FORMULA SHEET (tear off)

Transcription

1 FORMULA SHEET (tear off) N A = x C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) amu = x kg C = K K = C atm = 760 torr = 760 mm Hg 1 atm = bar pv = nrt R = L atm/mol K 1 L atm = J R = J/mol K 1 J= 1 kg m 2 /s 2 p A = X A p A [B] = k p B p A = X Bp A T b = K b m B T f = K f m B = [B]RT K p = K C (RT) n H = E + pv G rxn = G rxn + RT ln Q G = H - TS ln K = - G rxn/rt If ax 2 + bx + c = 0, then x = ( - b [b 2-4ac] 1/2 ) 2a K a K b = K w = 1.0 x (at T = 25 C) ph = pk a + log 10{[base]/[acid]}

2 GENERAL CHEMISTRY 2 SECOND HOUR EXAM JUNE 9, 2017 Name Solutions - Version 4 Panthersoft ID Signature Part 1 (24 points) Part 2 (48 points) Part 3 (48 points) TOTAL (120 points) Do all of the following problems. Show your work.. 2

3 Part 1. Multiple choice. Circle the letter corresponding to the correct answer. There is one and only one correct answer per problem. [4 points each] 1) Which of the following compounds is a strong soluble base? a) Ba(OH) 2 b) Co(OH) 2 A c) Fe(OH) 3 d) Both a and b e) Both a and b and c 2) Which of the following substances is classified as a salt? a) NaCl b) NaNO 3 D c) NaOH d) Both a and b e) Both a and b and c 3) In the reaction Al(H 2O) 6 3+ (aq) + H 2O( ) Al(H 2O) 5OH 2+ (aq) + H 3O + (aq) the Al(H 2O) 3+ 6 ion functions as a) the conjugate acid of H 2O b) the conjugate base of H 2O C c) a Bronsted acid d) a Bronsted base e) none of the above 4) Which of the following reactions goes essentially to completion in aqueous solution? a) The reaction of a strong acid with a strong base b) The reaction of a strong acid with a weak base E c) The reaction of a weak acid with a strong base d) Both a and b e) Both a and b and c 5) A small amount of a strong base is added to a buffer solution. The ph of the buffer solution a) will increase significantly b) will increase by a small amount B c) will be unchanged d) will decrease by a small amount e) will decrease significantly 6) moles of HClO 2 (chlorous acid, a weak monoprotic acid) is added to L of each of the following solutions. For which solution will the percent ionization of the chlorous acid be largest? a) A 1.0 x 10-3 M solution of HNO 2 (nitrous acid, a weak acid) b) A 1.0 x 10-3 M solution of HBr (hydrobromic acid, a strong acid) D c) A 1.0 x 10-3 M solution of NH 3 (ammonia, a weak base) d) A 1.0 x 10-3 M solution of KOH (potassium hydroxide, a strong base) e) Pure water Version 1: A D A E D C Version 2: A D B E C A Version 3: A D D E A B 3

4 Part 2. Short answer. 1) The ph of an aqueous solution is ph = 4.68 at T = 25 C. Find [H 3O + ], [OH - ], and the poh for the solution. [9 points] [H 3O + ] = 10 -ph = = 2.1 x 10-5 mol/l poh = ph = = 9.32 [OH-] = 10 -poh = = 4.8 x mol/l Version 1: poh = 9.76 [H 3O + ] = 5.8 x 10-5 M [OH - ] = 1.7 x M Version 2: poh = [H 3O + ] = 1.9 x 10-4 M [OH - ] = 5.2 x M Version 3: poh = [H 3O + ] = 6.9 x 10-4 M [OH - ] = 1.4 x M 2) The values for K a for four weak monoprotic acids are given below, at T = 25. C CH 3COOH K a = 1.8 x 10-5 HOCl K a = 3.8 x 10-8 HNO 2 K a = 4.5 x 10-4 C 6H 5OH K a = 1.3 x Based on this information answer the following questions by circling the correct answer. There is one and only one correct answer per question. [3 points each] Which of the following is the strongest acid? CH 3COOH HNO2 HOCl C 6H 5OH Which of the following is the strongest base? CH 3COO - NO 2 - OCl - C6H5O - Which of the following would be the best combination of substances for making a ph = 4.5 buffer CH3COOH and HNO 2 and HOCl and C 6H 5OH and NaCH3COO NaNO 2 NaOCl NaC 6H 5O 3) What are the conjugate acid and the conjugate base of the HPO 4 2- ion (correct formula and charge)? [3 points each] conjugate acid H 2PO 4- conjugate base PO 4 3-4

5 4) What is the ph of a saturated aqueous solution of manganese (II) hydroxide (Mn(OH) 2), at T = 25. C. The value for the solubility product of manganese (II) hydroxide at this temperature is K sp = 2.1 x [8 points] Mn(OH) 2(s) Mn 2+ (aq) + 2 OH - (aq) Ksp = [Mn 2+ ] [OH - ] 2 = 2.1 x Initial Change Equilibrium Mn 2+ 0 x x (x)(2x) 2 = 4x 3 = 2.1 x OH - 0 2x 2x x 3 = (2.1 x )/4 = 5.25 x [OH - ] = 2x = 2(3.74 x 10-5 ) = 7.49 x 10-5 x = (5.25 x ) 1/3 = 3.74 x 10-5 poh = - log 10(7.49 x 10-5 ) = 4.13 ph = poh = = ) A L solution at T = 25. C initially contains 4.0 x 10-4 M Ag + ion. How many grams of sodium chloride (NaCl, MW = g/mol) can be added to the solution before AgCl begins to precipitate? K sp(agcl) = 1.8 x [8 points] AgCl(s) Ag + (aq) + Cl - (aq) K sp = [Ag + ] [Cl - ] = 1.8 x NaCl(s) Na + (aq) + Cl - (aq) Initial Change Equilibrium Ag x x 10-4 Cl - 0 x x (4.0 x 10-4 )(x) = 1.8 x x = 1.8 x = 4.5 x x 10-4 since moles Cl - = moles NaCl moles NaCl = (1.000 L) (4.5 x 10-7 mol/l) = 4.5 x 10-7 moles NaCl grams NaCl = (4.5 x 10-7 moles NaCl) (58.44 g/mol) = 2.6 x 10-5 g NaCl (or 26 g NaCl) Version 1: 1.8 x 10-5 g Version 2: 1.3 x 10-5 g Version 3: 5.3 x 10-5 g 6) What is the oxidation number for phosphorus (P) in each of the following molecules? [2 points each] Na 3P -3 PH 3-3 P 4 0 P 2O

6 Part 3. Problems. 1) Find the ph of the following three solutions (all at T = 25. C) [8 points each] a) 6.44 g of potassium hydroxide (KOH, MW = g/mol), a strong soluble base, dissolved in water, to form a solution of final volume V = ml. KOH(s) K + (aq) + OH - (aq) moles KOH = 6.44 g 1 mol = mol g molarity KOH = mol = mol/l KOH L A strong soluble base, so [OH - ] = mol/l poh = - log 10[OH-] = - log 10(0.287) = 0.54 ph = poh = = Version 1: Version 2: Version 3: b) A M solution of propionic acid, a weak monoprotic acid, with K a = 1.4 x Let HA = weak monoprotic acid. Then HA(aq) + H 2O( ) H 3O + (aq) + A - (aq) K a = [H 3O + ] [A - ] = 1.4 x 10-5 Initial Change Equilibrium [HA] HA x x (x)(x) = 1.4 x 10-5 H 3O + 0 x x ( x) A - 0 x x Assume x < , then x 2 = 1.4 x 10-5 x 2 = (0.0800)(1.8 x 10-5 ) = 1.12 x 10-6 (0.0800) x = (1.12 x 10-6 ) 1/2 = 1.06 x 10-3 (so x is small) [H 3O + ] = x = 1.20 x 10-3 ph = - log 10[H 3O + ] = - log 10(1.06 x 10-3 ) = 2.98 Version 1: ph = 2.58 Version 2: ph = 2.77 Version 3: ph = 3.13 c) A solution containing M hydrochloric acid (HCl), a strong acid, and M ammonia (NH 3), a weak base, with K b = 1.8 x HCl(aq) + NH 3(aq) NH 4+ (aq) + Cl - (aq) Since we have equal concentrations of HCl and NH 3, the Reaction goes to completion. NH 4 + (aq) + H 2O( ) H 3O + (aq) + NH 3(aq) K a = (1.0 x )/K b = (1.0 x )/(1.8 x 10-5 ) = 5.56 x K a = [H 3O + ] [NH 3] = 5.56 x Initial Change Equilibrium [NH + 4 ] + NH x x (x) (x) = 5.56 x Assume x <<0.010 NH 3 0 x x ( x) H 3O + 0 x x 6

7 x 2 = 5.56 x x 2 = (0.0100) (5.56 x ) = 5.56 x (0.0100) x = (5.56 x ) 1/2 = 2.36 x 10-6 (so x is small) [H 3O + ] = 2.36 x 10-6 ph = - log 10[H 3O + ] = - log 10(2.36 x 10-6 ) = ) The following question concerns some of the solutions from problem 1 above. a) Is the solution in part b of problem 1 a buffer solution (yes/no and a justification for your answer)? [4 points] NO. The solution contains a significant concentration of weak acid (HA) but does not contain a significant concentration of conjugate base (A-). b) Is the solution in part c of problem 1 a buffer solution (yes/no and a justificantion for your answer)? [4 points] NO. The solution contains a significant concentration of weak acid (NH + 4 ) but does not contain a significant concentration of conjugate base (NH 3). 3) Balance the following unbalanced oxidation-reduction reactions [8 points each] a) SO 2(g) + Fe 3+ (aq) SO 2-4 (aq) + Fe 2+ (aq) (for acid conditions) ox SO 2(g) + 2 H 2O( ) SO 2-4 (aq) + 4 H + (aq) + 2 e - red Fe 3+ (aq) + e - Fe 2+ (aq) x 2 net SO 2(g) + 2 Fe 3+ (aq) + 2 H 2O( ) SO 4 2- (aq) + 2 Fe 2+ (aq) + 4 H + (aq) b) Cr(OH) 4- (aq) + ClO - (aq) CrO 4 2- (aq) + Cl - (aq) (for base conditions) ox - Cr(OH) 4 CrO 2-4 (aq) + 4 H + (aq) + 3 e - x 2 red ClO - (aq) + 2 H + (aq) + 2 e - Cl - (aq) + H 2O( ) x 3 2 Cr(OH) 4- (aq) + 3 ClO - 2- (aq) 2 CrO 4 (aq) + 3 Cl - (aq) + 2 H + (aq) + 3 H 2O( ) 2 OH - (aq) 2 OH - (aq) net, base 2 Cr(OH) 4 - (aq) + 3 ClO - (aq) + 2 OH - (aq) 2 CrO 4 2- (aq) + 3 Cl - (aq) + 5 H 2O( ) 7