Le Chatlier's principle can be used to decide whether the above equilibrium will be shifted left or right

Transcription

1 Problems, Chapter 17 (with solutions) NOTE: Unless otherwise stated, assume T = 25. C in all problems) 1) In which of these solutions will HNO2 ionize less than it does in pure water? a) 0.10 M NaCl b) 0.10 M NaOH c) 0.10 M KNO3 d) 0.10 M NaNO2 The ionization reaction for HNO2 is HNO2(aq) + H2O( ) H3O + (aq) + NO2 - (aq) Le Chatlier's principle can be used to decide whether the above equilibrium will be shifted left or right a) Added NaCl has no effect. b) Added KNO3 has no effect. c) Added NaOH increases OH - concentration, and so decreases H3O + concentration. The reaction will respond by shifting from left to right (and so more ionization). d) Added NaNO2 increases NO2 - concentration. The reaction will respond by shifting from right to left (and so less ionization). 2) (17.6) Determine the ph of the following solutions: a) A solution that is 0.20 M in NH3 (Kb(NH3) = 1.8 x 10-5 ) This is simply a solution of a weak base. The reaction taking place is NH3(aq) + H2O( ) NH4 + (aq) + OH - (aq) Kb = [NH4 + ] [OH - ] = 1.8 x 10-5 [NH3] NH4 + 0 x x OH - 0 x x NH x x (x) (x) = 1.8 x 10-5 ( x) 1

2 Assume x << 0.20 Then x 2 = 1.8 x 10-5 ; so x 2 = (0.20) (1.8 x 10-5 ) = 3.6 x 10-6 (0.20) x = (3.6 x 10-6 ) 1/2 = 1.9 x 10-3 poh = - log10(1.9 x 10-3 ) = 2.72 ph = = b) A solution that is 0.20 M in NH3 and 0.30 in NH4Cl. NH4Cl is a soluble ionic compound, so when added to water it undergoes the following reaction NH4Cl(s) NH4 + (aq) + Cl - (aq) Our solution initially contains 0.20 M NH3 and 0.30 M NH4 +. This is a solution of a weak base and its conjugate acid, and so is a buffer solution. The Henderson equation says ph = pka + log10{[base]/[acid]} We have a value for Kb for NH3, and so Ka for NH4 + is Ka = 1.0 x = 1.0 x = 5.6 x Kb 1.8 x 10-5 ph = pka + log10{[base]/[acid]} = - log10(5.6 x ) + log10(0.20/0.30) = ) Solve an equilibrium problem (using an ICE table) to calculate the ph of a solution that is M in HC2H3O2 and M in KC2H3O2. Note that Ka = 1.8 x 10-5 for HC2H3O2. Reaction is HC2H3O2(aq) + H2O( ) H3O + (aq) + C2H3O2 - (aq) Ka = [H3O + ] [C2H3O2 - ] = 1.8 x 10-5 [HC2H3O2] HC2H3O x x H3O x x C2H3O x x 2

3 (x) ( x) = 1.8 x 10-5 ( x) Assume x << Then x (0.125) = 1.8 x 10-5 ; so x = (0.195) (1.8 x 10-5 ) = 2.81 x 10-5 (0.195) (0.125) ph = - log10(2.81 x 10-5 ) = 4.55 Note that one could get the same result more quickly by using the Henderson equation ph = pk a + log 10 {[base]/[acid]} = - log 10 (1.8 x 10-5 ) + log 10 (0.125/0.195) = ) A buffer contains significant amounts of bromous acid (HBrO2) and sodium bromite (NaBrO 2 ). Write equations showing how this buffer neutralizes added acid and added base. Bromous acid is a weak acid, and bromite ion is the conjugate base of bromous acid, and so is a weak base. It forms from the ionization of sodium bromite NaBrO2(s) Na + (aq) + BrO2 - (aq) Consider what happens when a strong acid (HCl) or a strong base (NaOH) is added to a solution containing the above substances. HCl(aq) + BrO2 - (aq) HBrO2(aq) + Cl - (aq) effect of added strong acid NaOH(aq) + HBrO2(aq) H2O( ) + Na + (aq) + BrO2 - (aq) effect of added strong base The buffer converts HCl (a strong acid) into HBrO2 (a weak acid), and converts NaOH (a strong base) into BrO2 - (a weak base). By converting strong acid or base into weak acid or base the change in ph is minimized. 5) Use the Henderson-Hasselbalch equation to calculate the ph of a solution that is M in HClO and M in KClO. Ka = 2.9 x 10-8 for HClO. The Henderson equation is ph = pka + log10{[base]/[acid]} Also Ka(HClO) = 2.9 x 10-8 [HClO] = M [ClO - ] = M ph = - log10(2.9 x 10-8 ) + log10{(0.155)/(0.135)} =

4 6) (17.18) Which of the following solutions can act as a buffer? a) KCN/HCN c) NH3/NH4NO3 b) Na2SO4/NaHSO4 d) NaI/HI a) This contains a weak acid (HCN) and its conjugate base (CN - ), so a buffer. b) This contains a weak acid (HSO4 - ) and its conjugate base (SO4 2- ), so a buffer. c) This contains a weak base (NH3) and its conjugate acid (NH4 + ), so a buffer. d) This contains a strong acid (HI) and its conjugate base (I - ). The conjugate base of a strong acid has no acid/base properties, and so this is simply a solution of a strong acid, so not a buffer. 7) What mass of ammonium chloride (NH 4 Cl) should you add to 2.55 L of a M NH3 solution to obtain a buffer with a ph of 9.55? The Henderson equation is ph = pka + log10{[base]/[acid]} Kb(NH3) = 1.76 x 10-5 ; so Ka for NH4 + can be found by Ka Kb = 1.0 x 10-14, so Ka(C2H5NH3 + ) = 1.0 x = 5.68 x x = - log10(5.68 x ) + log10{[base]/[acid]} = log10{[base]/[acid]} So log10{[base]/[acid]} = = 0.30 Take the inverse log10, to get [base]/[acid] = = 2.00 And so [acid] = [base]/2.00 = (0.155 M)/2.00 = M M(NH4Cl) = g.mol So g(nh4cl) = 2.55 L soln mol g = g NH4Cl 1 L 1 mol 8) (17.27) A g sample of a monoprotic acid neutralizes 16.4 ml of a M solution of KOH. Find the molar mass of the acid. The neutralization reaction is (with HA = weak monoprotic acid) HA(aq) + KOH(aq) KA(aq) + H2O( ) 4

5 The moles of KOH is n(koh) = ( L) mol = x 10-3 mol L Since 1 mole of HA reacts with 1 mole of KOH, this is the number of moles of HA. Therefore MW = g = 201. g/mol x 10-3 mol 9) (17.30) In a titration experiment, 20.4 ml of a M solution of HCOOH neutralizes 19.3 ml of a solution of Ba(OH)2. What is the concentration of the Ba(OH)2 solution? The neutralization reaction is 2 HCOOH(aq) + Ba(OH)2(aq) Ba(HCOO)2(aq) + 2 H2O( ) The moles of Ba(OH)2 is moles Ba(OH)2 = L HCOOH mol HCOOH 1 mol Ba(OH)2 L soln 2 mol HCOOH = 8.50 x 10-3 mol Ba(OH)2 [Ba(OH)2] = 8.50 x 10-3 mol Ba(OH)2 = M Ba(OH) L soln 10) For the following titrations indicate whether the ph at the equivalence point of the titration will be much larger than 7.0, approximately 7.0, or much smaller than 7.0. a) Titration of CH3COOH (a weak acid) with KOH (a strong soluble base). The neutralization reaction is CH3COOH(aq) + KOH(aq) KCH3COO(aq) + H2O( ) At the equivalence point all of the reactants will have been converted into KCH3COO and water. K + has no acid/base properties, while CH3COO - is the conjugate base of CH3COOH, and so is a weak base. Therefore the ph will be much larger than 7.0. b) Titration of HCl (a strong acid) with KOH (a strong soluble base). The neutralization reaction is HCl(aq) + KOH(aq) KCl(aq) + H2O( ) 5

6 At the equivalence point all of the reactants will have been converted into KCl and water. Neither K + nor Cl - have any acid/base properties. Therefore the ph will be approximately ) Consider the titration of a 35.0 ml sample of M HBr with a M solution of KOH. Determine each quantity: a) The initial ph Since HBr is a strong acid initial [H3O + ] = initial [HBr] = M, so ph = - log10(0.175) = 0.76 b) The volume of added base required to reach the equivalence point The neutralization reaction is HBr(aq) + KOH(aq) KBr(aq) + H2O( ) Since this is a 1:1 acid:base reaction, it follows that at the equivalence point moles acid = moles base (35.0 ml) (0.175 M) = (x ml) (0.200 M) x = M (35.0 ml) = ml M c) The ph after the addition of 10.0 ml of base After 10.0 ml added base total volume = 35.0 ml ml = 45.0 ml initial moles acid = ( L) (0.175 M) = mol acid initial moles base = ( L) (0.200 M) = mol base moles excess acid = mol mol = mol So the concentration of excess acid is mol = M L Since the excess acid is a strong acid, ph = - log10(0.0916) =

7 d) The ph at the equivalence point At the equivalence point neutralization has occurred, and so what is present is a KBr solution, which has no acid-base properties. So the ph will be 7.0. e) A suitable indicator for the titration (see Table 17.3) We would like an indicator that changes color at ph ~ 7.0 (see Fig for an example of a titration curve for a strong acid titrated by a strong base). Suitable indicators would be methyl red ( ), chlorophenol blue ( ), bromothymol blue ( ), cresol red ( ), or phenolphthalein ( ) f) The ph after adding 5.0 ml of base beyond the equivalence point volume = volume at equivalence + excess volume = 35.0 ml = ml ml = 70.6 ml mol excess base = (0.005 L) (0.200 M) = mol concentration of excess base = mol = M L Since the base is strong, poh = - log10(excess base) = - log10(0.0142) = 1.84 And so ph = poh = = ) (17.45 a,c,e) Write balanced equations and solubility product expressions for each of the following compounds. a) CuBr(s) CuBr(s) Cu + (aq) + Br - (aq) Ksp = [Cu + ] [Br - ] c) Ag2CrO4(s) Ag2CrO4(s) 2 Ag + (aq) + CrO4 2- (aq) Ksp = [Ag + ] 2 [CrO4 2- ] e) AuCl3(s) AuCl3(s) Au 3+ (aq) + 3 Cl - (aq) Ksp = [Au 3+ ] [Cl - ] 3 13) Use the given molar solubilities in pure water to calculate Ksp for each compound. a) BaCrO4; molar solubility = 1.08 x 10-5 M BaCrO4(s) Ba 2+ (aq) + CrO4 2- (aq) Ksp = [Ba 2+ ] [CrO4 2- ] [Ba 2+ ] = 1.08 x 10-5 M [CrO4 2- ] = 1.08 x 10-5 M Ksp = (1.08 x 10-5 ) (1.08 x 10-5 ) = 1.17 x

8 b) Ag2SO3; molar solubiliy = 1.55 x 10-5 M Ag2SO3(s) 2 Ag + (aq) + SO3 2- (aq) Ksp = [Ag + ] 2 [SO3 2- ] [Ag + ] = 2 (1.55 x 10-5 M) = 3.10 x 10-5 M [SO3 2- ] = 1.55 x 10-5 M Ksp = (3.10 x 10-5 ) 2 (1.55 x 10-5 ) = 1.49 x c) Pd(SCN)2; molar solubility = 2.22 x 10-8 M Pd(SCN)2(s) Pd 2+ (aq) + 2 SCN - (aq) Ksp = [Pd 2+ ] [SCN - ] [Pd 2+ ] = 2.22 x 10-8 M [SCN - ] = 2 (2.22 x 10-8 M) = 4.44 x 10-8 M K sp = (2.22 x 10-8 ) (4.44 x 10-8 ) 2 = 4.38 x ) (17.56) The ph of a saturated solution of a metal hydroxide with formula MOH is Find Ksp for this compound. The solubility reaction for this compound is MOH(s) M + (aq) + OH - (aq) Ksp = [M + ] [OH - ] The poh of the solution is poh = ph = = 4.32 If we assume all of the hydroxide ion comes from the metal hydroxide, then [M + ] = [OH - ] = 10 -poh = = 4.8 x 10-5 M Ksp = [M + ] [OH - ] = (4.8 x 10-5 ) 2 = 2.3 x ) Calculate the molar solubility of copper II sulfide (CuS) in each liquid or solution. Note that Ksp(CuS) = 1.27 x Note that the reaction for the dissolution of CuS(s) is CuS(s) Cu 2+ (aq) + S 2- (aq) Ksp = [Cu 2+ ] [S 2- ] = 1.27 x a) pure water Cu 2+ 0 x x S 2-0 x x (x) (x) = x 2 = 1.27 x ; x = (1.27 x ) 1/2 = 1.13 x Molar solubility is 1.13 x M 8

9 b) 0.25 M CuCl2, which gives 0.25 M Cu 2+ Cu x x S 2-0 x x ( x) (x) = 1.27 x Assume x << 0.25, then (0.25) (x) = 1.27 x x = (1.27 x )/(0.25) = 5.08 x Molar solubility is 5.08 x M c) 0.20 M K 2 S, which gives 0.20 M S 2- Cu 2+ 0 x x S x x (x) ( x) = 1.27 x Assume x << 0.20, then (x) (0.20) = 1.27 x x = (1.27 x )/(0.20) = 6.35 x Molar solubility is 6.35 x M 16) (17.78) Which of the following ionic compounds will be more soluble in acid solution than in water: a) BaSO4 b) PbCl2 c) Fe(OH)3 d) CaCO3 To answer this question we need to write the solubility reaction, and then see what effect, if any, increasing H3O + concentration and decreasing OH - concentration will have, based on Le Chatlier's principle. a) BaSO4(s) Ba 2+ (aq) + SO4 2- (aq) Since SO4 2- will react with H3O + by the process SO4 2- (aq) + H3O + (aq) HSO4 - (aq) + H2O(l) then in acid solution SO4 2- will be removed by this process. Le Chatlier's principle predicts the system will respond by dissolving more BaSO4, so the solubility will increase. b) PbCl2(s) Pb 2+ (aq) + 2 Cl - (aq) Increasing H 3 O + concentration has no effect on this process, so no change. 9

10 c) Fe(OH)3(s) Fe 3+ (aq) + 3 OH - (aq) In acid solution OH - concentration will decrease. Le Chatlier's principle predicts the system will respond by dissolving more Fe(OH)3 to oppose this decrease. So solubility will increase. d) CaCO3(s) Ca 2+ (aq) + CO3 2- (aq) Since CO3 2- will react with H3O + by the process CO3 2- (aq) + H3O + (aq) HCO3 - (aq) + H2O(l) then in acid solution CO3 2- will be removed by this process. Le Chatlier's principle predicts the system will respond by dissolving more CaCO3, so the solubility will increase. 17) Calculate the molar solubility of copper II hydroxide (Cu(OH)2, Ksp = 2.2 x ) in a solution buffered at each of the following values for ph. a) ph = 4.0 b) ph = 7.0 c) ph = 9.0 Cu(OH)2(s) Cu 2+ (aq) + 2 OH - (aq) Ksp = [Cu 2+ ] [OH - ] 2 = 2.2 x So [Cu 2+ ] = 2.2 x [OH - ] 2 a) ph = 4.00 ; poh = = ; [OH - ] = = 1.00 x [Cu 2+ ] = molar solubility = 2.2 x = 2.2 M (1.0 x ) 2 b) ph = 7.00 ; poh = = 7.00 ; [OH - ] = = 1.00 x 10-7 [Cu 2+ ] = molar solubility = 2.2 x = 2.2 x 10-6 M (1.0 x 10-7 ) 2 c) ph = 9.00 ; poh = = 5.00 ; [OH - ] = = 1.00 x 10-5 [Cu 2+ ] = molar solubility = 2.2 x = 2.2 x M (1.0 x 10-5 ) 2 18) (17.99) Calculate the solubility (in g/l) of Ag2CO3. Ksp = 8.1 x The reaction is Ag2CO3(s) 2 Ag + (aq) + CO3 2- (aq) Ksp = [Ag + ] 2 [CO3 2- ] 10

11 Ag x 2x CO x x (2x) 2 (x) = 4x 3 = 8.1 x x = [ (8.1 x )/4 ] 1/3 = 1.27 x 10-4 So the molar solubility of Ag2CO3 is 1.27 x 10-4 mol/l. Since MW(Ag2CO3) = g/mol then the solubility in g/l is (1.27 x 10-4 mol/l) (275.8 g/mol) = g/l 19) (17.106) Cacodylic acid ((CH 3 ) 2 AsO 2 H) has an ionization constant K a = 6.4 x 10-7 at T = 25. C. a) Find the ph of 50.0 ml of a M solution of cacodylic acid. We will represent cacodylic acid as HA, and let A - = (CH 3 ) 2 AsO 2 - be the conjugate base of cacodylic acid. The equilibrium reaction is then HA(aq) + H2O( ) H3O + (aq) + A - (aq) Ka = [H3O + ] [A - ] = 6.4 x 10-7 H 3 O + 0 x x A - 0 x x HA x x (x) (x) = 6.4 x 10-7 (0.100 x) If we assume x << 0.100, then the above equation becomes [HA] x 2 = 6.4 x 10-7 x 2 = (0.100) (6.4 x 10-7 ) = 6.4 x 10-8 (0.100) x = (6.4 x 10-8 ) 1/2 = 2.5 x 10-4 ph = - log10(2.5 x 10-4 ) =

12 b) Find the ph of 25.0 ml of a M solution of (CH3)2AsO2Na, the sodium salt of the conjugate base of cacodylic acid. As before, let HA = cacodylic acid and A - = the conjugate base of cacodylic acid. To find Kb for the conjugate base, we can use the expression Ka Kb = 1.0 x (at T = 25. C) Kb = 1.0 x = 1.0 x = 1.6 x 10-8 Ka 6.4 x 10-7 When the sodium salt of the conjugate base of cacodylic acid is added to water, it will completely dissociate (since sodium salts are expected to be soluble in water) (CH3)2AsO2Na(s) Na + (aq) + (CH3)2AsO2 - The equilibrium reaction is then ( A - = (CH3)2AsO2 - ) A - (aq) + H2O( ) HA(aq) +OH - (aq) Kb = [HA] [OH - ] = 1.6 x 10-8 [A - ] HA 0 x x OH - 0 x x A x x (x) (x) = 1.6 x 10-8 (0.150 x) If we assume x << 0.150, the the above equation becomes x 2 = 1.6 x 10-8 x 2 = (0.150) (6.4 x 10-8 ) = 9.6 x 10-9 (0.150) x = (9.6 x 10-9 ) 1/2 = 9.8 x 10-5 poh = - log10(9.6 x 10-5 ) = 4.01 ph = = 9.99 c) The solutions in a and b are mixed. Find the ph of the resulting solution. When the solutions are mixed the total volume of solution will be V = 50.0 ml = 25.0 ml = 75.0 ml. The initial concentrations of cacodylic acid (HA) and conjugate base (A - ) will therefore be 12

13 Initial HA = (0.100 M) (50.0 ml) = M 75.0 ml Initial A - = (0.150 M) (25.0 ml) = M 75.0 ml We could do this problem as a normal equilibrium calculation, but since the solution contains a significant concentration of both a weak acid and its conjugate base, we can use the Henderson equation ph = pka + log10 ( [base]/[acid] ) = - log10(6.4 x 10-7 ) + log10( 0.050/0.067) = =