(2). Cu(NO 3 ) HCl CuCl HNO 3 all products soluble, no ppt

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1 CHEM 112 Solubility and Precipitation Rexn 25.0mL 0.010M HCl 25.0mL 0.10M NaOH 50.0mL M 25.0mL 0.050M 50.0mL M AgNO Fe(NO ) Cu(NO ) 2 (1). Fe(NO ) + HCl FeCl + HNO all products soluble, no ppt (). Fe(NO ) + NaOH Fe(OH) (s) + NaNO 2. Fe(OH) ppt. Ksp = 6. x Qsp =[Fe + ][OH - ] = 2. x 10-8 (2). Cu(NO ) HCl CuCl HNO all products soluble, no ppt (5). Cu(NO ) NaOH Cu(OH) 2 (s) + 2 NaNO 2. Cu(OH) 2 ppt.. Ksp = 1.6 x Qsp =[Cu 2+ ][OH - ] 2 = 6.25 x 10-5 (). AgNO + HCl AgCl(s) + HNO 2. AgCl ppt,. Ksp=1.8 x Qsp = [Ag + ][Cl - ]= 2.21 E-6 (6). 2AgNO + K 2 CrO Ag 2 CrO (s) +2 KNO 2. Ag 2 CrO ppt. Ksp = 9.0 x Qsp =[Ag + ] 2 [CrO -2 ] = 1.8 x 10-8 Calculate the solubility in mol/l and g/l for each of the following compounds in each of the given solutions listed on left column unless indicated otherwise pure water 0.01M HCl AgCl Mg(OH) 2 CaF 2 Ag PO 7. Ksp = S 2 S= 1. x 10-5 M = 1.9 x 10 - g/l 12. Common-ion effect Ksp =S(S+0.01) =0.01S S= 1.8 x 10-8 M = 2.58E-6 g/l 8. Ksp = S(2S) 2 =S S=1.55 x 10 - M = 9.06 E- g/l 1 WB + SA, 100 % neutralization: Mg(OH) 2 + 2H + Mg H 2 O ½ (.01MH + ) =.0050M = 0.29 g/l 9. Ksp = S(2S) 2 =S S = 2.1 x 10 - M = 1.67E-2 g/l 1. in 0.5 M KF Common-ion (F - ) effect S= 1.56 x M = 1.2 x 10-8 g/l = 1.28E-8 g/l Q. Fe(NO ) + NaOH Fe(OH) (s) + NaNO, Fe(OH) insoluble, possible ppt Fe(OH) (s) Fe + + OH - Ksp = 6. x 10-8 [Fe + ] = M *50mL = M [OH M * 25.0mL ] = = 0.0M 75mL 75.0mL 10 Ksp = (S) S=27S S =.68 x 10-6 M = 1.96 x 10 - g/l 15. in 0.1 M AgNO Common-ion (Ag + ) effect Ksp = (S+0.1) S = 1x 10 - S S= 1. x M = 5. E-15 g/l Qsp=[Fe + ][OH - ] = ( M)(0.0M) = 2.x10-8 > 6. x 10-8, Fe(OH) will ppt 8 Qsp 2.x10 29 Visibility of ppt: = =.8x10 > Ksp 6.x10 Fe(OH) ppt is visible with naked eyes

2 Q11. Fe [Fe(CN) 6 ] in H 2 O Fe [Fe(CN) 6 ] Fe + + [Fe(CN) 6 ] - Ksp = (S) (S) = 6912 S 7 =.0 x 10-1 S=.6 x 10-7 M Thermodynamics Worksheet (Show answer on this sheet, and work on separate paper) Dr. Ya-Ping Huang Name: Correction: work on 1e and answer of 2i. Reactions 1. Decomposition of 2. Habor process to. Ionization of water a fertilizer produce NH (g) a 2NH NO (s) 2N 2 (g) N 2 (g) + H 2 (g) H 2 O(l) H + (aq) + + H 2 O(g) + O 2 (g) 2NH (g) OH - (aq) b ΔH kj/ mol rexn /mol rexn 55.8 kj/mol rexn c q p for 500 g (kj) NH NO (s) NH (g) -15 kj H 2 O(l): 1550 kj d Δn e f g 25 C h ΔE kj/ mol rexn ΔS kj/mol.k Transition Temp ΔG kj/mol rexn kj/mol rexn 55.8kJ/mol rexn kj/mol.k kj/mol.k Does not exist = 65.1 K = C Does not exist (ΔG f ) kj i K eq.5 x x x 10-1 j Spontaneous? yes yes ph value for pure water: C 90 C k ΔG kj/mol rexn kj/mol rexn kj l K eq 7.08 x x 10-1

3 m Spontaneous? yes No ph value for pure water; 6.12 (1). 2NH NO (s) 2N 2 (g) + H 2 O(g) + O 2 (g) always spontaneous b. ΔH = 2ΔH f N 2 (g) + ΔH f H 2 O(g) + ΔH f O 2 (g) - 2ΔH f NH NO (s) = 2(0) +(-21.8) + 0 2(-65.6) = -26 kj/mol rexn c. q p = [500g/(80)]/2 x (-26) = kj (There are 2 moles NH NO for every mole reaction as written) e. ΔE = ΔH -ΔnRT = -26 kj (7)(8.1E-)(298.15) = -26 kj 17.5 = kj f. ΔS = 2S.N 2 (g) + S.H 2 O(g) + S O 2 (g) 2S.NH NO (s) = 2(191.5) +(188.7) (151.1) = J/mol.K = 1.006kJ/mol.K g. T tran = ΔH /ΔS = -26 kj/1.006kj/mol.k = -227 K, does not existent, (Transition temperature does not exist if T< 0 K.) This reaction is always spontaneous, favored by lower enthalpy, ΔH < 0 and increased entropy, ΔS > 0) h. ΔG = ΔH -T ΔS = -26kJ 298(1.006) = kj/mol rxn (or ΔG = 2ΔG f N 2 (g) + ΔG f H 2 O(g) + ΔG f O 2 (g) - 2ΔG f NH NO (s) = 2(0) +(-228.6) + 0 2(-18) = -56. kj/mol rexn) i. ΔG = logK logk = ΔG /( ) = -56.1/ = , K = 10 exp(+95.66) =.5 x j. reaction is spontaneous when ΔG < 0 k. ΔG = ΔH -T ΔS = -26 (00+27)(1.006) = kj l. ΔG = -2.0RTlogK logk = ΔG /-2.0(8.1E-)( ) = -82.7/ = K = 10exp(+75.85) = 7.08 x () H 2 O(l) H + (aq) + OH - (aq) always nonspontaneous b. ΔH = ΔH f H + (aq) + ΔH f OH - (aq) - ΔH f H 2 O(l) = 0 + (-20.0) (-285.8) = 55.8kJ/mol rexn c. q = [500g/(18)]mol x 55.8 = 1550 kj e. ΔE = ΔH -ΔnRT = 55.8kJ/mol rexn 0xRT = 55.8kJ/mol rexn f. ΔS = S.H + (aq) + S.OH - (aq) - S.H 2 O (l) = J/mol.K = kj/mol.k g. T tran = ΔH /ΔS = 55.8kJ/mol/ kj/mol.k = K, nonexistent h. 25 C :ΔG = ΔH -T ΔS =55.8kJ/mol - 298( ) = kj i. logk = ΔG /( ) = 79.82/ = -1.98, K = K w = 10E(-1.98) = 1.0 x 10-1

4 j. K w = [H + ][OH - ] = [H + ] 2 for pure water, [H + ] = x 10 = 1.02 x 10-7, ph = 6.99 k. 90 C: ΔG = ΔH -T ΔS =55.8kJ/mol - 6( ) = kj l. logk w = ΔG /-2.0(8.1E-)( )= 85.06/-2.0(8.1E-)(6) = -12.2, K = 10exp(-12.2) = 5.78 x 10-1 m. K w = [H + ][OH - ] = [H + ] 2, [H + ] = x 10 = 7.5 x 10-7, ph = 6.12 or K w = [H + ] 2 logk w = log([h + ] 2 ) 2log([H + ]) -pk w = -2 ph pk w = 2 ph ph=½ (pk w) = ½(-log(5.78x10-1 )) =½(12.2) = 6.12 Problem Set b: (2) q H2O = (.18 J/g o C)(100 ml)(1.02 g/ml)( o C) = 108 J q Cal = (2.0 J/ o C)( o C) = 79.2 J (Total heat evolved = q H2O + q Cal = 187 J) q total = q H2O + q Cal + q rexn = 0 q rexn = -187 J = nδh To find n, you have to find moles of each reagent and find which reactant is the limiting reagent. mol of CuSO = (0.00M)(0.050L) = mol x = molrexn 1molCuSO mol of NaOH = (0.600M)(0.0500L) = 0.00 mol x = molrexn 2molNaOH So NaOH is limiting reagent, and n = mole reaction q rexn = -187 J = nδh = ( mole reaction) ΔH ΔH = -187 J/( mole reaction) = J/mole rexn = kj/mole rexn Complete thermochemical equation includes balanced chemical equation and ΔH CuSO (aq) + 2NaOH(aq) Cu(OH) 2 (s) + Na 2 SO (aq) ΔH = kj/mol rxn. The thermite reaction, used for welding iron, is as following: 8 Al(s) + Fe O (s) 9 Fe(s) + Al 2 O (s) A. To calculate ΔH, the best method is using ΔH f : 8 Al(s) + Fe O (s) 9 Fe(s) + Al 2 O (s) ΔH = 9ΔH f, Fe(s) + ΔH f, Al 2 O (s) ΔH f, Fe O (s) 8 ΔH f, Al(s) = (-1676) (-1118) = -50 kj/mol rxn B. To find the energy released, given both reactants, it s a limiting reagent type of problem. The approach is to calculate the energy released by the reaction of each reactant. The smaller value is the answer.

5 a. ΔH based on 8.0 g Al(s): 1molAl 50kJ 8.0gAlx x x = 12kJ 27gAl 8molAl b. ΔH based on 20.0 g Fe O 1molFeO 50kJ 20.0gFeO x x x = 96. kj 22gFe O molfe O The answer is therefore 96. kj released. And Fe O (s) is the limiting reagent. c. Grams of Fe produced is based on the limiting reagent Fe O (s) 1molFeO 9molFe 55.8gFe 20.0g Fe O x x x = 1. gfe 22gFe O molfe O 1molFe Q5. Use the standard enthalpy of formation and bond energy to calculate ΔH for burning of 1mole of ethyl alcohol, C 2 H 5 OH(l). (- 167, kj/mol rxn ) Correction: based on bond energy, the value of ΔH is kj/mol rxn, not -128 kj/mol rxn Q7. Estimate ΔG and determine if the following reaction is spontaneous at 25, 1000 and 2000 C? N 2 (g) + O 2 (g) 2NO(g) [Ans]: ΔG = ΔH - TΔS ΔH = ΣΔH f, RHS - ΣΔH f, LHS = 2ΔH f, NO(g) (ΔH f N2(g) + ΔH f, O2(g) ) = 2(90.25) (0+0) = kj/mol rxn ΔS = ΣS, RHS - ΣS, LHS = 2S, NO(g) (ΔH f N2(g) + ΔH f, O2(g) ) = 2(210.7) ( ) = 2.9 J/K.mol rxn = 2.9x10 - kj/k.mol rxn ΔG = ΔH - TΔS At 25 C ( K) ΔG = kj/mol rexn (298.15)( 2.9x10 - kj/k.mol rxn) = kj/mol rxn (non-spontaneous) At 1000 C (127 K) ΔG = kj/mol rexn (127)( 2.9x10 - kj/k.mol rxn) = kj/mol rxn (non-spontaneous) At 2000 C (227 K) ΔG = kj/mol rexn (227)( 2.9x10 - kj/k.mol rxn) = kj/mol rxn (non-spontaneous) At 25 C ( K) ΔG can be calculated by ΔGH f ΔG = ΣΔG f, RHS - ΣΔG f, LHS = 2ΔG f, NO(g) (ΔG f N2(g) + ΔG f, O2(g) ) = 2(86.57) (0+0) = 17.1 kj/mol rxn, almost identical to the one calculated by ΔG = ΔH - TΔS Transition temperature, T transition = ΔH = ΔS 180.5kJ / mol. rxn = 729K = 6776 C 2.9x10 kj / K. mol. rxn The reaction is nonspontaneous below 6776 o C.

6 The reaction is spontaneous above 6776 o C. The reaction is at equilibrium at 6776 o C. Sample Test III.. a. q v = nδe; ΔE = q v /n = (-27.0 kj)/(1.000g/.08) = kJ/mol C 2 H O b. C 2 H O(g) + 5/2 O 2 (g) 2 CO 2 (g) + 2 H 2 O(l) c. ΔH = ΔE + ΔnRT = (-1.5)(8.1E-)(298.15) = = kj/mol rxn d. ΔH = 2 ΔH f, CO2 + 2 ΔH f, H2O - ΔH f, C2HO = 2 (-9.5) + 2 (-285.8) ΔH f, C2HO ΔH f, C2HO = = -16 kj/mol ** Please notice the values are slightly different than given in answer. Which of the buffer solutions will allow a higher molar solubility of Mg(OH)2? ph buffer or ph 10? [Ans]: Mg(OH) 2(s) Mg OH - 1. ph solution is acidic, the neutralization between H + and OH - (to produce H 2 O) will lower [OH - ]. Based on La Chatelier s principle, more OH - will be produced by the forward reaction, which dissolves Mg(OH) 2(s 2. ph 10 solution is basic, the extra OH - will exert the common ion effect and force reaction to go in the reverse direction, therefore reduce the solubility of Mg(OH) 2(s)