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Midterm

Test 2 Review.docx

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Department
Chemistry
Course Code
CAS CH 131
Professor
Mark Grinstaff

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Molarity- moles of solute / volume of solvent (L) Molality- moles of solute / mass of solvent (kg) If there is a 300 mL solution of 0.5 M MgCl , how2many moles of free ions are in solution? What would be the new concentration if 100 mL of 1.0 M MgCl was added 2nto the solution? 0.3 L * 0.5 M = 0.15 moles MgCl 2 3 ions * 0.15 moles = 0.45 moles of free ions 1.1 L * 1 M = 0.1 moles additional Finding the new concentration- 0.15 moles + 0.1 additional moles = 0.25 moles 300 mL + 100 mL = 400 mL or .4 L 0.25 moles / 0.4 L = 0.625 M What is the molality of a solution that has 0.2 moles of BaBr into 100 mL of di-ethyl-ether? 2 dissolved The density of di-ethyl-ether is 0.713 g*mL -1 -1 100 mL * 0.731 g*mL = 71.3 g = 0.0713 kg of diethyl ether 0.2 moles of solute / 0.0713 kg = Disolution Reaction- AB (s) -> A(aq)+ B (aq) Ca(NO ) 3 2(s)-> Ca (aq) 2NO 3(aq) Ionic Precipatation Reaction- A +(aq) BB (aq)-> AB (s) Ca 2+ + SS 2- -> CaS (aq) (aq) (s) Acid/ Base Reaction- - + HA + H O -2 A + H O 3 - + HNO + 3 O -> 2O 3 + H 3 Oxidation/ Reduction Reaction (REDOX) Reaction- + + M 1 M 2 -> M 1 + M 2 Which can occur in AQ. acidic or basic solutions + + Au + Cu -> Au + Cu Combustion Reaction- CH +42 O -> C2 + 2 H O 2 2 Ionic Precipatation Reactions: Ions with multiple charges usually give insoluble ions (ppt in solution) Ions with single charges usually give soluble ions in solution Which solids dissolve in solution of water? MgO - does not dissolve KI - yes CaBr 2 - yes Na 2O 3 - yes KNO 3 -yes BaSO 4 -no Reaction in solution- BaCl 2 (aq) K 2SO (aq) -> BaSO(s) + 2 KCl) K and Cl are spectator ions in solution and do not participate in the reaction Net Ionic Precipatation Reaction- 2+ 2- - Ba (aq)+ SO 4 (aq) -> BaSO 4 (s) Which occurs in AQ solution Acid/ Base Reactions: pH = -log H 10ncentration - pOH = -log OH 10ncentration + + acid donates an H ion in solution- excess H base removes an H ion in solution- excess OH - pH < 7.0 acid pH > 7 base Oxidation Reduction: LEO the lion goes GER LEO- loss of electrons oxidation GER- gain of electrons reduction Reducing Agent- Agent t1at gives up electrons to Species during a2reaction, reducing Species 2 Oxidizing Agent- Agent t2at takes electrons from Species , oxidizi1g Species 1 Rules for Redox Reactions: st nd 1) 1 and 2 column give up their electrons 2) 7 column takes 1 electron 2- 2- 3) Oxygen exists as O and Sulfer exists as S 4) Oxygen dominates over Sulfer (SO ) 42- + - 5) Hydrogen exists as H , unless it exists in LiH (where it is H) 6) Oxygen exists as O in H O (H-2-O2H) -1/2 7) Oxygen exists as O in KO 2 Oxidation Numbers for- Cation + anioin = formal charge Fe 2 ?32Fe 3+ 32- Cr 2 72-? 2 Cr 7O 2- = 2- N2 H ? 2N-4H + PO4 ? p 4O = 3- 2- SO 42-? 6+ 2- = 2- N 2 +4KCl -> NOCl + KNO 3 What is the balanced equation and what are the oxidation numbers for each species? What species underwent an oxidation or reduction? [2N 4O ] + [K Cl] -> [N O Cl] + [K N 3O ]- + 5+ 2- 3+ 5+ N goes from 4+ to N and N disproportion reaction Half Reactions: A way to break up a redox reaction into its individual components to balance correctly + - Na -> Na + e Fe 3+ + 3e -> Fe Redox Reactions- H 2 2 MnO 4--> Mn 2+ + O 2 In acidic conditions Assign Oxidation Numbers- Create Half Reactions- Balance the half reactions to combine together- Determine Conditions (acidic or basic) and balance to form water Combine totally and double check atoms, molecules, and charges Assign oxidation numbers to each species- - 7+ 2- - MnO 4 = [Mn 4O ] + - H 2 2 = [2H 2O] O 2 = 2O Half Reactions- 7+ - 2+ Mn + 5 e -> Mn 2 O -> 2 O + 2 e - Figure out how these relate and combine(hint: electrons)- 2 Mn 7+ + 10 e - -> 2 Mn 2+ 10 O - -> 10 O + 10 e - 2 Mn 7+ + 10 O - -> 2 Mn 2+ + 10 O Make sure that electrons cancel on either side of the reaction arrow + + in acidic conditions- there is an excess resource of H O or 3 2- this is what helps us to balance and convert O into H O 2 + 2- + 10H + 8O + 6H -> 8 H O 2 + 10 H comes from the 5 H O fro2 o2r previous results combining everything gives- - + 2+ 5 H 2 +22 MnO 4 + 6 H -> 2 Mn + 5 O +28 H O 2 Check to make sure the the charges balance What was oxidized and what was reduced? Colligative Properties: Properties that are dependent on the # of particles, and not the properties of the particles 4 types- Vapor pressure depression- Boiling point elevation- Freezing point depression- Osmotic pressure- ****Watch out for dissociating ions**** X1 + x2 = 1 Mole fraction of solvent + mole fraction of solute = 1 solubility of solute affects the degree of change in the colligative properties 1 mole of K F3CN can 6issolve in a variation of ways that will affect the mole fraction 4 moles dissolve= 3 K + 1 FeCN 63- + 3+ - 10 moles dissolve = 3 K + Fe + 6 CN Vapor pressure depression P1 = x * p1 0 At constant temperature New pressure of solvent = mole fraction of solvent * pure pressure of solvent Dp1(delta p 1 = p1 – p1 0 the result of adding a solute to solution and the change in solvent vapor pressure it causes deviations from ideal result in solvent/sollute repulsion and attraction positive deviation- solvent/solute repulsion creates an increase in pvapor pressure from the expected negative deviation- solvent/solute attraction creates a decrease in the vapor pressure from the expected Boiling Point Elivation- DT B = k *Bmolality (moles of solute/kg of solvent) Boiling Point (with solute) = Boiling Point (with pure solvent) + DT B Boiling depends on the pressure of the atmosphere of the environment and not temperature Boiling depends on the vapor pressure above the liquid, so a decrease in natural vapor pressure creates an increase in boiling point temperature more temperature has to be added to increase the amount of vapor above the liquid, so the boiling point can be reached Freezinhg Point Depression- DT F k * mFlality (moles of solute/kg of solvent) Freezing Point (with solute) = Freezing Point (with pure solvent) - DT F Formation of a solid starts to increase the solute concentration in solution This is not favorable and temperature has to be lowered beyond solution of pure solvent formation of a solid needs molecules to interact with each other and lower there EK addition of solute changes the normal interactions of a pure solvent and alters the freezing point Osmotic pressure- pi = solute concentration*r*t pi = row * G * h row – density G – acceleration due to gravity (10 m*s ) -1 H - height Follows the property that materials have a preference to move down their concentration gradient A semi permeable wall does not allow for the sollutes to move, so the solvent is the only material that can pass through the wall Solvent moves from pure solvent to solvent with solute through the wall This movement creates an increase in pressure within the solute solution At a certain pressure, there is not a net flow of solvent in any direction Volatile vs. non-volatile- Volatile effects on vapor pressure depression Volatile material can enter the gas phase, and contributes to the total pressure above the liquid Partial pressure (volatile) = pure pressure (volatile) * mole fraction (volatile) Total pressure above the liquid= partial pressure (volatile) + partial pressure (solvent) There is a solution of 232 g of acetone (58 g*mole ). The pressure above the liquid before the solute is added is 0.35 ATM. 105 g of pyruvate (87 g*mole ) is added into the solution. What is the vapor pressure above the liquid after pyruvate is added? identify what the equation is and set up the factors get everything into moles calculate if there is ionic dissociation within the solution realize that the vapor pressure must drop for this question for this question- pyruvate dissolves into solution but is not an ionic compound and does not break up further -1 Instead of pyruvate, what if 93.5 g of benzene (78 g*mole ) dissolved into the acetone solution. What is the total pressure above the liquid, assuming that benzene is volatile and has a pure vapor pressure of 0.5 ATM. Reduced vapor pressure = mole fraction solvent * pure pressure of solvent Reduced vapor pressure solute = mole fraction solute * pure pressure solute Total pressure = reduced vapor pressure of solute + reduced vapor pressure of solvent **only for a volatile solute** -1 o Glyceral (92 g*mole ) dissolved when added to 500 m
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