Sch4u Final Exam Review Solutions

SCH4U FINAL EXAM REVIEW SOLUTIONS

KINETICS: 1. It would require the simultaneous collision of 27 molecules, which is an unlikely event.

-5 - 1 - 116 mol C6.O2 - 5 - 1 - 1 2. 34� 10 mol O2 L s 4.06 10 mol CO 2 L s 25 mol O2 x y 3. Rate = [O2] [NO] . To determine the exponents x and y we construct a ratio of two rate laws using data from two different rows in our table x y x y -6 Rate k[ O2 ] [ NO] 3.29 10 k[ 0.00173] [ 0.00347] 1= 1 1 and = Rate k Ox NO y 6.55 10-6 k 0.00345x 0.00347 y 2 [ 2 ]2[ ] 2 [ ] [ ] From this ratio we cancel the rate constants, k, and the [0.00347]y terms to get x 3.29 10-6 [0.00173] = , Evaluating this ratio we get 0.4977 = (0.501)x. From 6.55 10-6 [0.00345]x this we conclude that x must be 1. To determine the value of y we substitute rows 1 and 3 of the table to get x y 3.29 10-6 [0.00173] [ 0.00347] = . We get 0.0628 = (0.250)y from which y = 2. 6.55 10-6 [0.00173]x[ 0.0139] y 2 Therefore the rate law is Rate = k[O2][NO] . We can now calculate the rate constant by substituting the data from any row in the table into the rate law. Using data from row 1 we get 3.29 x 10-6 mol L-1 s-1 = k(0.00173 mol L-1)(0.00347 mol L-1)2 k = 154 L2 mol-2 s-1. 4. The tool for solving this type of problem is the integrated rate equation for a first- order process. One form of this equation is ln[Q]o – ln[Q]t = kt. If we convert time to seconds we have all of the info needed. 4 194 minutes = 1.164 x 10 s. ln[Q]o – ln(8.54 x 10-6 mol L-1) = (3.64 x 10-4 s-1)(1.164 x 104 s) ln[Q]o = 4.24 – 11.67 = -7.43 -4 -1 [Q]o = 5.9 x 10 mol L . 5. O O N - - - -O- - - -N O k E 骣1 1 6. ln 1 =a 琪 - . All we need to do is convert the Celsius temperatures to k2 R桫 T 2 T 1 their Kelvin scale and insert the data to get 2.68 10-5 E 骣 1 1 ln =a 琪 - 10.72 10-5 8.314 J mol - 1 K - 1 桫 313 K 293 K -5 - 1 The reaction rate -1.386 =( - 2.62 10 kJ mol) Ea -1 Ea = 52.8 kJ mol 2

increased by a factor of four when the temperature was increased by 20oC. This is the same as doubling every 10oC. Therefore, reactions that double their rates with a ten degree rise in temperature have activstion energies of approximately 50 kJ. 7. Rate = k[(CH3)3CBr]

CHEMICAL EQUILIBRIUM

4 [NO2 ] [C2 H 6 ] (a) = K ; (b) = K 1. 2 3 c C H H c [N2 O] [ O 2 ] [ 2 4][ 2 ] 垐 1 1 2. (c) HI(g ) 噲 2 H2 ( g )+ 2 I 2 ( g ), Kc = 0.142 P4 K = NO2 3. p P2 P 2 N2 O O 2

4. (a) 3F2(g) + Br2(g) 噲垐 2BrF3(g) 2 [BrF3 ] (b) Kc = 3 [F2] [ Br 2 ]

Dng 80- 1 - 1 - 1.0 79 5. Kp= K c ( RT ) = (9.1创 10 )(0.0821 L atm mol K 298 K) = 3.7 10

For H2(g) + O2(g) 噲垐 H2O(g) .

The equation that we want Kp for has all of its coefficients one half of those given with the value of Kc. That means that we must take the square root of the value of Kc first and then convert to Kp. (9.1 x 1080)1/2 = 3.2 x 1040. Δngas = 1 mol – (1 mol + 0.5 mol) = -0.5 mol

Dng 40- 1 - 1 - 0.5 39 Kp= K c ( RT ) = (3.2创 10 )(0.0821 L atm mol K 298 K) = 6.47 10

2+ 3 臌轾Zn 6. Kc = 3+ 2 臌轾Fe

7. We set up an equilibrium table to keep track of all the concentrations andtheir changes. first we must accurately transfer data from the problem to the first brow of the table. For the change line we insert x’s and the stoichiometric coefficients. since the reaction goes in the forward direction, the 2x has a minus sign and the other x values are positive. The third line is the sum of the first two and are the terms inserted into the equilibrium law.

2HI 噲垐 H2(g) + I2(g) initial concentration 0.200 0.00500 0.00500 Change in -2x +x +x concentration equilibrium 0.200 – 2x 0.00500 + x 0.00500 + x concentrations Setting up the equilibrium law we get: 3

[H2][ I 2 ] [ 0.00500+x][ 0.00500 + x] -4 2= 2 = 8.4 10 [HI] [ 0.200- 2x] We can take the square root of both sides of the equation to get [0.00500 + x] = 2.9 10-2 [0.200- 2x] Multiplying both sides by 0.200 – 2x results in 0.00500 + x = 5.8 x 10-3 – (5.8 x 10-2)x x + 0.058x = 0.00580 – 0.00500 x = 7.6 x 10-4 Using this value of x we calculate that [HI] = 0.200 – 0.002 = 0.198 mol L-1

8. 2HBr(g) 噲垐 H2(g) + Br2(g) initial concentration 0.200 0.0 0.0500 Change in -2x +x +x concentration equilibrium 0.200 – 2x 0.0 + x 0.0500 + x concentrations Setting up the equilibrium law we get:

[H2][ Br 2 ] ( 0.0+x)( 0.0500 + x) -5 2= 2 = 3.3 10 [HBr] ( 0.200- 2x)

( x)(0.0500) -5 We will assume that x << 0.0500 and that 2x << 0.200 . 2 = 3.3 10 (0.200) Solving this equation gives us a value x = 6.6 x 10-8 and [HBr] = 0.200 M. 9. (a) decrease (b) increase (c) no change (d) increase (e) no change

ACIDS AND BASES;

+ - 1. NH3 (acid) – NH4 (base) ; NH3 (base) – NH2 (acid) 2. (a) pH = -log[0.25] = 0.60 (b) pOH = -log(.25) = 0.60; pH = 14 - pOH = 13.40 (c) pOH = -log(.0020) = 2.7; pH = 11.30

3. (a) HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) (overall) + - H (aq) + OH (aq)  H2O(l) (chemical reaction) -1 (b) nHCl = cV = (0.45 mol L )(0.0250 L) = 0.01125 mol -1 nNaOH = cV = (0.35 mol L )(0.0300 L) = 0.01050 mol. Since this is a 1:1 reaction [H+] = 0.1125 mol – 0.01050 mol = 7.5 x 10-4 mol. [H+] = n/V = .0075/0.0750 L) = 0.01 mol L-1 pH = -log(0.01) = 2 (rounded!!). 4. Aniline has been identified as a weak base. Since it is the only solute, we will use the Kb and start our problem by writing the chemical equation. Most weak bases we see in this course are related to ammonia and we see that if the C6H5 in aniline was replaced by a hydrogen atom it would have the formula of ammonia, NH3. The advantage is that we know the equation for the reaction of ammonia with + - water. NH3(aq) + H2O  NH4 (aq) + OH (aq) . We can use that as a template for our ionization reaction, or any other nitrogen 4

base. For the second part of the problem, to convert from the Kb to the pKb -14 pKb = -log Kb and to determine Ka involves KaKb = 1.0 x 10 . + - C6H5NH2(aq) + H2O C6H5NH3 (aq) + OH (aq) -10 pKb = -log(1.5 x 10 ) = 9.82 -14 1.0 10 -5 -5 Ka = = 6.7 10 ; pKa = -log(6.7 x 10 ) = 4.17 1.5 10-10 5. First we identify that the category that this problem fits into is one where a weak acid is the only solvent. second, we will write the ionization reaction for propanoic acid and third, we write the equilibrium law. + - C2H5COOH  H + C2H5COO The equilibrium law is: 轾- 轾 + 2 臌C2 H 5 COO 臌 H x K = = a initial concentration [C2 H 5 COOH] [ ] -5 We will substitute the value of Ka (1.34 x 10 ) and the initial concentration of propanoic acid (0.620 M) into the equation x2 1.34� 10-5 [0.620M ] x2= (0.620 M )( 1.34 10- 5 ) -3 轾 + 轾 - x =2.88� 10 = 臌 H 臌 C2 H 5 COO Finally we calculate the remaining unionized propanoic acid by subtracting the amount that does ionize. [C2H5COOH] = 0.620 – 0.00288 = 0.617 pH = -log(2.88 x 10-3) = 2.54

6. When dissolved in water KNO2 undergoes the reaction + - KNO2  K (aq) + NO2 (aq) + - If we add H to the NO2 we get HNO2 and recognize it as a weak acid. If we add OH- to K+ we get KOH that is recognized as a strong base. Therefore K+ will not - affect the pH of the solution because it is a very weak conjugate acid, but the NO2 will. It will act as a base (it must be a conjugate base since it came from an acid - HNO2). We will then write the reaction of NO2 with water as - - NO2 + H2O 噲垐 HNO2 + OH The equilibrium law for this reaction is HNO轾 OH- [ 2 ]臌 K = b 轾NO- 臌 2 - The initial concentration of NO2 is 0.100 M, and a small amount, x, reacts with water. We then calculate that - - [NO2 ] = 0.100 – x and [HNO2] = [OH ] = x - To simplify the calculations we assume that x is very small so that [NO2 ] = 0.100. -4 We look up Ka = 7.1 x 10 for HNO2. Kb -11 2 2 = Kw/Ka = 1.4 x 10 = x /0.100 x 5

= (1.4 x 10-11)(0.100) = 1.4 x 10-12 x = 1.2 x 10-6 = [OH-]; pOH = 5.93; pH = 14.00 – pOH = 14.00 – 5.93 = 8.07 + 7. Yes, NH4 reacts with water. pH = 4.93 (same method as #6). + - 8. CH3NH2(aq) + H2O  CH3NH3 (aq) + OH (aq) and 轾+ 轾 - 臌CH3 NH 3 臌 OH Kb = [CH3 NH 2 ] + - CH3NH2(aq) + H2O  CH3NH3 (aq) + OH (aq) (2.5 x 10-3 – x) x x xx 4.4� 10-4 2.5� 10-3 x x2 + (4.4 x 10-4)x – (1.1 x 10-7) = 0 -b� b2 4 ac -4.4幢 10-4 1.9 � 10 - 7 - 4(1)( 1.1 10 - 7 ) x = = 2a 2(1) x = 1.8 x 10-4 and x = -6.2 x 10-4. Only the positive values has any meaning so -3 -4 -3 [CH3NH2] = 2.5 x 10 – 1.8 x 10 = 2.3 x 10 M + - -4 [CH3NH3 ] = [OH ] = 1.8 x 10 M pOH = -log(1.8 x 10-4) = 3.74; pH = 14.00 - 3.74 = 10.26 9. The buffer solution contains both the weak acid CH3COOH and its conjugate base - CH3COO . We can use either Ka or Kb, whichever is handy. In our tables we find -6 Ka = 1.8 x 10 for CH3COOH, so the simplest approach is to use the equation for the ionization of the acid. + - CH3COOH 噲垐 H + CH3COO 轾H+ 轾 CH COO - 臌臌 3 -5 Ka = =1.8 10 [CH3 COOH] + - CH3COOH 噲垐 H + CH3COO initial concentration 0.090 0.0 0.11 Change in -x +x +x concentration equilibrium 0.090 – x  0.090 +x 0.11 + x  0.11 concentrations For buffer solutions the quantity x will be very small, so it is safe to make the simplifying assumptions. ( x)(0.11+ x) ( x)( 0.11) K = � 1.8 10-5 a (0.090- x) ( 0.090) Solving for x gives us (0.090)创 1.8 10-5 x = =1.5 10-5 and [H+] = 1.5 x 10-5 M. (0.11) pH = -log(1.5 x 10-5) = 4.82. 10. We begin with the chemical equation and the Kb expression. + - NH3 + H2O 噲垐 NH4 + OH 6

轾NH+ 轾 OH - 臌4 臌 -5 Kb = =1.8 10 [NH3 ] + The key here is to use this Kb expression to calculate [NH4 ]. We obtain [OH-] from the given pH: pOH = 14.00 - 9.10 = 4.90 [OH-] = 1.26 x 10-5 M. + Because Kb is small and the common ion NH4 is present, the equilibrium concentration of NH3 will essentially equal its initial concentration. [NH3] = 0.25 M.

+[NH3 ] -5 (0.25) 轾NH4 =Kb = ( 1.8� 10) -5 0.36 M 臌 - (1.26 10 ) 臌轾OH + -1 Moles of NH4 = (0.200 L)(0.36 mol L ) = 0.071 mol -1 Mass of NH4Cl = nM = (0.071 mol)(53.5 g mol ) = 3.82 g + - 11. C2H5COOH 噲垐 H + C2H5COO 轾OH-=K initial concnetration 创 7.4 10 -10 = 0.048 6.0 10 - 6 M . 臌 b �( ) The reaction of propanoic acid with KOH is - KOH + C2H5COOH 噲垐 C2H5COO + H2O 1. Determine the equivalence point volume by converting ml of propanoic acid to mL of KOH using basic stoichiometric steps

0.0800 mol C2 H 5 COOH 1 mol KOH 1 L KOH 25.00 mL C2 H 5 COOH= 16.7 mL KOH 1 L C2 H 5 COOH 1 mol C 2 H 5 COOH 0.1200 mol KOH The equivalence point is expected at 16.7 mL 2. The pH at the start of the titration is simply a solution containing a weak acid and the hydrogen ion concentration is calculates, using the assumptions as 轾H+=K ( initial concentration ) 创 1.34 10 -5 = 0.080 1.0 10 - 3 M 臌 a � and the pH = -log(0.001) = 3.00 3. This part of the problem is actually a simple limiting reactant problem. Start by calculating the moles of each substance we have. -1 -3 nKOH = cV = (0.01000 L KOH)(0.1200 mol L KOH) = 1.200 x 10 mol KOH n = -1 -3 C2 H 5 COOH (0.02500 L)(0.0800 mol L ) = 2.00 x 10 mol C2H5COOH

We have more moles of C2H5COOH than KOH so that all of the KOH is used up - -3 forming an equal number of moles of C2H5COO (1.200 x 10 ). Subtracting the -3 consumed KOH from the original moles of C2H5COOH leaves us with 0.80 x 10 mol C2H5COOH. 轾C H COO- 轾 H + 轾 1.200 10 -3 轾 H + 臌2 5 臌-5 臌臌 K = =1.34� 10 a 轾0.80 10-3 [C2 H 5 COOH] 臌 Solving this for [H+] = 8.9 x 10-6M and the pH = -log(8.9 x 10-6) = 5.05 4. To calculate the pH at the equivalence point we need to know what the solution contains. since we have reacted exactly equal moles of KOH and C2H5COOH neither is present in any major amount. The solution is simply one that contains the conjugate base of propanoic acid. We need to calculate Kb for this salt as -14 -5 -10 Kb = Kw/Ka = (1.0 x 10 )/(1.34 x 10 ) = 7.5 x 10 7

The concentration of the conjugate base is calculated by dividing the moles of the original acid by the total volume at the equivalence point. - -3 [C2H5COO ] = (2.00 x 10 mol)(0.02500 L + 0.0167 L) 0.048M. the hydroxide ion concentration can now be determined as 轾OH-=K initial concnetration 创 7.4 10 -10 = 0.048 6.0 10 - 6 M 臌 b �( ) . The pOH = 5.22 and the pH = 8.78 at the equivalence point. 5. To select an indicator we need to change its colour right at the equivalence point. Therefore we look at the table of indicators and try to find one that has a good colour change right around pH 8.78. In other words, the pKin should be close to 8.78. There are several in this range including bromothymol blue and phenolphthalein.

SOLUBILITY EQUILIBRIUM:

2+ 2- 1. CaSO4(s) 噲垐 Ca (aq) + SO4 (aq) initial concentration 0.0 0.0 Change in +x +x concentration equilibrium +x +x concentrations -5 2+ 2- 2 Ksp = 2.4 x 10 = [Ca ][SO4 ] = (x)(x) = x x = 4.9 x 10-3 M . d = 1.0 g mL-1 4.9 10-3 mol CaSO1 mL solution 136 g CaSO 0.67 g CaSO m = 4.创 4= 4 . 1000 mL 1 g solution 1 mol CaSO4 1000 g solution Since we want grams per 100 mL we divide the numerator and denominator by 10 to get 0.067 g CaSO4 per 100 mL.

2. Same as above with common ion effect!! 2+ 2- CaSO4(s) 噲垐 Ca (aq) + SO4 (aq) initial concentration 0.0 0.14 Change in +x +x concentration equilibrium +x x + 0.14 concentrations -5 2+ 2- 2 Ksp = 2.4 x 10 = [Ca ][SO4 ] = (x)(x + 0.14) = x + 0.14 x If you recall that the solubility in the presence of a common ion is always less than in distilled water, we can safely assume that (x + 0.14) = 0.14. 2.4 x 10-5 = (x)(0.14) x = 1.7 x 10-4 M. 1.7 10-4 mol CaSO1 mL solution 136 g CaSO 0.023 g CaSO 4.创 4= 4 and 1000 mL 1 g solution 1 mol CaSO4 1000 g solution

0.0023 g CaSO4 per 100 mL. 8

2+ 2- 3. CaC2O4(s) 噲垐 Ca + C2O4 and 2+ 2- -9 Ksp = [Ca ][C2O4 ] = 2.3 x 10 Our final solution will have a total volume of 75.0 mL + 115 mL = 190 mL. We can calculate the final concentration of calcium ions as -5 c1V1 = c2V2 and (4.5 x 10 M)(75.0 mL) = c2(190 mL) c2 2+ -5 =[Ca ] = 1.78 x 10 Mcalcium ions For the oxalate ions we calculate c1V1 = -6 c2V2 and (1.3 x 10 M)(115.0 mL) = c2(190 mL) c2 2- -7 =[C2O4 ] = 7.87 x 10 Moxalate ions Now we can enter the two molarities into the Ksp expression to get -5 -7 -11 Q = (1.78 x 10 Mcalcium ion)(7.87 x 10 Moxalate ion) = 1.4 x 10 We compare Q to Ksp and find that Q is much less than the value of Ksp and we conclude that a precipitate will not form.

THAT’S ALL FOLKS!!!!!