Index UIL Science Chemistry Problems. indexIndex Oxidation Rules Oxidation Rules Oxidation Rules Oxidation Rules Oxidation Number/State Prob D/R Oxidation.

index UIL Science Chemistry Problems

indexIndex Oxidation Rules Oxidation Rules Oxidation Rules Oxidation Rules Oxidation Number/State Prob D/R Oxidation Number/State Prob D/R Oxidation Number/State Prob D/R Oxidation Number/State Prob D/R Neutralization Reaction Info Neutralization Reaction Info Neutralization Reaction Info Neutralization Reaction Info Neutralization Rxn Problems D/R/S Neutralization Rxn Problems D/R/S Neutralization Rxn Problems D/R/S Neutralization Rxn Problems D/R/S Equilibrium Constant Kc Info Equilibrium Constant Kc Info Equilibrium Constant Kc Info Equilibrium Constant Kc Info Equilibrium Constant Problems R/S Equilibrium Constant Problems R/S Equilibrium Constant Problems Equilibrium Constant Problems Weak Acid/ Weak Base Info Weak Acid/ Weak Base Info Weak Acid/ Weak Base Info Weak Acid/ Weak Base Info Weak Acid/Base Problems D/R/S Weak Acid/Base Problems D/R/S Weak Acid/Base Problems D/R/S Weak Acid/Base Problems D/R/S Light & Energy – Electrons Info Light & Energy – Electrons Info Light & Energy – Electrons Info Light & Energy – Electrons Info Light & Energy – Electrons Problems Light & Energy – Electrons Problems Light & Energy – Electrons Problems Light & Energy – Electrons Problems Solubility & K sp Info D/R/S Solubility & K sp Info D/R/S Solubility & K sp Info D/R/S Solubility & K sp Info D/R/S Solubility & K sp Problems D/R/S Solubility & K sp Problems D/R/S Solubility & K sp Problems D/R/S Solubility & K sp Problems D/R/S Reaction Rates Info Reaction Rates Problems

index Assigning Oxidation Numbers Rules Rules 1. Uncombined elements =0 2. Sum of oxidation numbers In a compound = 0 In an ion = charge of the ion 3. Group 1 = +1 Group 2 = +2 Group 13 = +3 4. F = -1 H = +1

index Assigning Oxidation Numbers Rules cont’d 5. O = - 2 ( except in peroxides O = -1, superoxides: O = 1/2 ) 6. In binary compounds only (electrically neutral) Group 17 = -1 Group 16 = -2 Group 15 = -3 (Follow the rules in order always restart at rule 1) Shortcut: H = +1, sum = charge, O = -2 will get you the answer in most problems.

index Oxidation Number/State Problems 1 D1 2008 21. What is the oxidation number assigned to the phosphorous atom in sodium phosphate? a)0 b)-3 c)+7 d)+5 e)+1 District 1 2008

index Oxidation Number/State Problems 1 21. What is the oxidation number assigned to the phosphorous atom in sodium phosphate? 1. Write the formula of the compound or just phosphate Na + PO 4 3- Na 3 PO 4 2. Use the oxidation rules. (apply first rule that applies, always restart at rule 1) Rule 5: oxygen is -2 (4 oxygens = -8) Rule 2: sum of poly. Ion = charge. Phosphorous plus -8 must equal -3 therefore P = +5 (1X – 8 = -3 solve for X) District 1 2008

index Oxidation Number/State Problem 2 D2 2008 21. The oxidation state of chromium in potassium dichromate is _____. a)3.5 b)2.5 c)12 d)6 e)5

index Oxidation Number/State Problem 2 D2 2008 21. The oxidation state of chromium in potassium dichromate is _____. K 2 Cr 2 O 7

index Neutralization Reactions Info Acid + base = ionic compound + water (HOH) Acid + base = ionic compound + water (HOH) H from acid combines with OH from base to make water. The anion (-) from the acid combines with cation (+) from the base to make the ionic compound. H from acid combines with OH from base to make water. The anion (-) from the acid combines with cation (+) from the base to make the ionic compound. Molarity (M) = #moles/liters of solution Molarity (M) = #moles/liters of solution pH = -log [H + ] [H + ] = 10 -pH pH = -log [H + ] [H + ] = 10 -pH pOH = - log [OH] [OH - ] = 10 -pOH pOH = - log [OH] [OH - ] = 10 -pOH pH + pOH = 14 pH + pOH = 14 If the concentration of the H + or OH - is greater than 1x 10 -6 M we can ignore the contribution by water. If the concentration of the H + or OH - is greater than 1x 10 -6 M we can ignore the contribution by water.

index Neutralization Info Strong acids = HCl (hydrochloric), HBr (hydrobromic), HI (hydroiodic), HNO 3 (nitric), H 2 SO 4 (1 st H only) (sulfuric), HClO 4 (perchloric). Strong acids = HCl (hydrochloric), HBr (hydrobromic), HI (hydroiodic), HNO 3 (nitric), H 2 SO 4 (1 st H only) (sulfuric), HClO 4 (perchloric). Strong bases = (group 1 hydroxides and Ca, Ba, & Sr) NaOH, KOH, LiOH, Ca(OH) 2, Ba(OH) 2, Sr(OH) 2 Strong bases = (group 1 hydroxides and Ca, Ba, & Sr) NaOH, KOH, LiOH, Ca(OH) 2, Ba(OH) 2, Sr(OH) 2 Dilution equation: C 1 V 1 =C 2 V 2 where C = concentration in molarity (M), molality (m), or % and V = volume

index Naming Acids Steps for Naming Acids: 1.Determine the ionic name. 2.Using the ending of the negative ion determine which prefix/suffix to use. -ide = hydro- -ic acid -ate = -ic acid -ite = -ous acid 3. add the suffix and prefix to the root of the anion name (chlor, fluor, brom, nitr, sulfur, phosphor 4.If ionic name had Per- or hypo- add it to the acid name.

index Acids ending in -ide 1.“-ide” ending is changed to “hydro- -ic” Formula-ideHydro- -ic HCl Hydrogen Chloride Hydrochloric acid HBr Hydrogen bromide Hydrobromic acid HF Hydrogen fluoride Hydrofluoric acid

index Acids ending in -ate 1.“-ate” ending is changed to “-ic” acid Formula-ate -ic HClO 3 Hydrogen chlorate Chloric acid HNO 3 Hydrogen nitrate Nitric acid H 2 CO 3 Hydrogen carbonate Carbonic acid H 2 SO 4 Hydrogen sulfate Sulfuric acid

index Acids ending in -ate 1.“-ate” ending is changed to “-ic” acid Formula-ate -ic H 3 PO 4 Hydrogen phosphate Phosphoric acid CH 3 COOH Hydrogen acetate Acetic acid HC 2 H 3 O 2 Acetic acid HClO 4 Hydrogen perchlorate Perchloric acid

index Acids ending in -ite 1.“-ite” ending is changed to “-ous” acid Formula-ite -ous HNO 2 Hydrogen nitrite Nitrous acid HClO 2 Hydrogen chlorite Chlorous acid H 2 SO 3 Hydrogen sulfite Sulfurous acid HClO Hydrogen hypochlorite Hypochlorous acid

index Neutralization Reaction 1 D1 2008 29. 15.0 mL of a 0.250 M H 2 SO 4 solution exactly neutralizes 20.0 mL of an NaOH solution. What is the molarity of the NaOH solution? a) 0.667 M b) 0.375 M c) 0.188 M d) 0.0938 M e) 0.250 M District 1 2008 Q29

index Neutralization Reaction 1 15.0 mL of a 0.250 M H 2 SO 4 solution exactly neutralizes 20.0 mL of an NaOH solution. What is the molarity of the NaOH solution? 1. Write the reaction 2. Balance it 3. Calculate the number of moles of acid and convert mL to L, then use the coefficients of the reaction to convert to moles of NaOH, divide by volume and covert to L Shortcut: calculate moles of acid, double it since it will need two OH to neutralize the two H, calculate molarity of NaOH

index Neutralization Reaction 2 D1 2008 35. Calculate the pH of a solution prepared by adding 80.0 mL of 0.100 M NaOH to 100 mL of 0,100 M HNO 3 solution. a) 2.24 b) 2.08 c) 2.16 d) 2.02 e) 1.95 District 1 2008 Q35

index Neutralization Reaction 2 Calculate the pH of a solution prepared by adding 80.0 mL of 0.100 M NaOH to 100 mL of 0.100 M HNO 3 solution. 1. This a limiting reagent problem where you need to calculate the amount of excess reactant. 2. Write the balanced equation 3. Calculate the moles of each. 4. Determine which one is limiting by taking the number of moles and dividing by the coefficient. (NaOH is Limiting reagent) 5. Use ICE (Initial, change, equilibrium) method to solve for excess acid. 6. Since nitric acid is strong [HNO 3 ] = [H + ] 7. Calculate molarity then pH

index Neutralization Reaction 2 Calculate the pH of a solution prepared by adding 80.0 mL of 0.100 M NaOH to 100 mL of 0.100 M HNO 3 solution. I0.00800 mol 0.0100 mol0 0 C-0.00800 -0.00800 + 0.00800 +0.00800 E 0 0.00200 0.00800 0.00800

index Neutralization Reactions 3 D2 2008 29. One millimole of HCl would neutralize exactly _______ millimole(s) of barium hydroxide? a) ½ b) 2 c) 3 d) 1 e) 3/2

index Neutralization Reactions 3 D2 2008 29. One millimole of HCl would neutralize exactly _______ millimole(s) of barium hydroxide? 1. Write the formula for barium hydroxide. Ba 2+ OH - = Ba(OH) 2 2. Write the balanced chem. reaction 3. Covert moles of HCl to moles of Ba(OH) 2 using coefficients of the reaction. Shortcut: Since only 1H and 2OH’s. 1 H will only neutralize half of it.

index Neutralization Reaction 4 D2 2008 35. What is the pH at the point in a titration at which 20.0 mL of 1.00 M KOH has been added to 25.0 mL of 1.00 M HBr? a) 0.95 b) 2.71 c) 1.22 d) 1.67 e) 3.84

index Neutralization Reaction 4 35. What is the pH at the point in a titration at which 20.0 mL of 1.00 M KOH has been added to 25.0 mL of 1.00 M HBr? 1. Write the balanced equation. 2. Calculate moles of each. 3. Determine limiting reagent. (KOH is LR) 4. Use ICE to determine amount of excess reactant. 5. Calculate molarity and pH.

index Neutralization Reaction 4 35. What is the pH at the point in a titration at which 20.0 mL of 1.00 M KOH has been added to 25.0 mL of 1.00 M HBr? I 0.0250 mol 0.0200 mol 0 0 C - 0.0200 mol - 0.0200 mol + 0.0200mol +0.0200 mol E 0.0050 mol 0 0.0200 0.0200

index Neutralization Reaction 5 D1 2009 28. What is [Cl - ] in the solution formed by mixing 50 mL of 0.20 M hydrochloric acid (aq) with 50 mL of 0.30 M sodium hydroxide (aq)? a) 0.25 M b) 0.30 M c) 0.10 M d) 0.20 M e) 0.50 M

index Neutralization Reaction 5 28. What is [Cl - ] in the solution formed by mixing 50 mL of 0.20 M hydrochloric acid (aq) with 50 mL of 0.30 M sodium hydroxide (aq)? 1. Not really a neutralization, but looks like one. 2. Write the formula of hydrochloric acid. 3. Calculate moles of HCl, Write dissociation reaction for HCl and covert mol HCl to moles of chloride 4. Since no additional Cl - from NaOH just calculate new molarity based on new volume of solution. Shortcut: Since HCl has 1 Cl then [HCl] = [Cl - ] than just use dilution equation C 1 V 1 =C 2 V 2

index Neutralization 6 R 2009 28. How many milliliters of 0.250 M HCl are needed to neutralize 5.00 g of Na 2 CO 3 ? 2 HCl + Na 2 CO 3  2 NaCl + CO 2 + H 2 O a) 377 mL b) 189 mL c) Impossible to calculate d) 94.3 mL e) 754 mL

index Neutralization 6 R 2009 28. How many milliliters of 0.250 M HCl are needed to neutralize 5.00 g of Na 2 CO 3 ? 2 HCl + Na 2 CO 3  2 NaCl + CO 2 + H 2 O 1. Determine moles of sodium carbonate by dividing by molar mass. 2. Convert moles of sodium carbonate to moles HCl using coefficients of the reaction 3. Calculate volume of HCl using concentration.

index Neutralization 7 S 2009 28. It required 25.0 mL of 0.333 M sodium hydroxide solution to completely neutralize 15.0 mL of sulfuric acid solution. What was the molarity of the sulfuric acid solution? a) 0.200 M b) 0.278 M c) 0.555 M d) 1.11 M e) 0.430 M

index Neutralization 7 S 2009 28. It required 25.0 mL of 0.333 M sodium hydroxide solution to completely neutralize 15.0 mL of sulfuric acid solution. What was the molarity of the sulfuric acid solution? 1. Calculate the number moles of NaOH used. 2. Write the balanced reaction. 3. Convert moles NaOH to moles H 2 SO 4 using the coefficients of the reaction. 4. Calculate the molarity of the solution

index Equilibrium Constants Info 1.No matter what the starting concentrations, at equilibrium the mass action expression is always equal to a certain number, the equilibrium constant (K c ) for that reaction at a specified temp. (known as equilibrium law for that reaction.) 2.The reaction quotient (Q) is also equal to the value of the mass action expression. At equilibrium Q = K c. If Q > K c then reaction is on the right side of equation moving toward the left. If Q< K c reaction is on the left moving toward the right. When a reaction starts with only reactants (no products), Q will increase as it approaches equilibrium. Q is a number that changes as the concentrations of the reactants and products change (as the reaction proceeds toward equilibrium) 3.Q shows where the reaction is in reference to equilibrium, it usually uses the starting concentrations or whatever concentrations are available 4.K c is a constant the equilibrium expression uses only the concentrations at equilibrium 5.For equilibrium reactions only gases and aqueous species count. Solids and liquids ignored. 6.Equilibrium constants change with temperature.

index Equilibrium Constants Info Predicting the Equilibrium Law:

index Equilibrium Constants Info Manipulating Equations for Chemical Equilibria So K c ’ = 1/K c If Reaction 1 is combined with Reaction 2 to make reaction 3 then K 3 = K 1 xK 2 Significance of the Magnitude of K: 1.K is very large: reaction proceeds far toward completion, The position of equilibrium lies far toward the products 2.K ≈ 1: the concentrations of reactants and products are nearly the same. 3.K is very small: Extremely small amounts of products are formed, position of equilibrium lies far toward the reactants

index Equilibrium Constants Info K p = equilibrium constant using partial pressure of gases T in Kelvin (+273K) R = Gas Constant

index Equilibrium Constant Problem 1 D1 2008 34. In which of the following cases will the least time be required to arrive at equilibrium a) Kc is a very small number b) Cannot tell since the time to arrive at equilibrium does not depend on K c c) K c is about 1 d) Cannot tell without knowing the value of K c e) K c is a very large number

index Equilibrium Constant Problem 1 D1 2008 34. In which of the following cases will the least time be required to arrive at equilibrium Cannot tell since the time to arrive at equilibrium does not depend on K c

index Equilibrium Constant Problem 2 D1 2010 39. Given: H 3 O + (aq) + NO 2 - (aq)  HNO 2 (aq) + H 2 O K c = 3.78 x 10 3 at T = 298 K Calculate Kc for the reaction: NO 2 - (aq) + H 2 O (l)  HNO 2 (aq) + OH - (aq) a) 0.30 x 10 -11 b) 0.83 x 10 -9 c) 0.57 x 10 -9 d) 0.15 x 10 -10 e) 0.38 x 10 -10

index Equilibrium Constant Problem 2 D1 2010 39. Given: H 3 O + (aq) + NO 2 - (aq)  HNO 2 (aq) + H 2 O K c = 3.78 x 10 3 at T = 298 K Calculate Kc for the reaction: NO 2 - (aq) + H 2 O (l)  HNO 2 (aq) + OH - (aq) 1. Need to add a reaction to it that we know the K c for that changes H 3 O + to OH - like 2H 2 O = H 3 O + + OH -. 2. When you add reactions the K c of the new reaction will equal the K’s multiplied together.

index Equilibrium Constant Problem 2 D1 2010 39. Given: H 3 O + (aq) + NO 2 - (aq)  HNO 2 (aq) + H 2 O K c = 3.78 x 10 3 at T = 298 K Calculate Kc for the reaction: NO 2 - (aq) + H 2 O (l)  HNO 2 (aq) + OH - (aq) When adding reactions, like substances on opposite sides of the arrow cancel.

index Equilibrium Constant Problem 3 D2 2010 39. K c = 1.506 x 10 -5 at T = 100 C for the reaction A (g)  2B (g). What is K c for the reaction: 2B(g)  A (g) at T = 100 C? a) 119500 b) 93000 c) 66400 d) 26600 e) 46500

index Equilibrium Constant Problem 3 D2 2010 39. K c = 1.506 x 10 -5 at T = 100 C for the reaction A (g)  2B (g). What is K c for the reaction: 2B(g)  A (g) at T = 100 C? The Kc of the reverse of a reaction is 1/Kc of the forward reaction.

index Equilibrium Constant Problem 4 R 2009 34. What is the form of the equilibrium constant for the reaction 2 HgO (s)  2 Hg (l) + O 2 (g) ? a) Kc = [Hg] 2 [O 2 ] b) Kc = [O 2 ] c) None of the other answers is correct d) Kc = [O 2 ]/[HgO] 2 e) Kc = [Hg] 2 [O 2 ]/[HgO] 2

index Equilibrium Constant Problem 4 R 2009 34. What is the form of the equilibrium constant for the reaction 2 HgO (s)  2 Hg (l) + O 2 (g) ? Since we ignore liquids and solids use only the oxygen gas when writing the equilibrium constant expression. K c = [O 2 ]

index Equilibrium Constant Problem 5 S 2009 34. Given the reaction 2 CO (g) + O 2 (g)  2 CO 2 (g) which is the relationship between K c and K p ? a) K p = K c (RT) -2 b) K p = K c c) K p = K c RT d) K p = K c (RT) -1 e) K p = K c (RT) 2

index Equilibrium Constant Problem 5 S 2009 34. Given the reaction 2 CO (g) + O 2 (g)  2 CO 2 (g) which is the relationship between K c and K p ?

index Equilibrium Constant Problem 6 R 2010 34. If K c = 0.040, at 450 C, what is the K p for PCl 5 (g)  PCl 3 (g) + Cl 2 (g) ? a) 6.7 x 10 -4 b) 5.2 x 10 -2 c) 0.40 d) 2.4 e) 0.64

index Equilibrium Constant Problem 6 R 2010 34. If K c = 0.040, at 450 C, what is the K p for PCl 5 (g)  PCl 3 (g) + Cl 2 (g) ?

index Equilibrium Constant Problem 7 S 2010 34. The equilibrium constant, K c, for the gaseous reaction HCHO (g)  H 2 (g) + CO (g) has the numerical value 0.50 at 600°C. A mixture of HCHO, H 2, and CO is introduced into a flask at 600C. After a short time, analysis of a small sample of the reaction mixture shows the concentration to be [HCHO] = 1.5, [H 2 ] = 0.5 and [CO] = 1.0 moles/liter. Which of the following statements is true? a) The reaction mixture is not at equilibrium, but will move toward equilibrium by using up more HCHO b) The reaction mixture is not at equilibrium, but no further reaction will occur. c) The reaction mixture is at equilibrium d) The forward rate of this reaction is the same as the reverse rate at those concentrations e) The reaction mixture is not at equilibrium, but will move toward equilibrium by forming more HCHO

index Equilibrium Constant Problem 7 S 2010 34. The equilibrium constant, K c, for the gaseous reaction HCHO (g)  H 2 (g) + CO (g) has the numerical value 0.50 at 600°C. A mixture of HCHO, H 2, and CO is introduced into a flask at 600C. After a short time, analysis of a small sample of the reaction mixture shows the concentration to be [HCHO] = 1.5, [H 2 ] = 0.5 and [CO] = 1.0 moles/liter. Which of the following statements is true? 1. Calculate Q (reaction quotient) and compare it to the value of K c 2. Since Q < Kc, (too much HCHO) reaction is not at equilibrium and will move towards equilibrium by using up HCHO

index Weak Acid – Weak Base Info 1. K w = [H + ][OH - ] = 1 x 10 -14 (at 25° C) 2. Acid ionization constant, (K a ) shows the strength of acids (the amount of dissociation) (usually only for weak acids) 1. Base ionization constant (K b ) Equilibrium constants (K c, K a, K sp, K b ) only involve aqueous (aq) and gases (g). No solids, no liquids.

index Weak Acid – Weak Base Info 1. Weak Acids 2. Conjugate Base of a weak acid will produce OH - 3. K a x K b = K w ( K w = 1 x 10 -14 ) for a conjugate acid –base pair like HF and F -

index Weak Acid – Weak Base Info 1. Weak Bases 2. Conjugate acid of a weak base will produce H +

index Weak Acid – Weak Base Info 1. Ionic compounds with conjugate acids or conjugate base of strong acids or bases will produce neutral solutions. Ex: NaCl, KNO 3, 2. Ionic compounds with only conjugate bases of weak acids will produce basic solutions. Examples: KF, NaCH 3 COO, 3. Ionic compounds with only conjugate acids of a weak base will produce acidic solutions. Examples: NH 4 Cl 4. Ionic compounds with both – pH is determined by relative strengths 5. The higher K a the stronger the acid the higher the K b the stronger the base.

index Weak Acid – Weak Base Problem 1 D2 2008 34. Consider: H 3 O + (aq) + OH - (aq) ↔ 2 H 2 O If the initial concentrations are [H 3 O + (aq)] = 0.1 M [OH - (aq)] = 0.7 M at equilibrium, [OH - (aq)] will be _________. a) 0.5 M b) 0.6 M c) 0.8 M d) 0.7 M e) 0.0 M

index Weak Acid – Weak Base Problem 1 D2 2008 34. Consider: H 3 O + (aq) + OH - (aq) ↔ 2 H 2 O If the initial concentrations are [H 3 O + (aq)] = 0.1 M [OH - (aq)] = 0.7 M at equilibrium, [OH - (aq)] will be _________. 1. Write the reaction, use ICE (initial, change, & equilibrium) to determine equilibrium values. Since volume doesn’t change we can use molarity. 2. Use the equilibrium expression, K w = 1 x 10 -14 = [H 3 O + ][OH - ] (equilibrium concentrations) to solve for X.

index Weak Acid – Weak Base Problem 1 D2 2008 34. Consider: H 3 O + (aq) + OH - (aq) ↔ 2 H 2 O If the initial concentrations are [H 3 O + (aq)] = 0.1 M [OH - (aq)] = 0.7 M at equilibrium, [OH - (aq)] will be _________. Since X can’t be.7 it must be.1 therefore 0.7-X = 0.6 0.6 M

index Weak Acid – Weak base 2 D1 2009 35. For a given weak acid, HA, the value of K a ________. a) None of the other answers are correct b) Cannot be less than 10 -7 c) Will change with pH d) Does not change with temperature e) Cannot be greater than 10 -7

index Weak Acid – Weak base 2 D1 2009 35. For a given weak acid, HA, the value of K a ________. a) None of the other answers are correct b) Cannot be less than 10 -7 c) Will change with pH d) Does not change with temperature e) Cannot be greater than 10 -7 A. None of the other answers are correct

index Weak Acid – Weak Base 4 35. Given the following acid ionization constants, K a HF7.2 x 10 -4 HNO 2 4.5 x 10 -4 CH 3 COOH1.8 x 10 -5 HClO3.5 x 10 -8 HCN4.0 x 10 -10 Which is the weakest acid? a) HClO b) HNO 2 c) HF d) CH 3 COOH e) HCN

index Weak Acid – Weak Base 4 35. Given the following acid ionization constants, K a HF7.2 x 10 -4 HNO 2 4.5 x 10 -4 CH 3 COOH1.8 x 10 -5 HClO3.5 x 10 -8 HCN4.0 x 10 -10 Which is the weakest acid? HCN has the smallest K a

index Weak Acid – Weak Base 5 R 2009 35. 0.50 moles of HCN are added to a liter of water. What is the pH if K a for HCN = 4.0 x 10 -10 ? a) 5.35 b) 4.85 c) 9.40 d) 4.35 e) 4.69

index Weak Acid – Weak Base 5 R 2009 35. 0.50 moles of HCN are added to a liter of water. What is the pH if K a for HCN = 4.0 x 10 -10 ? 1. Write the balanced dissociation reaction for HCN 2. Calculate the molarity of HCN, since volume doesn’t change you can use molarity. 3. Use ICE to determine equilibrium values (use X for the change) 4. Write the equilibrium expression and use the value of Ka to solve for X 4. Use this to determine molarity of H 3 O +, then calculate pH

index Weak Acid – Weak Base 5 R 2009 35. 0.50 moles of HCN are added to a liter of water. What is the pH if K a for HCN = 4.0 x 10 -10 ? *Simplification can be made can be made if [HA] initial ≥ 400 K a

index Weak Acid/Weak Base 6 R 2009 40. A solution is prepared by adding 2.0 moles of acetic acid and 1.0 mole of sodium acetate to enough water to make 1.0 L of solution. Finally, that solution was diluted to a final volume of 5.0L. What is the pH of the final solution? For acetic acid K a = 1.8 x 10 -5 a) 2.6 b) 4.0 c) 4.4 d) 5.0 e) 7.5

index Weak Acid/Weak Base 6 R 2009 40. A solution is prepared by adding 2.0 moles of acetic acid and 1.0 mole of sodium acetate to enough water to make 1.0 L of solution. Finally, that solution was diluted to a final volume of 5.0L. What is the pH of the final solution? For acetic acid K a = 1.8 x 10 -5 1. Calculate the molarity of acetic acid and sodium acetate. 2. Write the balanced dissociation equation for acetic acid. Use ICE to determine the equilibrium values. 3. Write the equilibrium expression and use the value of Ka to solve for X 4. Use this to solve for molarity then pH

index Weak Acid/Weak Base 6 R 2009 40. A solution is prepared by adding 2.0 moles of acetic acid and 1.0 mole of sodium acetate to enough water to make 1.0 L of solution. Finally, that solution was diluted to a final volume of 5.0L. What is the pH of the final solution? For acetic acid K a = 1.8 x 10 -5 *Simplification can be made can be made if [HA] initial ≥ 400 K a

index Weak Acid/Weak Base 7 S 2009 35. K a for HX is 1.31 x 10 -3 at 25 o C. What are [H 3 O + ] and [HX] for a 0.0200 M solution of HX? [H 3 O + ] [HX] a) 4.5 x 10 -3 1.6 x 10 -2 b) 5.1 x 10 -3 1.5 x 10 -2 c) 5.3 x 10 -3 1.5 x 10 -2 d) 5.7 x 10 -3 1.4 x 10 -2 e) 6.0 x 10 -3 1.4 x 10 -2

index Weak Acid/Weak Base 7 S 2009 35. K a for HX is 1.31 x 10 -3 at 25 o C. What are [H 3 O + ] and [HX] for a 0.0200 M solution of HX? 2. Write the balanced dissociation equation for HX. Use ICE to determine the equilibrium values. 3. Write the equilibrium expression and use the value of Ka to solve for X. [HX]/Ka = 15.2 simplification doesn’t work. 4. Use this to solve for molarity.

index Weak Acid/Weak Base 7 S 2009 35. K a for HX is 1.31 x 10 -3 at 25 o C. What are [H 3 O + ] and [HX] for a 0.0200 M solution of HX?

index Weak Acid/Weak Base 8 R 2010 40. Given that K w = 3.995 x 10 -13 at some temperature, what is the pH of pure water at the temperature? a) 6.2 b) 6.6 c) 7.0 d) 7.4 e) 7.8

index Weak Acid/Weak Base 8 R 2010 40. Given that K w = 3.995 x 10 -13 at some temperature, what is the pH of pure water at the temperature? Since K w = [H 3 O + ][OH - ]

index Light & Energy – Electrons Info 1. Speed of light (c) = 3.00 x 10 8 m/s 2. Wavelength = λ, frequency = ν (hz, 1/s) c = λ ×ν 1m = 10 9 nm 1m = 10 10 A (angstrom) 3. Energy: E = hν = hc/λ h = Planck’s constant 4. spectrum: (low freq/energy to high freq/energy) radio (short-TV-micro)-Ir-Vis-UV-Xrays-Gamma Rays ROY G BIV 5. Line Spectrum of Hydrogen (Finds wavelength of light emission from hydrogen)

index Light & Energy – Electrons Info 6. Bohr’s equation – calculates energy of orbits of hydrogen 7. De Broglie’s equation – calculates the wavelength of electrons 1J = 1N  m 1N = 1Kg m/s 2

index Principle Quantum number (n) Main energy level (shell) 1.Allowable value: positive integer (1,2,3,4…) 2.As n   energy  3.n 2 = # orbitals in that shell (since 1 orbital will hold 2 e - ) 4. 2n 2 = # of electrons in that shell 5.n = # of sublevels or subshells

index Angular momentum quantum number (l) Subsidiary quantum number Subshell or sublevel, Shape of the orbital 1.Allowable value: l = 0….(n-1) all positive integers between 0 and n-1 2.The value of l = shape l=0 “s” shape l=1 “p” shape l=2 “d” shape l = 3 “f” shape 3.As l   energy 

index Magnetic Quantum number (m) Which orbital – orientation around the nucleus 1.Allowable values: m = ( -l …0…+l) all integers between –l and +l including zero. 2.“s” subshell – 1 orbital (2 e - ) 3.“p” subshell – 3 orbitals (6 e - ) 4.“d” subshell – 5 orbitals (10 e - ) 5.“f” subshell – 7 orbitals (14 e - ) 6.“g” subshell – 9 orbitals (18 e) (not occupied in ground state)

index Spin Quantum number 1.Allowable value: + ½ or – ½ 2.Electrons exist in two states of spin. One spins one way the other spins the other way. 3.An orbital will hold 2 e - - to be in the same orbital they must have opposite spins (Pauli Exclusion Principle) 4.Spin is usually represented by an up or down arrow.  - spin in one direction  - spin in the other direction

index Rules of Electrons 1.Each row on the periodic table represents a shell. 2.Outer shell (ring) can have a max of 8 e - 3.Shell #1 (n=1): 1 subshell (s), 1orbital, & can have a max of 2 e - 4.Shell #2 (n=2): 2 subshells (s,p), 4 orbitals, & can have a max of 8 e - 5.Shell #3 (n=3): 3 subshells (s,p,d), 9 orbitals, & can have a max of 18 e - 6.Shell #4 (n=4): 4 subshells (s,p,d,f), 16 orbitals, & can have a max of 32 e -

index Some Patterns in the Rules: 1.# of orbitals in a subshell: 1, 3, 5, 7 2.# of electrons in a subshell: 2, 6, 10, 14 3.n = shell number = row in the periodic table 4.# of subshells = n 5.# of orbitals in a shell = n 2 (1, 4, 9, 16…) 6.# of electrons in a shell = 2n 2 (2, 8, 18, 32)

index s n=row d n = (row# -1) p n = row# f n = (row# - 2)

index Light & Energy – Electrons Info

index Light & Energy – Electrons Problem 1 D1 2008 23. The number of valence electrons in one tin atom is ______ a) 0 b) 4 c) 2 d) 14 e) 8

index Light & Energy – Electrons Problem 2 D1 2008 25. ________equals Planck’s constant times the frequency of the electromagnetic radiation a) Velocity b) Wavelength c) Intensity d) Quantum e) Energy

index Light & Energy – Electrons Problem 3 D1 2008 26. When n=4, l = 2, and m l = -1, what atomic orbital type is this? a) 4d b) 2p c) 2d d) 4p e) 5d

index Light & Energy – Electrons Problem 4 D2 2008 23. The number of valence electrons in a gallium atom is _____ a) 13 b) 1 c) 2 d) 31 e) 3

index Light & Energy – Electrons Problem 5 D2 2008 25. The higher the energy of electromagnetic radiation, the ____. a) Higher the velocity of light b) Greater its mass c) Longer its wavelength d) Lower its frequency e) Shorter its wavelength

index Light & Energy – Electrons Problem 6 D2 2008 26. Is n=5, l =-3, m l =-2, m s = + ½ a valid set of quantum numbers for a hydrogen atom? a) Yes b) No, because l cannot be negative c) No, because m cannot be negative d) No, because n cannot be as large as 5 e) No, because l must be equal to n-1

index Light & Energy – Electrons Problem 6 D2 2008 26. Is n=5, l =-3, m l =-2, m s = + ½ a valid set of quantum numbers for a hydrogen atom? No, because l cannot be negative

index Light & Energy – Electrons Problem 7 D1 2010 24. What is the maximum number of spatial orbitals there can be with n=2 and l=2? a) 1 b) 5 c) 2 d) None e) 3

index Light & Energy – Electrons Problem 7 D1 2010 24. What is the maximum number of spatial orbitals there can be with n=2 and l=2? None. Acceptable values of l = 0…(n-1)

index Light & Energy – Electrons Problem 8 D1 2010 23. _____light has the highest energy per photon. a) Infrared b) Red c) Ultraviolet d) Green e) Blue

index Light & Energy – Electrons Problem 9 D2 2010 23. What is the frequency of photons with wavelength = 6580 A? a) 4.6 x10 15 /s b) 8.0 x 10 14 /s c) 3.0 x 10 14 /s d) 4.6 x 10 14 /s e) 8.7 x 10 14 /s

index Light & Energy – Electrons Problem 9 D2 2010 23. What is the frequency of photons with wavelength = 6580 A?

index Light & Energy – Electrons Problem 10 D2 2010 24. The number of spatial orbitals with the quantum numbers n=6, l=2, and m l = -1 a) 3 b) 1 c) 9 d) 5 e) 7

index Light & Energy – Electrons Problem 11 D1/D2 2010 25. The Lanthanide and actinide series (inner transition metals) consists of elements filling what orbitals? a) g b) d c) p d) s e) f

index Light & Energy – Electrons Problem 12 R 2009 23. What is the wavelength of an electron traveling one tenth the speed of light? a) 7.3 x 10 -7 m b) 1.6 x 10 -19 m c) 3.0 x 10 8 m d) 2.4 x 10 -15 m e) 2.4 x 10 -11 m

index Light & Energy – Electrons Problem 12 R 2009 23. What is the wavelength of an electron traveling one tenth the speed of light? 1. Use De Broglie equation.

index Light & Energy – Electrons Problem 13 S 2009/R 2010 23/22. The number of unpaired electrons in the lowest energy electron configuration of an isolated sulfur atom? a) 1 b) 3 c) 4 d) 0 e) 2

index Light & Energy – Electrons Problem 13 S 2009/R 2010 23. The number of unpaired electrons in the lowest energy electron configuration of an isolated sulfur atom? 1s 2 2s 2 2p 6 3s 2 3p 4 1s 2 2s 2 2p 6 3s 2 3p 4 3p 2

index Light & Energy – Electrons Problem 14 S 2009 25. The n and l quantum numbers of the highest energy electrons of an element are n=4 and l=2. a) Nonmetal b) Noble gas c) d-transition metal d) Representative element e) f –transition metal

index Light & Energy – Electrons Problem 14 S 2009 25. The n and l quantum numbers of the highest energy electrons of an element are n=4 and l=2. n = 4, l=2 is 4d therefore d-transition metal

index Light & Energy – Electrons Problem 15 R 2010 24. For a subsidiary quantum number value of 3, what are the possible values for the angular momentum quantum number, m l ? a) +2, +1, 0, -1, -1 b) None of the other choices are correct c) +3, +2, +1, 0, -1, -2, -3 d) +3, +2, +1, 0 e) +2, +1, 0

index Light & Energy – Electrons Problem 15 R 2010 24. For a subsidiary quantum number value of 3, what are the possible values for the angular momentum quantum number, m l ? a) +2, +1, 0, -1, -1 b) None of the other choices are correct c) +3, +2, +1, 0, -1, -2, -3 d) +3, +2, +1, 0 e) +2, +1, 0 Subsidiary quantum number = l l = 0….(n-1) = 0, +1, +2

index Light & Energy – Electrons Problem 15 R 2010 29. A maximum of _____ electrons having a given spin can occupy any given molecular orbital. a) 0 b) ½ c) 4 d) 2 e) 1

index Light & Energy – Electrons Problem 15 R 2010 29. A maximum of _____ electrons having a given spin can occupy any given molecular orbital. a) 0 b) ½ c) 4 d) 2 e) 1 Pauli exclusion principle says no two electrons can have the same 4 quantum numbers (can’t have two electrons in the same orbital with the same spin.

index Light & Energy – Electrons Problem 16 S 2010 23. In the Bohr atom, what is the wavelength of light emitted when the electron goes from the n=4 level to the n=3 level? Rydberg’s constant is 1.097x10 7 m -1 a) 9100 nm b) 1900 nm c) 3000 nm d) 4200 nm e) 6200 nm

index Light & Energy – Electrons Problem 16 S 2010 23. In the Bohr atom, what is the wavelength of light emitted when the electron goes from the n=4 level to the n=3 level? Rydberg’s constant is 1.097x10 7 m -1

index Light & Energy – Electrons Problem 17 S 2010 24. How many spatial orbitals exist for a g sub-energy level? a) 5 b) 1 c) 9 d) 7 e) 3

index Light & Energy – Electrons Problem 17 S 2010 24. How many spatial orbitals exist for a g sub-energy level? a) 5 b) 1 c) 9 d) 7 e) 3 1, 3, 5, 7, 9, 11… s = 1, p = 3, d = 5, f = 7, g = 9

index Light & Energy – Electrons Problem 18 S 2010 25. If the n and l quantum numbers of the last electron added in the Aufbau sequence for an element are n=4 and l=3, the element is a(n) ________. a) Noble gas b) Representative element c) Non-metal d) f-transition metal e) d-transition metal

index Light & Energy – Electrons Problem 18 S 2010 25. If the n and l quantum numbers of the last electron added in the Aufbau sequence for an element are n=4 and l=3, the element is a(n) ________. a) Noble gas b) Representative element c) Non-metal d) f-transition metal e) d-transition metal n = 4, l = 3 is 4f therefore f-transition metal

index Solubility & K sp Info 1. Dissolved ionic compounds 100% dissociate. 2. When two solutions are mixed you must check the new possible compounds (double replacement reaction) to see if they are soluble or insoluble using the solubility rules. If they are both soluble then no reaction (NR) occurs. 3. Only one of the products can be insoluble. If one is, then the solution turns cloudy from the formation of the precipitate. 4. Precipitate – tiny solid particles formed when formation of an insoluble compound occurs

index Solubility & K sp Info Solubility Product Constant, K sp (another equilibrium constant) 1. All ionic compounds will dissolve at least a little bit (even the ones we consider insoluble) 2. The Ksp tells us how soluble an ionic compound is. Higher Ksp = more soluble. 3. Saturated solution = no more will dissolve. 4. When an ionic compound is dissolved and forms a saturated solution an equilibrium is set up. Precipitate will form if: ion product (Q sp ) > K sp (supersaturated) No Precipitate will form if: ion product = K sp (saturated) ion product < K sp (unsaturated)

index Solubility Rules Solubility Rules of Ionic compounds 1.All compounds w/ group 1metals or NH 4 + (ammonium) are soluble. 2. All compounds with nitrate, acetate, perchlorate are soluble. 3.All compounds with chloride are soluble except AgCl, PbCl 2, Hg 2 Cl 2 (mercury I) (Usually Iodides and bromides as well) 4.All sulfates are soluble except Ba 2+, Ca 2+, Pb 2+, Hg 2+ 5.All compounds that contain PO 4 3-, CO 3 2-, SO 3 2-, and S 2- are insoluble except those of group 1 and NH 4 + 6.All hydroxides and all metal oxides are insoluble except Group 1, NH 4 +, Ca 2+, Sr 2+, and Ba 2+

index Dissociation Reaction Ionic compounds  ions

index Net Ionic Equations Net ionic equation Na + & NO 3 - : spectator ions Total ionic

index Solubility & K sp Prob 1 R 2009 21 Which reactants will generate a precipitated product? a) AgNO 3 + CH 3 COONa b) FeCl 2 + Na 2 SO 4 c) (NH 4 ) 2 SO 4 + KOH d) BaCl 2 + Na 2 SO 4 e) NaH + H 2 O

index Solubility & K sp Prob 1 R 2009 21 Which reactants will generate a precipitated product? a) AgNO 3 + CH 3 COONa = Sodium nitrate (sol) + silver acetate (sol) b) FeCl 2 + Na 2 SO 4 = sodium chloride (sol) + iron (II) sulfate (sol) c) (NH 4 ) 2 SO 4 + KOH all group 1 and NH 4 are sol d) BaCl 2 + Na 2 SO 4 = sodium chloride (sol) + barium sulfate (insol) e) NaH + H 2 O (hydrides are very reactive with water) 1. Switch the ions (double replacement reaction) 2. Use solubility rules to see which one is insoluble D because all sulfates are soluble except Ba, Ca, Pb (II), & Hg (II).

index Solubility & K sp Prob 2 R 2009 37. What is the concentration of Pb 2+ and F - in a saturated solution of PbF 2 ? K sp = 3.7 x 10 -8 [Pb 2+ ] [F - ] a) 2.1 x 10 -3 4.2 x 10 -3 b) 1.9 x 10 -4 3.8 x 10 -4 c) 2.1 x 10 -3 2.1 x 10 -3 d) 3.3 x 10 -3 6.6 x 10 -3 e) 9.6 x 10 -5 1.9 x 10 -4

index Solubility & K sp Prob 2 R 2009 37. What is the concentration of Pb 2+ and F - in a saturated solution of PbF 2 ? K sp = 3.7 x 10 -8 1. Write dissociation reaction. 2. Write equilibrium expression. 3. Use ICE to determine equilibrium values in terms of X. Assume that you start with zero product. 4. Use equilibrium expression and value of Ksp to solve for X. 5. Solve for concentrations.

index Solubility & K sp Prob 2 R 2009 37. What is the concentration of Pb 2+ and F - in a saturated solution of PbF 2 ? K sp = 3.7 x 10 -8

index Solubility & K sp Prob 3 S 2009 22. Write the balanced, total ionic equation for the reaction of calcium hydroxide with nitric acid using integers. What is the sum of the stoichiometric coefficients of the ions on the left side of the balanced equation? a) Three b) Four c) Seven d) Six e) five

index Solubility & K sp Prob 3 S 2009 22. Write the balanced, total ionic equation for the reaction of calcium hydroxide with nitric acid using integers. What is the sum of the stoichiometric coefficients of the ions on the left side of the balanced equation? 7

index Solubility & K sp Prob 4 S 2009 37. If a solution is 1.00 x 10 -5 M in manganese nitrate and 1.50 x 10 -3 M in aqueous ammonia, will manganese hydroxide precipitate? K b (ammonia) = 1.8 x 10 -5 K sp (manganese hyrdoxide)= 2.5 x 10 -13 a) No, because Q sp >K sp b) No, because K sp >Q sp c) Yes, because K sp >Q sp d) Yes, because Q sp >K sp e) Yes, because Q sp =K sp

index Solubility & K sp Prob 4 S 2009 37. If a solution is 1.00 x 10 -5 M in manganese nitrate and 1.50 x 10 -3 M in aqueous ammonia, will manganese hydroxide precipitate? K b (ammonia) = 1.8 x 10 -5 K sp (manganese hyrdoxide)= 2.5 x 10 -13 1. Determine the concentration of OH formed by ammonia by writing the dissociation reaction for ammmonia, using ICE to determine the equilibrium values in terms of X, using the value of K b to solve for X, use X to determine the M 2. Write the dissociation reaction for Manganese hydroxide. Write the K sp expression for reaction. Use the starting concentrations to calculate Q sp. Compare Q sp with K sp to see if it precipitates

index Solubility & K sp Prob 4 S 2009 37. If a solution is 1.00 x 10 -5 M in manganese (IV) nitrate and 1.50 x 10 -3 M in aqueous ammonia, will manganese (IV) hydroxide precipitate? K b (ammonia) = 1.8 x 10 -5 K sp (manganese hyrdoxide)= 2.5 x 10 -13 *Simplification can be made can be made if [HA] initial ≥ 400 K a No, because K sp >Q sp

index Solubility & K sp Prob 5 R 2010 36. How many mL of a 0.0010 M chloride solution must be added to a 100 mL solution of 7.2 x 10 -5 M Ag + solution for AgCl to begin to precipitate? K sp of AgCl = 1.8x10 -10 a) Need to know the solution pH to solve this problem b) 2.6 mL c) 0.25 mL d) 0.030 mL e) 4.1 mL

index Solubility & K sp Prob 5 R 2010 36. How many mL of a 0.0010 M chloride solution must be added to a 100 mL solution of 7.2 x 10 -5 M Ag + solution for AgCl to begin to precipitate? K sp of AgCl = 1.8x10 -10 1. Write the dissociation reaction for AgCl. 1. Write the dissociation reaction for AgCl. 2. Write K sp expression for AgCl 3. Use K sp expression and value of K sp to calculate the M of Cl - to make a saturated solution. 4. Use this molarity to calculate how much Cl to add

index Solubility & K sp Prob 5 R 2010 36. How many mL of a 0.0010 M chloride solution must be added to a 100 mL solution of 7.2 x 10 -5 M Ag + solution for AgCl to begin to precipitate? K sp of AgCl = 1.8x10 -10

index Solubility & K sp Prob 6 S 2010 36. Magnesium hydroxide, which has a solubility product constant of 1.5 x 10 -11, dissolves in water when NH 4 Cl is added to the water because __________ a) Mg 2+ forms a very stable complex ion with ammonia b) Hydroxide ion reacts with chloride ion to form the weak acid HClO c) Hydroxide ion is converted to NH 4 OH by reaction with NH 4 + d) Part of the magnesium hydroxide is oxidized to form a different species. e) Magnesium chloride is a salt and completely ionized in water solution.

index Solubility & K sp Prob 6 S 2010 36. Magnesium hydroxide, which has a solubility product constant of 1.5 x 10 -11, dissolves in water when NH 4 Cl is added to the water because __________ a) Mg 2+ forms a very stable complex ion with ammonia b) Hydroxide ion reacts with chloride ion to form the weak acid HClO c) Hydroxide ion is converted to NH 4 OH by reaction with NH 4 + d) Part of the magnesium hydroxide is oxidized to form a different species. e) Magnesium chloride is a salt and completely ionized in water solution. Since NH 4 + is the conjugate acid of a weak base it will react with some OH - to establish an equilibrium

index Solubility & K sp Prob 7 D1 2008 36. The value of K sp for copper (I) iodide is 1.0 x 10 -12. What is the solubility for copper (I) iodide in grams per liter? a) 1.0 x 10 -12 b) 9.0 x 10 -6 c) 1.9 x 10 -4 d) 1.0 x 10 -6 e) 1.0 x 10 -4

index Solubility & K sp Prob 7 D1 2008 36. The value of K sp for copper (I) iodide is 1.0 x 10 -12. What is the solubility for copper (I) iodide in grams per liter? 1.

index Rate Laws Info

index Solution Concentration Info

index Types of Colloids Sol: solid in liquid(Paints, mud) Gel: solid network in liquid(gelatin, Jello) Liquid emulsion: liquid in liquid(milk, mayo) Foam: gas in a liquid(Shaving/whipped cream) Solid aerosol: solid in a gas(smoke, dust) Liquid aerosol: liquid in a gas (fog, cloud, aerosols) Solid emulsion: liquid in a solid (cheese, butter)

indexConcentration The amount of solute dissolved in a specific quantity of solvent or solution. Different methods 1.Molarity (mol/L) 2.Molality (mol/kg) 3.Mass percent (%) 4.Parts per million (ppm) 5.Parts per billion (ppb)

Index UIL Science Chemistry Problems. indexIndex Oxidation Rules Oxidation Rules Oxidation Rules Oxidation Rules Oxidation Number/State Prob D/R Oxidation.

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Index UIL Science Chemistry Problems. indexIndex Oxidation Rules Oxidation Rules Oxidation Rules Oxidation Rules Oxidation Number/State Prob D/R Oxidation.

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Presentation on theme: "Index UIL Science Chemistry Problems. indexIndex Oxidation Rules Oxidation Rules Oxidation Rules Oxidation Rules Oxidation Number/State Prob D/R Oxidation."— Presentation transcript:

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