CHM152 Exam #2 Review

CHM152 Exam #2 Review Spring 2014 (Ch. 16. 17) KEY 1. For each of the following acid / base reactions, determine if each of the following will have more products or reactants (Ka HSO4- = 1.3x10-2, Kb NH3 = 1.8x10-5, Ka HF = 6.8x10-4):  NH4+ (aq) + F- (aq)___ NH3 (aq) + HF (aq) More Products More reactants  OH- (aq) + HNO3 (aq) ___ H2O (l) + NO3- (aq) More Products More reactants  HSO4- (aq) + NH3 (aq) ___ NH4+ (aq) + SO42- (aq) More Products More reactants  2H2O (l) ___ H3O+ (aq) + OH- (aq) More Products More reactants 2. For the acid/base reactions determine the conjugate acid/base pairs for each.     NH4+ / NH3 (acid / base), F- / HF (base / acid) OH- / H2O (base / acid), HNO3 / NO3 (acid / base) HSO4- / SO42- (acid / base), NH3 / NH4+ (base / acid) H2O / H3O+ (base / acid), H2O / OH- (acid / base) The equilibrium of a Bronsted/Lowry acid/base reaction will favor the side of the equation with the weaker acid and weaker base. We can use Ka and Kb values to determine the weaker acid/base. 3. Explain how NH3 is both a Bronsted/Lowry base and a Lewis base. It is a Bronsted/Lowry base because it is able to accept protons. It is a Lewis base because it is able to donate electrons. 4. Determine the pH, pOH, [H3O+], and [OH-] for the following solutions:  0.0051M HBr pH = -log(0.0051) = 2.29 , pOH = 14.00 – 2.29 = 11.71 [H+] = 10-2.29 = 0.0051M, [OH-] = 10-11.71 = 1.9x10-12M  0.010M NaOH pOH = -log(0.010) = 2.00, pH = 14.00 – 2.00 = 12.00 [OH-] = 10-2.00 = 0.10M, [H+] = 10-12.00 = 1.0x10-12M  0.010M Sr(OH)2 pOH = -log(0.010 x 2) = 1.70, pH = 14.00 – 1.70 = 12.30 [OH-] = 101.7- = 0.020M, [H+] = 10-12.30 = 5.0x10-13M 5. Determine whether the following salts are acidic, basic, or neutral:  LiCl nuetral  Al(NO3)3 acidic  BaF2 basic  NH4I acidic 6. What is the pH of a 0.100 M formic acid solution with a Ka = 1.8 x 10-4? HCOOH + H2O ⇌ H3O+ + HCOO0.100M 0M 0M +x -x -x 0.100M + x -x -x pH = -log[H3O+] = -log(4.24x10-3) = 2.37 x2  1.8x10 4 0.100  x x 2  1.8x10 5 x  4.24x10 3 x2  1.8x10 4 0.100 x 2  1.8x10 5 4.24x10 3 x100  4.2% OK assumption 0.100 7. What is the percent dissociation? 4.2% 8. If the concentration of the formic acid was 4.0M instead, explain how the pH, [OH-], and % dissociation would be different (higher/lower/same). The pH would be lower (more acidic/higher [H3O+]), [OH-] would be lower, % dissociation would be lower (high initial concentration = lover % dissociation for same acid or base). 9. Calculate the pH if 2.56 grams of sodium fluoride is dissolved in 4.50 liters of water. Ka for hydrofluoric acid is 7.1 x 10-4 2.56g NaF 1molNaF  6.097x10 2 molNaF 41.99g NaF F- + H2O -2 1.353x10 M -x 1.353x10-2M – x ⇌ OH- HF 0M 0M +x +x x x pH = 14-log[OH+] = 14-log(4.37x10-7) = 7.64 6.097x10 2 molNaF  1.353x10 2 MNaF 4.50L 1.0x10 14 K a  Kb  K w  1.0x10 14  1.41x10 11 7.1x10 4 x2  1.41x10 11 2 1.353x10  x x 2  1.91x10 13 x2  1.41x10 11 2 1.353x10 7 x 2  1.91x10 13 x  4.37x10 10. What is the initial concentration of benzoic acid (C6H5COOH) in a solution with pH = 2.59 (assume the benzoic acid was added to the water with no other ions initially)? The acid dissociation constant for this monoprotic acid is 6.5 x 10-5. HA + H2O ? (y) -x y-2.57x10-3 ⇌ H3O+ + A0M 0M +x +x -3 2.57x10 2.57x10-3 0.104M - 2.57x10-3M = 0.101M or 0.10M [H3O ]  10pH  10-2.59  2.57x10 3 (2.57x10 3 )2  6.5x10 5 y  2.57x10 3 y  0.104M 11. Explain the common-ion effect. Weak acid or weak base’s % dissociation will decrease if they are placed in solutions with one of the products of dissociation. For example, The % dissociation of HCN will be decrease if additional CN- is added to the solution (as NaCN, for example). HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN-(aq) 12. Calculate the percent ionization of a 1.00M solution of HF (pKa = 3.17). HF + H2O ⇌ H3O+ + FI 1.00M C -x E 1.00-x 0 +x +x 0 +x +x K a  10 3.17  6.76x10 4 (x)(x)  6.76x10 4 1.00 x  6.76x10 4  2.6x10 2 2.6x10 2 x100  2.6% 1.00 13. Calculate the percent ionization if there is also a 0.50M concentration of NaF in solution. HF + H2O ⇌ H3O+ + FI 1.00M 0 0.500M C -x +x +x E 1.00-x +x 0.500+x (x)(0.500)  6.76x104 1.00 x  1.35x10-3 1.35x10-3 x100  0.14% 1.00 14. Calculate the percent ionization if the pH of the solution before the HF was added was 0.30. HF + H2O ⇌ H3O+ + FI 1.00M 0.500M 0 C -x +x +x E 1.00-x 0.500+x +x [H3O  ]  10 0.30  0.50M (0.500)(x)  6.76x10 4 1.00 x  1.35x10-3 1.35x10-3 x100  0.14% 1.00 15. You are titrating 10.00mL of 1.00M of HCl with 2.00M NaOH.  What would be the pH at ½ the equivalence point? NaOH + HCl → H2O + NaCl vol of NaOH@eq. 0.01000L HCl 1.00 molHCl 1 molNaOH 1 L NaOH  0.00500L NaOH 1 L HCl 1 molHCl 2.00molNaOH 1  2.00mol  mol of NaOH@ eq. (0.00250L)   0.00500molNaOH 2  1L   1.00mol  mol of HCl. (0.01000L)   0.0100molNaOH H2O + NaCl  1L  1 0.00500L NaOH vol of NaOH@ eq.  0.00250L NaOH 2 2 HCl + NaOH ⇌ B 0.0100mol 0.00500mol Δ -0.00500mol -0.00500mol A 0.0050mol 0 This makes sense since only half of the acid should be left at ½ eq. If you realize this, you do not have to use a change table.   mol HCl 0.0050mol    log pH  - log[H  ]  - log   log(0.40)  0.40 1   0.002500L  0.01000L   total volume @ 2 eq   What would be the pH at the equivalence point? 1.00 mol HCl 1 mol NaOH 1 L NaOH  0.00500L NaOH 1 L HCl 1 mol HCl 2.00mol NaOH  2.00mol  mol of NaOH@ eq.  (0.00500L)   0.0100mol NaOH  1L  vol of NaOH@ eq.  0.01000L HCl     H+ + OH- ⇌ H2O B 0.0100mol 0.0100mol Δ -0.0100mol -0.0100mol A 0 0 There will only be a neutral salt at the equivalence point of a Strong Acid/Strong Bade titration. pH = 7  What would be the pH after adding 0.75mL of NaOH after the equivalence point?  2.00mol  Total mol of NaOH  (0.00575L)   0.0115mol NaOH  1L  H+ + OH- ⇌ H2O B 0.0100mol 0.0115mol Δ -0.0100mol -0.0100mol A 0 0.0015mol After the equivalence point, you are just adding extra OH ions to the solution 0.0015mol   pH  14 - log   12.98  0.01000L  0.00500L  0.00075L  16. You are titrating 10.00mL of 1.00M of benzoic acid (Ka is 6.4x10-5) with 2.00M NaOH. What would be the pH after 2.50 mL has been added? HA + OH- ⇌ H2O + AB 0.0100mol 0.00500mol 0 Δ -0.00500mol -0.00500mol +0.00500mol A 0.00500mol 0 +0.00500mol [A - ] pH  pKa  log [HA] pH  4.19  log 0.00500mol  4.19 0.00500mol  What would be the pH after adding a total of 5.00 mL NaOH? Think about what is left at the equivalence point of a weak acid/strong base titration. You could use a change table if you are unsure. But it’s faster if you don’t have to.  acid(HA)@start mol of conjugatebase(A - ) ateq. mol of  1.00mol  mol of HA.  (0.01000L)   0.0100mol HA atstart  molof A @eq  1L  1.00 mol HA 1 molNaOH 1 L NaOH vol of NaOH@eq. 0.01000L HA  0.00500L NaOH 1 L HAl 1 mol HA 2.00molNaOH [ A- ]  molof A0.0100mol A  0.6667M totalvolume@ eq. 0.01000L  0.00500L A- + H2O ⇌ OH- + HA I 0.6667M 0 0 C -x +x +x E 0.6667-x x x Kb  (x)(x)  1.56x10 10 0.6667 Kw  1.56x10 10 Ka x  (1.56x10 10 )(0.6667)  1.0x10 5 pH = 14-log[OH-] = 14-log (1.0x10-5) = 9.01  What would be the pH after adding a total of 5.75 mL NaOH? HA + OH- ⇌ H2O + AB 0.0100mol 0.0115mol 0 Δ -0.0100mol -0.0100mol +0.0100mol A 0 0.0015mol +0.0100mol Although you have two bases after the equivalence point, assume that the concentration is going to have a much greater effect on pH than the weak base.  2.00mol  Total mol of NaOH  (0.00575L)   0.0115mol NaOH  1L  0.0015mol   pH  14 - log   12.98  0.01000L  0.00500L  0.00075L  17. What is the pH of 100.0mL solution containing 0.30 M NH3 and 0.30 M NH4NO3 (Kb of NH3 is 1.8x10-5)? pKb  logKb  - log1.8x10-5  4.745 pKa  14  pKa  14  4.745  9.255 pH  pKa  log  - [A ] 0.30M  9.255  log  9.26 [HA] 0.30M Since this has the components of a buffer solution, we can use Henderson-Hasslebalch What is the pH after adding 1.00mL of 2.00M NaOH 0.30 mol  0.030 molsof each 1L 2.00 mol molsof NaOH added 0.00100L  0.00200 molsof NaOH 1L molsof NH3 andNH4   0.1000L NaOH + NH4+ → NH3 B 0.00200mol 0.030mol 0.030mol Δ -0.00200mol -0.00200mol +0.00200mol A 0 0.028mol 0.032mol pH  pKa  log [A- ] 0.032mol  9.225  log  9.28 [HA] 0.028mol  What is the pH after adding 2.00mL of 1.00M HCl? HCl B 0.00200mol Δ -0.00200mol A 0 molsof HCl added 0.00200L + NH3 → NH4+ 0.030mol 0.030mol -0.00200mol +0.00200mol 0.028mol 0.032mol pH  pKa  log 1.00 mol  0.00200 molsof HCl 1L [A- ] 0.028mol  9.225  log  9.17 [HA] 0.032mol 18. You have 60.00mL of a buffer solution consisting of 0.900M CH3COOH and 1.10M NaCH3COO. Kb = 5.6x10-10  Determine the pH of the solution after adding 10.11mL of 2.23M HBr. 0.06000L CH 3 COOH 0.01011L HBr 0.900mol CH 3 COOH  0.0540mol CH 3 COOH 1 L CH 3 COOH 2.23mol HBr  0.0225mol HBr 1 L HBr HBr B 0.0225mol Δ -0.0225mol A 0  CH3COO0.0660mol -0.0225mol 0.0435mol + 1.10mol CH 3 COO  0.0660mol CH 3 COO 1 L CH 3 COO - 1.0x10 14  1.8 x105 5.6 x1010 → CH3COOH + Br[A - ] 0.0435 pH  pKa  log  4.74  log  4.44 0.0540mol [HA] 0.0860 +0.0225mol 0.0860mol How much total HBr would you have to add until the buffer was used up? 0.06000L CH 3 COO -  Ka  0.06000L CH 3 COO - 1.10mol CH 3 COO 1 mol HBr 1 L HBr  0.0296L 1 L CH 3 COO 1 mol CH 3 COO - 2.23 mol HBr What would be the pH when at the point (that you calculated above?) HBr + CH3COOB 0.0660mol 0.0660mol Δ -0.0660mol -0.0660mol A 0 0mol → CH3COOH + Br0.1200mol CH 3 COOH 0.0540mol  1.34M (0.06000L  0.0296L) +0.0660mol 0.1200mol CH3COOH + H2O ⇌ H3O+ + CH3COOI 1.34M 0 0 C -x +x +x E 1.34M-x x x (x)(x) 1.34 x  (1.8x10 -5 )(1.34)  4.9 x10 3  1.8 x10 5 pH  log(4.9x10 3 )  2.31 19. If you start over with fresh buffer (same conditions/volumes as original), what would be the pH of the solution after adding 20.00mL of 2.23M KOH? 0.06000L CH 3 COOH 0.02000L NaOH 0.900mol CH 3 COOH  0.0540mol CH 3 COOH 1 L CH 3 COOH 0.06000L CH 3 COO - 1.10mol CH 3 COO  0.0660mol CH 3 COO 1 L CH 3 COO - 2.23mol NaOH  0.0446mol NaOH 1 L NaOG NaOH B 0.0446mol Δ -0.0446mol A 0  + CH3COOH 0.0660mol -0.0446mol 0.0214mol → CH3COO- + Na+ [A- ] 0.0986 pH  pKa  log  4.74  log  5.40 0.0540mol [HA] 0.0214 +0.0446mol 0.0986mol After adding 30.00mL of 2.23 KOH? 0.03000L NaOH 2.23mol NaOH  0.0669mol NaOH 1 L NaOG NaOH B 0.0669mol Δ -0.0660mol A 0.0009 mol + CH3COOH 0.0660mol -0.0660mol 0 mol pOH   log[OH  ]   log( 0.01)  2.0 → CH3COO- + Na+ 0.0009mol NaOH 0.0540mol  0.01M NaOH (0.06000L  0.03000L) +0.0660mol 0.1200mol pH  14.0 - 2.0  12.0 20. If you wanted to make a buffer with a pH of 3.90 using the same buffer as before with a total concentration of 2.00M, what would the concentration of each part of the buffer need to be? pH  pKa  log 0.145  x (2.0 - x) [A- ] [HA] 3.90  4.74  log 0.29  0.15x  x x (2.0 - x) 0.29 1.15x  1.15 1.15  0.84  log x (2.0 - x) [A ]  x  0.25M  x   log  (2.0- x)   10 0.84  10 [HA]  2.00M - 0.25M  1.75M 21. *If 0.00056 moles of AgNO3 is added to 1.00L of 0.0020M NaCl, will AgCl precipitate form? (Ksp AgCl = 1.8x10-10). Explain. AgCl(s) ⇌ Ag+(aq) + Cl-(aq) Ksp = [Ag][Cl] Qsp = (0.00056)(0.0020) = 1.1x10-6 since Q > K will shift to the left (solid AgCl), so it will precipitate! What would the minimum concentration of NaCl need to be to form a precipitate? Ksp = [Ag][Cl] = (0.00056)[Cl-] = 1.8x10-10 [Cl-] = 3.2x10-7M 22. Ni(OH)2 is an insoluble compound in water. How would the molar solubility be affected in each if the following situations?    NaOH is added – Decrease solubility (common ion) NaCl is added – No effect on solubility. Does not react with Ni2+ or OHHCl is added – Increase solubility. H+ reacts with OH-. 23. Assume you have a saturated solution of nickel (II) phosphate with some solid in the container (Ksp = 4.74 x 10-32). What is it’s molar solubility? Determine the concentration of the Ni2+ and PO43- ions in the solution. Ni3(PO4)2 ⇌ 3Ni2+ + 2PO43Ksp = [Ni2+]3[PO43-]2 4.74x10-32 = (3x)3(2x)2 =(27x3)(4x2) I 0 0 C +3x +2x 4.74x10-32 = 108x5 x = 2.13x10-7 x = molar solubility = E 3x 2x 2.13x10-7M [Ni2+] = (3)(2.13x10-7) = 6.39x10-7M [PO43-] = (2)(2.13x10-7) = 4.26x10-7M 24. What would be it’s molar solubility be if nickel (II) phosphate if 0.100moles of Na2PO4 was added to 1.00L of the solution above? What would be the concentrations of the Ni2+ and PO43ions? Ni3(PO4)2 + H2O ⇌ 3Ni2+ + 2PO43(27x3)(0.0100) I 0 0.100 C +3x +2x E 3x 0.100+2x Ksp = [Ni2+]3[PO43-]2 4.74x10-32 = (3x)3(0.100)2 = 4.74x10-32 = 0.270x3 x = 5.60x10-11 x = molar solubility = 5.60x10-11M [Ni2+] = (3)(5.60x10-11) = 1.68x10-10M [PO43-] = 0.100+(2)( 5.60x10-11) ≈0.100M 25. How much (in g) barium sulfate (BaSO4, MW=233.39) will dissolve in 500. mL of water. For BaSO4: Ksp = 1.1 x 10-10 Ba2 SO4 + H2O ⇌ Ba2+ + SO42I 0 0 C +x x E x x Ksp = [Ba2+][SO42-] 1.1x10-10 = x2 x = 1.04x10-5 = molar solubility = 1.04x10-5M 0.500L H2 O 1.04x10 5 molBaSO4 233.39g BaSO4  1.2x10 3 g BaSO4 1 L H2 O 1 molBaSO4