Answer key - Anoka-Ramsey Community College

Aspaas Chem 1062 S05 EXAM 3 Name____________________I. Multiple choiceChoose the best answer from the choices given, and clearly circle the letter of yourchoice. You may need to consult the tables on page 2 for some answers. (3 pts each)1. Which of the following is the strongest acid in water?a. HBrOb. HBrO 2c. HBrO 3d. HBrO 4e. Not enough information to say.2. For the following equilibrium, which of the following statements is true?HF + H 2 O ! F – + H 3 O +a. The equilibrium lies predominately to the right.b. The equilibrium lies predominately to the left.c. This isn’t an equilibrium at all, since HF is a strong acid.d. This reaction will not occur at all.e. None of these statements are true.3. Which of the following is a diprotic acid?a. HC 2 H 3 O 2b. H 2 CO 3c. HCHO 2d. C 6 H 5 NH 2e. HC 3 H 3 O 34. Which of the following solutions will have the highest pH?a. 0.1 M Na 2 CO 3b. 0.1 M KClc. 0.1 M NaNO 3d. 0.1 M NaCle. 0.1 M NH 4 Cl5. Which of the following solution mixtures is a buffer?a. HCl / NaClb. KH 2 PO 4 / KNO 3c. HC 2 H 3 O 2 / NaC 2 H 3 O 2d. HOCN / NaOHe. HOCN / HCNPage 3 of 9

Aspaas Chem 1062 S05 EXAM 3 Name____________________II. ConceptualPredict whether aqueous solutions of the following salts will be acidic, basic, or neutral.(3 pts each)6. Na 2 S basic7. Cu(NO 3 ) 2 acidic8. KClO 4 neutral9. CH 3 NH 3 Cl acidic10. NH 4 F acidicAn unidentified acid is titrated by a strong base, and its titration curve is shown below:pH8.923.8532 mL 64 mLvolume strong basesolution11. (3 pts) Is the acid strong or weak? weak12. (3 pts) Is the acid monoprotic, diprotic, or triprotic? monoprotic13. (4 pts) The identity of the acid can be determined from the values on this curve.Which acid is it?pH at halfway point = pKa of acid = 3.85; K a = 1.41 x 10 -4 (Pyruvic acid)Page 4 of 9

Aspaas Chem 1062 S05 EXAM 3 Name____________________III. ProblemsChoose 3 of the following problems to complete. If you answer more than three, only thefirst three will be graded. (20 points each)I would like questions numbered _____, _____, and _____ to be graded.14. a. Calculate [H 3 O + ], [HS - ], and pH for a 0.10 M H 2 S solution.H 2S + H 2O = HS - + H 3O +0.10 M 0 0!x +x +x0.10 ! x x xK a= 8.9 " 10 !8 =c aK a=x 20.10 ! x0.10>> 100!88.9 " 108.9 " 10 !8 # x20.10x = (0.10)(8.9 " 10 !8 ) = 9.43 " 10 !5 M = [H 3O + ] = [HS ! ]pH = ! log(9.43 " 10 !5 ) = 4.03b. Calculate [S 2– ] in this solution.HS ! + H 2O = S 2! + H 3O +9.43 " 10 -5 M 0 9.43 " 10 -5!x +x +x9.43 " 10 !5 ! x x 9.43 " 10 !5 + xK a= 1.2 " 10 !13 = x(9.43 " 10!5 + x)9.43 " 10 !5 ! xc a9.43 " 10!5= >> 100!13K a1.2 " 101.2 " 10 !13 = x(9.43 " 10!5 )9.43 " 10 !5 = x[S 2! ] = 1.2 " 10 !13 MPage 5 of 9

Aspaas Chem 1062 S05 EXAM 3 Name____________________15. a. Use the quadratic equation method to calculate the pH of a 0.0015 M solutionof NaCN.CN ! + H 2O = HCN + OH !0.0015 M 0 0!x +x +x0.0015 ! x x xK b=c bK b=K w1.0 " 10!14=K a, HCN4.9 " 10 = 2.04 " !10 10!5 =0.0015= 73.5 < 100!52.04 " 10(0.0015 ! x)(2.04 " 10 !5 ) = x 2x 2 + 2.04 " 10 !5 x ! 3.06 " 10 !8 = 0x 20.0015 ! xx = !2.04 " 10!5 ± (2.04 " 10 !5 ) 2 + 4(3.06 " 10 !8 )2x = 1.65 " 10 !4 = [OH ! ]pOH = 3.78pH = 10.22b. Perform the same calculation, this time using the approximation method.2.04 ! 10 "5 =x 20.0015x = (0.0015)(2.04 ! 10 "5 ) = 1.749 ! 10 "4pOH = 3.76pH = 10.24c. Do the answers differ? Explain why or why not, in terms of the ionization andequilibrium that is occurring here.Yes, the answers are slightly different. By assuming the equilibrium CN -concentration is equal to the initial CN - concentration, we’re dividing by alarger number and x becomes slightly larger. Since c b / K b < 100, thisassumption cannot be made.Page 6 of 9

Aspaas Chem 1062 S05 EXAM 3 Name____________________16. a. Calculate the molar concentration of pure water, assuming its density is1.00 g/mL.1.00 g1 mL!1000 mL1 L! 1 mol = 55.51 mol/L18.016 gb. Use this value along with the value of K w to calculate the K a of H 2 O,the K a of H 3 O + , and the K b of OH – . (Hint: no ICE tables are required!)K a(H 2O): H 2O + H 2O = HO ! + H 3O +K a= [H 3 O+ ][OH ! ][H 2O]= K w1.0 " 10!14=[H 2O] 55.51K a(H 3O + ) : H 3O + + H 2O = H 2O + H 3O +K a= [H 2 O][H 3 O+ ][H 3O + ]= 1.80 " 10 !16= [H 2O] = 55.51 = 5.551 " 10 1K b(OH ! ) : OH ! + H 2O = H 2O + OH !K b= [H 2 O][OH! ][OH ! ]= [H 2O] = 5.551 " 10 1c. How do the K a values you calculated compare in size to the K a values on thetable on page 2? What does this mean about the relative strengths of H 3 O + andH 2 O as acids?K a of H 2 O is much smaller than any other K a on the table, therefore water isa weak acid.K a of H 3 O + is much larger than any other K a on the table, thereforehydronium is a relatively strong acid.K b of OH - is much larger than any other K b on the table, therefore hydroxideis a relatively strong base.Page 7 of 9

Aspaas Chem 1062 S05 EXAM 3 Name____________________17. A buffer solution with a total volume of 100.0 mL is 0.100 M in acetic acid and0.100 M in sodium acetate.a. Calculate the pH of this solution.c a0.100= >> 100"5K a1.7 ! 10pH = pK a+ log [A" ][HA]pK a= " log(1.7 ! 10 "5 ) = 4.77pH = 4.77 + log .100.100 = 4.77(note pH = pK awhen acid and base are at equal concentration)b. Calculate the pH of this same solution after 20.0 mL of 0.100 M HNO 3 isadded.Initial moles of all components0.100 moln HA= 0.100 L ! = 0.0100 mol HAL0.100 moln A" = 0.100 L ! = 0.0100 mol A "L0.100 moln HNO3= 0.0200 L ! = 0.00200 molLNew total volume = 0.100 L + 0.0200 L = 0.120 LMoles of acid and base after HNO 3is added:n A"= 0.0100 mol " 0.00200 mol = 0.008 moln HA= 0.0100 mol + 0.00200 mol = 0.012 mol0.008 mol[A " ] =0.120 L = 0.0667 M[HA] =0.012 mol0.120 L = 0.100 MpH = 4.77 + log 0.06670.100 = 4.59Page 8 of 9

Aspaas Chem 1062 S05 EXAM 3 Name____________________-END OF EXAM-Page 9 of 9