Saturday Study Session 1 1 st Class Reactions. Things to remember Solids, liquids, and gases can NOT...

Saturday Study Session 11st Class Reactions

Things to remember

• Solids, liquids, and gases can NOT be broken into ions

• SOLUTION – if it says a solution, then it CAN be broken into ions if it is soluble in water.

• Only ionic compounds can become separate ions in a solution.

• The 6 strong acids (HCl, HBr, HI, HNO3, HClO4, H2SO4) and the strong bases (group 1 + OH-and Ba, Sr, Ca + OH-) are always written as separate ions.

Electrolytes

Formula Equation (Molecular Equation)

• Gives the overall reaction stoichiometry but not necessarily the actual forms of the reactants and products in solution.

• Reactants and products generally shown as compounds.

• Use solubility rules to determine which compounds are aqueous and which compounds are solids.

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

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Complete Ionic Equation

• All substances that are strong electrolytes are represented as ions.

Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)

AgCl(s) + Na+(aq) + NO3-(aq)


HCl --- Strong electrolyte100% dissociation

Net Ionic Equation

• Includes only those solution components undergoing a change.– Show only components that actually react.

Ag+(aq) + Cl-(aq) AgCl(s)

• Spectator ions are not included (ions that do not participate directly in the reaction).– Na+ and NO3

- are spectator ions.


Precipitant

White Precipitant

AgCl, Mg(OH)2,BaSO4

Opening activity– Solid calcium oxide is added to water

CaO + H2O Ca(OH)2 molecular equation

CaO + H2O Ca2+ + 2 (OH)- net ionic

– Potassium chlorate(s) is heated2KClO3 2KCl + 3O2 molecular and ionic

– Calcium carbonate(s) is heatedCaCO3 CaO + CO2 molecular and ionic

– Sodium hydroxide(aq) is heated 2NaOH Na2O + H2O molecular

2Na+ + 2(OH)- Na2O + H2O ionic

Aluminum metal is added to solution of copper II chloride2Al + 3CuCl2 2 AlCl3 + 3Cu molecular

2Al + 3Cu2+ 2Al3+ + 3 Cu ionic

– Fluorine gas is bubbled into solution of sodium bromide

F2 + 2NaBr 2NaF + Br2 molecular

F2 + 2Br- 2F- + Br2

– Solutions of Hydrofluoric acid is added to ammonium hydroxideHF + NH4OH NH4F + H2O molecular

HF + NH4OH NH4 + + F- + H2O ionic

– Butane is burned in Air2C4H10 + 13O2 8CO2 + 10H2O molecular and ionic

Stoichiometry

• Exercise 1A 500.0-g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution?

Stoichiometry

• Exercise 1A 500.0-g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution?

1.57 M

moles of solute = Molarity = liters of solution

M

Problem #1

• Step 1 - find molecular mass of K3PO4

• 212.274 g/mol

• Step 2 - convert from g of K3PO4 to mol • 500.0 g / 212.274 g/mol = 2.355 mol

• Step 3 - divide mol by liters to get molarity• 2.355 mol / 1.50 L = 1.57 M

• Exercise 2What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution?

• Exercise 2What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution?

60.0 mL

• M1V1 = M2V2• (2 M)(X) = (0.800M)(150 mL)• X = (.800M)(150mL)

(2.00M) X = 60.0 mL

10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).

– What precipitate will form?lead(II) phosphate, Pb3(PO4)2

– What mass of precipitate will form?1.1 g Pb3(PO4)2


(Part I)Exercise 3

All group I metals are solubleAll nitrates are soluble2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb2(PO4)2(s)

6Na+ + 2PO4-3 + 3Pb+2 + 6NO3

- → 6Na+ + 6NO3- + Pb3(PO4)2

2PO4-3 + 3Pb+2 → Pb3(PO4)2

• R 2PO4-3 + 3Pb+2 → 1Pb3(PO4)2

• I .0030 .0040 0• C -2x -3x +1x• E .000033 excess 0 0.0013

• 0.0040 -3x = 0 therefore x = 0.001333333• 0 + 1x = 0.0013333 mol Pb3(PO4)2

• Molecular mass of Pb3(PO4)2 equals 811.55 g/mol

• 0.0013 mol x 811.55 g/mol = 1.1 g of Pb3(PO4)2


– What is the concentration of nitrate ions left in solution after the reaction is complete?

Nitrate 0.27 M


(Part II)Exercise 4

• Step 1 – find moles of NO3- present

• 0.200 M Pb(NO3)2 = 2 NO3- x 0.200 M = 0.400 M

• 0.400 mol/L x .0020 L = .00080 mol of NO3-

• Step 2 – find new volume after mixing solutions

• 20.0 ml + 10.0 ml = 30.0 ml or 0.003 L

• Step 3 - find molarity of NO3-• 0.0008 mol / .003 L = 0.27 M


– What is the concentration of phosphate ions left in solution after the reaction is complete?

0.011 M

Excess was .000033 mol / .03 L = .011 M


(Part III)

Which of the following ions form compounds with Pb2+ that are generally soluble in water?

a) S2–

b) Cl–

c) NO3–

d) SO42–

e) Na+


Which of the following solutions containsthe greatest number of ions?

a) 400.0 mL of 0.10 M NaCl.b) 300.0 mL of 0.10 M CaCl2.

c) 200.0 mL of 0.10 M FeCl3.

d) 800.0 mL of 0.10 M sucrose.


A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution?

a) Add water to the solution.b) Pour some of the solution down the sink drain.c) Add more sodium chloride to the solution.d) Let the solution sit out in the open air for a couple

of days.e) At least two of the above would decrease the

concentration of the salt solution.


After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of 38 percent. The correct value for the percentage of water in the hydrate is 51 percent. Which of the following is the most likely explanation for this difference? a. Strong initial heating caused some of the hydrate sample to spatter out of the crucible.b. The dehydrated sample absorbed moisture after heating.c. The amount of the hydrate sample used was too small.

d. The crucible was not heated to constant mass before use.e. Excess heating caused the dehydrated sample to decompose.

Acid–Base Reactions (Brønsted–Lowry)“Bro-Pro”

Brønsted deals with Protons

• Acid—proton donor• Base—proton acceptor• Conjugate Acid – formed when base gains a H+

• Conjugate Base – formed when acid loses a H+

• HF + NH3 NH4+ + F-

Acid Base Conjugate Conjugate

Acid Base


For the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of 0.500 M sulfuric acid to reach the endpoint?

1.00 mol NaOH

2NaOH + 1H2SO4 → Na2SO4 + 2H2O

It takes 2 mol of OH- to react with 1 mol 2H+(.500M)(1.00L) = .500 mol H+ so .500 mol x 2 =1.00 mol NaOH


Titration – Laboratory Procedure

Redox reaction or acid baseNeutralization reaction

To determine the concentrationof a solution

Redox Characteristics

• Transfer of electrons• Transfer may occur to form ions• Oxidation – increase in oxidation state (loss of

electrons); reducing agent• Reduction – decrease in oxidation state (gain of

electrons); oxidizing agent


Reaction of Sodium and Chlorine


Find the oxidation states for each of the elements in each of the following compounds:

• K2Cr2O7

• CO32-

• MnO2

• PCl5

• SF4

Find the oxidation states for each of the elements in each of the following compounds:

• K2Cr2O7

• CO32-

• MnO2

• PCl5

• SF4


K = +1; Cr = +6; O = –2C = +4; O = –2Mn = +4; O = –2P = +5; Cl = –1S = +4; F = –1

Which of the following are oxidation-reduction reactions? Identify the substance oxidized and reduced.

a)Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

b)Cr2O72-(aq) + 2OH-(aq) 2CrO4

2-(aq) + H2O(l)

c)2CuCl(aq) CuCl2(aq) + Cu(s)

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Free Response2003B Answer the following questions that relate to chemical reactions. (a) Iron(III) oxide can be reduced with carbon monoxide according to the following equation. Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) A 16.2 L sample of CO(g) at 1.50 atm and 200.°C is combined with 15.39 g of Fe2O3(s). (i) How many moles of CO(g) are available for the reaction?

(ii) What is the limiting reactant for the reaction? Justify your answer with calculations. (iii) How many moles of Fe(s) are formed in the reaction?

. A 0.345 g sample of anhydrous BeC2O4, which contains an inert impurity, was dissolved in sufficient Water to produce 100.0 mL of solution. A 20.0 mL portion of this solution was titrated with KMnO4(aq). The balanced equation for the reaction that occurs is: 16H+(aq) + 2 MnO4

-(aq) + 5 C2O42-(aq) 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l)

The volume of the 0.0150 M KMnO4(aq) required to reach the equivalence point of the titration was 17.80 mL i. Identify the the substance oxidized and the substance reduced in the titration reaction. ii. For the titration at the equivalence point, calculate the number of moles of each of the following that reacted: a. MnO4

- b. C2O42-

iii. Calculate the total number of moles of C2O4

2- that were present in the 100.0 mL of prepared solution. iv. Calculate the mass percent of the BeC2O4 in the impure 0.345 g sample

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Saturday Study Session 1 1 st Class Reactions. Things to remember Solids, liquids, and gases can NOT...

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