thus, a total of 6 (not 12) electrons are transferred In every redox reaction there are two halves. the oxidation the reduction We call these the half reactions.

2AuCl4− + 3Cu → 2Au + 8Cl− + 3Cu2+ From time to time, AP will ask you to write out the half reactions. Pull out the two substances involved in oxidation, then balance, first the oxidation atom, then the other atoms, then oxygen, then hydrogen − then include the number of electrons The oxidation half reaction is Cu → Cu e− reduction AuCl4− + 3e− → Au + 4Cl− this corresponds with what we had determined by the previous method Au reduced 3e− gained, Cu oxidized 2e− lost

NO3− + Al → Al3+ + NO In an acidic solution

SO32− + MnO4− → SO42− + Mn2+ In an acidic solution

Zn + NO3− → Zn2+ + N2 In an acidic sol’n

(in acidic solution) Zn + NO3− → Zn2+ + N2
Identify which elements are oxidized and reduced 0 + +5,-2 → N reduced 5e-/atom, Zn oxidized 2e- Balance the number of redox atoms by inspection Zn + 2NO3− → Zn2+ + N2 Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 5Zn + 2NO3− → 5Zn2+ + N2 balance any other elements (other than H and O) - none balance oxygen by adding water as necessary 5Zn + 2NO3− → 5Zn2+ + N2 + 6H2O balance hydrogen by adding H+ as necessary 12H+ + 5Zn + 2NO3− → 5Zn2+ + N2 + 6H2O recheck by confirming that ion charge is balanced − = +10

Write out the two half reactions
Write out the two half reactions. 12H+ + 5Zn + 2NO3− → 5Zn2+ + N2 + 6H2O

Zn + NO3− → Zn2+ + N2 (in acidic)
oxidation Zn → Zn e− note that charge balances in half reactions as well reduction 2NO3− → N2 this may involve using water and H+ ions 12H+ + 2NO3− + 10e− → N2 + 6H2O this corresponds with what we had determined by the previous method N reduced 5e- gained per N, but the N2 requires 10e- total, Zn oxidized 2e- lost 117

Br2 + AsO2− → Br− + AsO43− In a basic solution
Identify which elements are oxidized and reduced Balance the redox atoms by inspection balance the redox atoms based on total electrons transferred balance any other elements (other than H and O) balance O by adding water as necessary balance H by adding H+ as necessary if the reaction occurs in basic solution, convert to basic by adding an equal number of OH− ions to both sides to cancel out the H+ ions recheck by confirming that charge is balanced

Br2 + AsO2-1 → Br-1 + AsO4-3 “pretend” in acid, then convert to basic
Identify which elements are oxidized and reduced 0 + +3, -2 → , -2 Br reduced 1e-/atom, As oxidized 2e- balance those atoms based on total electrons transferred Br2 + AsO2-1 → 2Br-1 + AsO4-3 balance any other elements (other than H and O) - none balance O by adding water as necessary 2H2O + Br2 + AsO2− → 2Br− + AsO43− balance H by adding H+1 as necessary 2H2O + Br2 + AsO2− → 2Br− + AsO43− + 4H+ if the reaction occurs in basic solution, convert to by adding an equal number of OH-1 ions to both sides to cancel out the H+1 ions 4OH− + 2H2O + Br2 + AsO2− → 2Br− + AsO43− + 4H+ + 4OH− 4OH− + 2H2O + Br2 + AsO2− → 2Br− + AsO43− + 4H2O 4OH− + Br2 + AsO2− → 2Br− + AsO43− + 2H2O recheck by confirming that charge is balanced −5 = −5

Br2 → BrO3− + Br− In basic solution
Identify which elements are oxidized and reduced This is a disproportionation reaction - the same element is both oxidized and reduced. Balance the number of redox atoms by inspection Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) balance any other elements (other than H and O) balance oxygen by adding water as necessary balance hydrogen by adding H+ as necessary since the reaction occurs in basic solution, convert to by adding an equal number of OH− ions to both sides to cancel out the H+ ions recheck by confirming that ion charge is balanced

Br2 → BrO3− + Br− “pretend” in acid, then convert to basic
Identify which elements are oxidized and reduced 0 → +5, This is a disproportionation reaction - the same element is both oxidized and reduced. Br reduced 1e-/atom, and Br oxidized 5e-/atom Balance the number of redox atoms by inspection − they are already Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 6Br2 → 2BrO3− + 10Br− balance any other elements (other than H and O) − none balance oxygen by adding water as necessary 6 Br2 + 6 H2O → 2 BrO3− + 10Br− balance hydrogen by adding H+ as necessary 6 Br H2O → 2 BrO3− Br− H+ recheck by confirming that ion charge is balanced since the reaction occurs in basic solution, convert to by adding an equal number of OH− ions to both sides to cancel out the H+ ions 12OH− + 6 Br2 + 6H2O → 2 BrO3− + 10Br− + 12H+ + 12OH− 12OH− + 6 Br2 + 6H2O → 2 BrO3− + 10Br− + 12H2O 12OH− + 6 Br2 → 2 BrO3− + 10Br− + 6H2O recheck by confirming that charge is balanced −12 = −12

In basic solution Br2 → BrO3− + Br− Write out the half reactions which also should be balnced as in a basic solution.

Br2 → BrO3− + Br− (in basic) Write out the half reactions.
oxidation Br2 → 2 BrO3− + 10e− Br2 + 6 H2O → 2 BrO3− H+ + 10e− (pretending acidic) Br2 + 12OH− → 2 BrO3− + 6 H2O + 10e− reduction Br2 + 2e− → 2Br− this corresponds with what we had determined by the previous method N reduced 5e− gained per N, but the N2 requires 10e− total, Zn oxidized 2e− lost

CN− + MnO4− → CNO− + MnO2 basic sol’n Identify which elements are oxidized and reduced Balance the number of redox atoms by inspection Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) balance any other elements (other than H and O) balance oxygen by adding water as necessary balance hydrogen by adding H+ as necessary since the reaction occurs in basic solution, convert to by adding an equal number of OH− ions to both sides to cancel out the H+ ions recheck by confirming that ion charge is balanced

CN− + MnO4− → CNO− + MnO2 3CN− + 2MnO4− + H2O → 3CNO− + 2MnO2 + 2OH−
basic sol’n Identify which elements are oxidized and reduced You may worry about what to do with the oxidation numbers - consider N to be -3 and watch the C change +2, , -2 → +4, -3, , -2 Mn reduced 3e-, and C oxidized 2e- Balance the number of redox atoms by inspection − they are already Adjust the number of redox atoms so the number of electrons lost equals the number of electrons gained. (This is the number of electrons transferred.) 3CN− + 2MnO4− → 3CNO− + 2MnO2 balance any other elements (other than H and O) − already done balance oxygen by adding water as necessary 3CN− + 2MnO4− → 3CNO− + 2MnO2 + H2O balance hydrogen by adding H+ as necessary 3CN− + 2MnO4− + 2H→ 3CNO− + 2MnO2 + H2O since the reaction occurs in basic solution, convert to by adding an equal number of OH− ions to both sides to cancel out the H+ ions 2OH− + 3CN− + 2MnO4− + 2H+ → 3CNO− + 2MnO2 + H2O + 2OH− 3CN− + 2MnO4− + 2H2O → 3CNO− + 2MnO2 + H2O + 2OH− 3CN− + 2MnO4− + H2O → 3CNO− + 2MnO OH− recheck by confirming that ion charge is balanced −5 = − 5

CN− + MnO4− → CNO− + MnO2 oxidation CN− → CNO− + 2e−
basic sol’n oxidation CN− → CNO− e− CN− + H2O → CNO− e− + 2H+ (pretending acidic) CN− + 2OH− → CNO− e− + H2O reduction MnO4− + 3e− → MnO2 MnO4− + 4H e− → MnO2 + 2H2O (pretending acidic) MnO4− + 2H2O + 3e− → MnO OH−

Pull out the two half reactions, and balance (basic) Pb(OH)42− + ClO− → PbO2 + Cl−
127

Pull out the two half reactions, and balance (basic) Pb(OH)42− + ClO− → PbO2 + Cl−
oxidation Pb(OH)42− → PbO2 + 2H2O + 2e− reduction H2O + ClO− + 2e− → Cl− + 2OH− 128

Pull out the two half reactions, and balance (acidic) each one individually. Label which one is oxidation and which is reduction. I2 + OCl− → IO3− + Cl− 129

I2 + 6H2O → 2IO3− + 12H+ + 10e− 2H+ + OCl− + 2e− → Cl− + H2O oxidation
Pull out the two half reactions, and balance (acidic) each one individually. Label which one is oxidation and which is reduction. I2 + OCl− → IO3− + Cl− oxidation I2 + 6H2O → 2IO3− + 12H+ + 10e− reduction 2H+ + OCl− + 2e− → Cl− + H2O 130

Pull out the two half reactions, and balance (basic) H2O2 + Cl2O7 → O2 + ClO2−
131

Pull out the two half reactions, and balance (basic) H2O2 + Cl2O7 → O2 + ClO2−
redution 3H2O + Cl2O7 + 8e− → 2ClO2− + 6OH− oxidation 2OH− + H2O2 → O2 + H2O + 2e− 132

Unit B Chapter 4 Electrolytes

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