x 2 = 6.8 x 10-4 x = 2.61 x 10-2 M = [H + ] ph = -log (2.61 x 10-2 ) ph = 1.58

Transcription

1 CHM1046 (Ref #625001) Chapter 17 and Chapter 18 Exam 4 - Worksheet Acid-Base Equilibria and Solubility Equilibria: 1. You are asked to prepare a ph = 3.00 buffer starting from 1.50 L of 1.00 M of hydrofluoric acid (HF) and an excess of sodium fluoride (NaF). a. What is the ph of the hydrofluoric acid solution prior to adding sodium fluoride? Weak acid calculation (need to do ICE). HF (aq) H + (aq) + F - (aq) K a = 6.8 x 10-4 I C -x + x + x E 1.00 x x x K a = 6.8 x 10-4 = = x 2 = 6.8 x 10-4 x = 2.61 x 10-2 M = [H + ] ph = -log (2.61 x 10-2 ) ph = 1.58 b. How many grams of sodium fluoride should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium fluoride is added. Given: ph = 3.00 and [HF] = 1.00 M Need [F - ] Since this is a buffer we can use the Henderson-Hasselbach equation: ph = pk a + log 3.00 = log log = = = [F - ] = (0.681) ([HF]) = (0.681) (1.00) = M NaF Na + + F - [NaF] = M Need grams of NaF (0.681 moles NaF/1 L) (1.50 L) (41.99 g NaF/ mol NaF) = 42.9 g NaF Page 1 of 7

2 2. What is the ratio of HCO 3 - to H 2 CO 3 in blood of ph 7.4? How is this ratio affected in an exhausted marathon runner whose blood H is 7.1? H 2 CO 3 H + + HCO 3 - K a = 4.3 x 10-7 ph = pk a + log log = = 1.03 = = 10.7 at ph 7.1: log = = 0.73 = = How many milliliters of M NaOH are required to titrate each of the following solutions to the equivalence point: a ml of M CH 3 COOH At the equivalence point, the number of moles of CH 3 COOH equals number of moles of NaOH. Number of moles of CH 3 COOH: (35.0 ml) ( moles/1000 ml) = x 10-3 moles CH 3 COOH. Number of moles of NaOH has to equal x 10-3 moles. To calculate volume: (volume) x ([NaOH]) (2.975 x 10-3 moles NaOH) (1000 ml/ moles) = 35.0 ml b ml of M HF At the equivalence point, the number of moles of HF equals number of moles of NaOH. Number of moles of HF: (40.0 ml) ( moles/1000 ml) = 3.6 x 10-3 moles HF. Number of moles of NaOH has to equal 3.6 x 10-3 moles. To calculate volume: (volume) x ([NaOH]) (3.6 x 10-3 moles NaOH) (1000 ml/ moles) = 42.4 ml Page 2 of 7

3 4. A 35.0 ml sample of M acetic acid (CH 3 COOH) is titrated with M NaOH solution. Calculate the ph after the following volumes of base have been added: a. 0 ml Initial ph, calculation is of weak acid (need to do ICE) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) K a = 1.8 x 10-5 I C -x + x + x E x x x b ml K a = 1.8 x 10-5 = = x 2 = 2.7 x 10-6 x = 1.64 x 10-3 M = [H + ] ph = -log (1.64 x 10-3 ) ph = 2.78 Not yet at equilibrium (number of moles of NaOH less than number of moles of CH 3 COOH) # moles CH 3 COOH: (35.0 ml) (0.150 mol/1000 ml) = moles # moles of NaOH: (17.5 ml) (0.150 mol/ 1000 ml) = moles When NaOH is added: CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) Before: mol 0 0 Addition: mol mol mol After: mol mol [CH 3 COOH] = ( mol/ L) = M [CH 3 COO - ] = ( mol/ L) = M ph = pk a + log = ph = 4.74 Page 3 of 7

4 c ml CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) Before: mol 0 0 Addition: mol mol mol After: 0 mol mol [CH 3 COOH] = ( mol/0.070 L) = M CH 3 COO - (aq) CH 3 COOH (aq) + OH - (aq) K b = (K w /K a ) = 5.56 x I C - x + x + x E x x x K b = 5.56 x = x 2 /(0.075 x) = x 2 /0.075 x 2 = 4.17 x x = 6.45 x 10-6 = [OH - ] poh = 5.19 ph = For manganese (II) hydroxide, Mn(OH) 2 the K sp = 1.6 x a. Calculate the ph of this solution. Mn(OH) 2 (s) Mn 2+ (aq) + 2 OH - (aq) K sp = 1.6 x K sp = 1.6 x = [Mn 2+ ] [OH - ] 2 = (x)(x 2 ) x = 5.43 x 10-5 M = [Mn(OH) 2 ] Now need to calculate [OH - ]: (5.43 x 10-5 mol Mn(OH) 2 / 1 L) (2 mol OH - /1 mol Mn(OH) 2 ) = 1.09 x 10-4 M = [OH - ]\ poh = 3.96 ph = 10.0 Page 4 of 7

5 b. Calculate the molar solubility of Mn(OH) 2 in a solution that contains M NaOH. Initial [OH - ] = M Mn(OH) 2 (s) Mn 2+ (aq) + 2 OH - (aq) I C + x + 2x E x x K sp = 1.6 x = (x) ( x) 2 = x(0.020) 2 [Mn(OH) 2 ] = 4.0 x M 6. Will Ag 2 SO 4 precipitate when 100 ml of M AgNO 3 is mixed with 10 ml of M Na 2 SO 4 solution? Show your work. 2 AgNO 3 (aq) + Na 2 SO 4 (aq) 2 NaNO 3 (aq) + Ag 2 SO 4 (?) To determine if Ag 2 SO 4 will precipitate at these conditions, we need to calculate Q: Ag 2 SO 4 (s) 2Ag + (aq) + SO 4 2- (aq) Q = [Ag + ] 2 [SO 4 2- ] To calculate [Ag + ]: (0.100 L) (0.050 mol AgNO 3 /1L)(1 mol Ag + / 1 mol AgNO 3 ) = moles Ag + Total volume = 110 ml = L [Ag + ] = (0.005 moles/0.110 L) = M To calculate [SO 4 2- ]: (0.010 L) (0.050 mol Na 2 SO 4 /1L)(1 mol SO 4 2- / 1 mol Na 2 SO 4 ) = 5.0 x 10-4 moles SO 4 2- Total volume = 110 ml = L [SO 4 2- ] = (5.0 x 10-4 moles/0.110 L) = M Q = [Ag + ] 2 [SO 4 2- ] = (0.045) 2 (0.0045) = 9.11 x 10-6 Q < K sp Precipitate will not form. Page 5 of 7

6 Thermodynamics: 1. For the following reaction, calculate S universe and determine whether the reaction is spontaneous, nonspontaneous, or at equilibrium: H 2 (g) + I 2 (g) 2HI (g) S univ = S sys + S Surr S Surr = -( H sys )/T Values from the Appendix: H f (H 2 ) = 0 kj/mol H f (I 2 ) = kj/mol H f (HI) = 25.9 kj/mol H rxn = 2(25.9 kj/mol) [(62.25 kj/mol) + 0 kj/mol] = kj/mol = x 10 4 J/mol S Surr = -( H sys )/T = (-1.05 x 10 4 J/mol)/(273 K) = 38.5 J/mol K Values from the Appendix: S (H 2 ) = J/mol K S (I 2 ) = J/mol K S (HI) = J/mol K S rxn = 2(206.3 J/mol K) [(260.6 J/mol K) J/mol K] = J/mol K S univ = (21.03 J/mol K+ (38.5 J/mol K) = 59.5 J/mol K process is spontaneous 2. Calculate the standard free-energy change for the following reaction at 298 K, given that H = kj/mol and S = 24.7 J/mol K. Is this reaction spontaneous under these conditions? N 2 (g) + O 2 (g) 2 NO (g) G = H - T S = (180.7 kj/mol) (298 K) (2.47 x 10-2 kj/mol K) G = kj/mol G is positive, the reaction is not spontaneous under these conditions. Page 6 of 7

7 3. Calculate G at 298 K for a mixture of 1.0 atm N 2, 3.0 atm H 2, and 0.50 atm NH 3 being used in the following process: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) For this reaction, H = kj/mol and S = J/mol K. G = G + RT ln Q We first need to calculate G to calculate G, we need to find Q: G = H - T S = (-92.4 kj/mol) (298 K)( kj/mol K) G = kj/mol kj.mol G = kj/mol Q = = = 9.3 x 10-3 G = G + RT ln Q = (-33.3 kj/mol) + [(8.314 x 10-3 kj/mol K)(298 K) ( ln 9.3 x 10-3 )] G = (-33.3 kj/mol) + (-11.6 kj/mol) = kj/mol 4. The value of K a for nitrous acid (HNO 2 ) at 25 C is 4.5 x a. Write the chemical equation for the equilibrium. HNO 2 (aq) H + (aq) + NO 2 - (aq) K a = 4.5 x 10-4 b. By using the value of K a, calculate G for the dissociation of nitrous acid in aqueous solution. G = - RT ln K = - (8.314 J/mol K) (298 K) (ln 4.5 x 10-4 ) = J/mol = kj/mol c. What is the value of G at equilibrium? At equilibrium G = 0 d. What is the value of G when [H + ] = 5.0 x 10-2 M, [NO 2 - ] = 6.0 x 10-4 M, and [HNO 2 ] = 0.20 M? Q = = = 1.5 x 10-4 G = G + RT ln Q = ( kj/mol) + [(8.314 x 10-3 kj/mol K)(298 K) ( ln 1.5 x 10-4 )] G = ( kj/mol) + (-21.8 kj/mol) = -2.7 kj/mol Page 7 of 7