Some values of Ksp. Ni(OH) x Sulfates BaS x Sulfides CaS 8 x 10-6 NiS 3.0 x 10-21

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1 Before you begin, make sure that your exam has all 10 pages. There are 32 required problems (3 points each, unless noted otherwise) and two extra credit problems (3 points each). Stay focused on your exam. YOU MUST: Put your name and student ID on the bubble sheet correctly. Put all your answers on the bubble sheet; nothing on this exam will be used for grading. Sign the statement on the last page of the exam. Turn in both the exam and bubble sheet when you are done. Good Luck! Some values of Ksp Carbonates BaCO x 10-9 NiCO x 10-9 Hydroxides Fe(OH) x Fe(OH) x Ni(OH) x Sulfates BaS x Sulfides CaS 8 x 10-6 NiS 3.0 x page 1

2 1. According to the Brønsted-Lowry definition, a base A. increases the H 3 O + concentration in an aqueous solution. B. increases the OH - concentration in an aqueous solution. C. is a proton acceptor. D. is a proton donor. E. is an electron-pair donor. Brønsted acids are proton donors and bases are proton acceptors 2. Which of the following is never a Brønsted-Lowry acid in an aqueous solution? A. hydrogen chloride, HCl(g) B. dihydrogen sulfide, H 2 S(g) C. ammonium chloride, NH 4 Cl(s) D. hydrogen fluoride, HF(g) E. magnesium oxide, MgO(s) Each of the others are recognized either as strong or as weak Brønsted-Lowry acids 3. Which salt forms a 0.10 M aqueous solution with the highest ph? A. NaNO 3 B. NH 4 Cl C. FeCl 3 D. Ca(ClO 4 ) 2 E. NaF NaNO 3 and Ca(ClO 4 ) 2 form neutral salts. NH + 4 is a weak acid and Cl - is the conjugate base of HCl NH 4 Cl is acidic. Hydrated Fe 3+, Fe 3+ (H 2 O) 6, is also acidic. F -, the conjugate base of a weak acid, is a weak base. NaF forms a basic solution. 4. Consider the following series of acids: CH 3 CO 2 H, ClCH 2 CO 2 H, Cl 2 CHCO 2 H, Cl 3 CCO 2 H Which statement explains the trend the pk A s (left to right). A. pk A s increase due to inductive effects generated by the substitution of H by Cl. B. pk A s decrease due to inductive effects generated by the substitution of H by Cl. C. pk A s increase due to the larger number of resonance structures possible in the anion. D. pk A s decrease due to the larger number of resonance structures possible in the anion. E. Both B and D. The greater electronegativity of Cl relative to H stabilizes the conjugate base anion by an inductive effect; pk a s decrease as more Cl is substituted for H. Resonance also stabilizes the anion, but the resonance structures are the same in all members of the series, and thus resonance does not contribute to the trend. page 2

3 5. At 15 C, the water ionization constant, K w, is What is the H 3 O + concentration in neutral water at this temperature? A M K w = [H 3 O + ][OH - ] B M For neutral water, [H 3 O + ] = [OH - ] C M K w = [H 3 O + ] 2 [H 3 O + ] = sqrt(k w ) D M [H 3 O + ] = sqrt(4.5 x ) = 6.7 x 10-8 E. 2.2 M 6. If you mix equal molar quantities of HF (K a = ) and KCN (K b = ), the resulting solution will be A. acidic because K a of HF is greater than K b of CN -. HF is a stronger acid than B. basic because K a of HF is greater than K a of HCN. CN - is a base. The solution C. basic because K a of HF is greater than K b of CN -. becomes acidic, since both D. basic because K b of F - is less than K b of CN -. are present in equal E. neutral because the weak acid neutralizes the weak base. amounts. HF wins. 7. Determine the equilibrium constant for the reaction below. (K a (HF) = , K a (NH 4 + ) = ) HF(aq) + NH 3 (aq) NH 4 + (aq) + F - (aq) A B C D E K is given by K a (HF)/K a (NH 4 + ) = ( )/( ) = 1.3 x Which of the following chemical equations corresponds to K a3 for phosphoric acid? A. HPO 2-4 (aq) + H 2 O(l) PO 3-4 (aq) + H 3 O + (aq) 3 rd Ionization, K a3 B. PO 3-4 (aq) + H 2 O(l) HPO 2-4 (aq) + OH - (aq) K b3 = K w /K a3 C. H 3 PO 4 (aq) + H 2 O(l) H 2 PO - 4 (aq) + H 3 O + (aq) 1 st Ionization, K a1 D. H 3 PO 4 (aq) + 3 H 2 O(l) PO 3-4 (aq) + 3 H 3 O + (aq) Overall Rxn, K a1 K a2 K a3 E. H 2 PO - 4 (aq) + H 2 O(l) HPO 2-4 (aq) + H 3 O + (aq) 2 nd Ionization, K a2 page 3

4 9. Boric acid (H 3 BO 3 ) has a pk a value of What is the K b of sodium borate (NaH 2 BO 3 ) at 25 o C? A B C D E pk b = pk w - pk a = = 4.86 K b = = 1.4 x In the following acid-base reaction equilibrium C 9 H 7 NH + + C 6 H 5 COO - C 9 H 7 N + C 6 H 5 COOH A. C 9 H 7 NH + is the conjugate acid of C 9 H 7 N. B. C 6 H 5 COO - is the conjugate acid of C 6 H 5 COOH. C. C 6 H 5 COOH and C 6 H 5 COO - represent a conjugate acid-base pair D. Both A and B E. Both A and C 11. Suppose that K < 1 for the equilibrium in problem 10. A. C 6 H 5 COO - is the weaker base and C 9 H 7 NH + is the stronger acid B. C 6 H 5 COO - is the weaker base and C 9 H 7 NH + is the weaker acid C. C 9 H 7 N is the weaker base and C 6 H 5 COOH is the weaker acid D. C 9 H 7 N is the stronger base and C 6 H 5 COOH is the stronger acid E. Both B and D 12. (4 pts) For the following three reactions, K > 1 HF (aq) + C 6 H 5 COO - (aq) F - (aq) + C 6 H 5 COOH (aq) HC 9 H 7 O 4 (aq) + C 6 H 5 COO - (aq) C 6 H 5 COOH (aq) + C 9 H 7 O 4 - (aq) HF (aq) + C 9 H 7 O 4 - (aq) F - (aq) + HC 9 H 7 O 4 (aq) Among the three acids and three conjugate bases, which are the strongest acid and base? A. HC 9 H 7 O 4 and C 6 H 5 COO - K > 1, therefore strong acids & bases are on left. B. HF and C 6 H 5 COO - The strongest acid has the weakest conj. base. C. HF and C 9 H 7 O 4 - The weakest acid has the strongest conj. base. D. C 6 H 5 COOH and F - Identify strongest and weakest acids. E. C 9 H 7 CO 4 - and F - HF, always on the left, is the strongest acid. C 6 H 5 OO -, always on the left, is the strongest base. page 4

5 13. Two solutions of the same weak acid (pk A = 6) were made to have different concentrations (10-4 M and 0.1 M). The 10-4 M solution was found to have a A. smaller ph than the 0.1 M solution As more acid is added, the [H 3 O + ] will increase B. larger ph than the 0.1 M solution () and the ph will decrease. The ph of the 10-4 M C. the same ph as the 0.1 M solution solution will be larger (smaller [H 3 O + ]). D. Not enough information is provided to determine if a difference in ph exists. 14. Two buffer solutions were made with equal molar mixtures of a weak acid its conjugate base, and differed only by the total concentration of acid plus base. The buffer solution with the larger total acid plus base concentration had A. a ph that was larger than the buffer with the smaller total concentration. B. a ph that was smaller than the buffer with the smaller total concentration. C. a ph that was the same as the buffer with the smaller total concentration. D. Not enough information is provided to make this decision. ph is determined by the [conj. base]/[acid] ratio. Doubling both concentrations will leave the ph unchanged. 15. Each of the following mixtures can produce a buffer solution EXCEPT A. HClO 4 and NaClO 4 HClO 4 is the only strong acid. All the B. HF and NaF others are weak acids and will form buffers. C. NaHCO 3 and Na 2 CO 3 D. Na 2 HPO 4 and Na 3 PO 4 E. NH 4 Cl and NH Which of the following mathematical expressions is the Henderson-Hasselbalch equation? A. p Ka = ph + log [ conjugate base] [ acid] B. - OH ph = p Ka + log + HO 3 C. ph = p Ka + log [ acid] [ conjugate base] D. p Ka = ph - log [ acid] [ conjugate base] E. ph = p Ka + log [ conjugate base] [ acid] : E. Definition of the HH equation. page 5

6 17. What is the ph of a solution that results from adding 25 ml of 0.15 M HCl to 25 ml of 0.52 M NH 3? (K b of NH 3 = ) A B C D E Rxn of HCl with NH 3 is complete to from 0.15/2 M NH + 4 and 0.37/2 M NH 3. Use HH eqn: ph = pka + log [NH 3 ]/[ NH + 4 ] = 14 + log( ) + log(0.37/0.15) = Which of the following acid-base pairs is most suited to make a buffer with a ph = 7.68? pk A pk B A. HCO 2 H/NaHCO B. H 2 CO 3 /NaHCO C. HOCl/NaOCl The HOCl/NaOCl acid- D. H 3 BO 3 /NaH 2 BO conjugate base pair has E. NH 4 Cl/NH the pk a closest to ml of a buffer solution formed with NH 4 Cl and NH 3 has a ph = 9.64 and the [NH 3 ] = 0.5 M. What is the number of grams of NH 4 Cl required to make this solution? (MW NH3 = 17.0 g/mol, MW NH4Cl = 53.5 g/mol) (K b of NH 3 = ) A g Find [NH + 4 ], then moles of NH + 4, then the mass of NH 4 Cl. B g K a = [H 3 O + ][NH 3 ]/[NH + 4 ] [NH + 4 ] = [H 3 O + ][NH 3 ]/K a C g [NH + 4 ] = [H 3 O + ][NH 3 ]K b /K w D g [NH + 4 ] = ( )(0.5)(1.8x10-5 )/(10-14 ) = M E g mole/l x 0.5 L x 53.5 g/mole = 5.52 g 20. If the ratio of acid to base in a buffer increases by a factor of 10, the ph of the buffer A. decreases by 1 B. decreases by 10 C. increases by 10 D. increases by 1 E. remains unchanged When the acid/base ratio increases by 10, the ph will decrease by 1. log([base]/[acid]) = log(0.1) = What is the ph of the buffer that results when 11 g of NaCH 3 CO 2 is mixed with 85 ml of 1.0 M CH 3 CO 2 H and diluted with water to 1.0 L? (K a of CH 3 CO 2 H = ; MW of CH 3 CO 2 Na = g/mol) A B C D E Moles acetate = (11 g)/(82.04 g/mol) = Moles Acetic acid = L x 1 M = ph = pka + log[(moles base)/(moles acid)] = -log(1.8 x 10-5 ) + log(0.13/0.085) = 4.95 page 6

7 22. Which one of the following conditions is always met at the equivalence point of the titration of a monoprotic weak base with a strong acid? A. The ph of the solution is equal to B. The volume of acid added from the buret equals the volume of base titrated. C. The molarity of the acid equals the initial molarity of the weak base. D. The percent ionization of the acid equals the percent ionization of the base. E. The moles of acid added from the buret equals the initial moles of weak base. At the equivalence point moles weak base = moles strong acid. 23. Which one of the following conditions is always true for a titration of a weak acid with a strong base? A. A colored indicator with a pk a less than 7 should be used. B. If a colored indicator is used, it must change color rapidly in the weak acid s buffer region. C. Equal volumes of weak acid and strong base are required to reach the equivalence point. D. The equivalence point occurs at a ph equal to 7. E. The equivalence point occurs at a ph greater than 7. At the equivalence point, the solution consists of the conjugate base of the weak acid that was present initially. The solution has a ph > A 25.0 ml sample of M NH 3 (aq) is titrated with M HCl(aq). What is the ph at the equivalence point? (K b of NH 3 = ) A B C D E The titration generates a weak acid solution (NH + 4 ). First, calculate the [NH + 4 ] at the equivalence point, and then carry out an ICE calculation of the NH + 4 solution. Moles of NH 3 = moles of NH + 4 = moles of HCl added = x 0.02 M = 5 x 10-4 moles, Liters 0.01 M HCl added = 5 x 10-4 moles/(0.01 moles/l) = 0.05 L, Final volume = L = L; [NH + 4 ] = 5 x 10-4 moles/0.075 = M, [H 3 O + ] = sqrt(k a x ) = sqrt(5.56 x x ) = 1.92 x 10-6 ph = Which of the following equations is the solubility product for magnesium iodate, Mg(IO 3 ) 2? A. B. C. D. E. K K K K sp sp = [Mg ][I ] [O ] = [Mg ][I ] [3O ] = [Mg ][IO ] 2+ - sp 3 = [Mg ] [IO ] sp 3 K = [Mg ][IO ] Follows directly from the definition of K sp sp 3 page 7

8 26. The solubility of SrSO 4 in water is g in 1.0 L at 25 C. What is the value of K sp for SrSO 4? A B C D E MW(SrSO 4 ) = x = g/mole [Sr 2+ ] = (0.107 g)/[( g/mole)*(1 L)] = 5.82 x 10-4 M = X K sp = [Sr 2+ ][SO 4 2- ] = X 2 = (5.82 x 10-4 ) 2 = 3.4 x What is the molar solubility of Fe(OH) 3 (s) in a solution that is buffered at ph 2.75 at 25 C? The K sp of Fe(OH) 3 is at 25 C. A mol/l Fe(OH) 3 Fe 3+ (aq) + 3OH - (molar solubility equals [Fe 3+ ]) B mol/l K sp = [Fe 3+ ][OH - ] 3 ; poh = 14 ph = 11.25; [OH - ] = C mol/l [Fe 3+ ] = K sp /[OH - ] 3 = /( ) 3 = /( ) D mol/l [Fe 3+ ] = = E mol/l : E 28. The following anions can be separated by precipitation as silver salts: Cl -, Br -, I -, CrO If Ag + is added to a solution containing the four anions, each at a concentration of 0.10 M, in what order will they precipitate? Compound K sp K sp [Ag + ] [Ag + ] (M) AgCl [Ag + ][Cl - ] K sp /[Cl - ] Ag 2 CrO [Ag + ] 2 [CrO 2-4 ] (K sp /[CrO 2-4 ]) AgBr [Ag + ][Br - ] K sp /[Br - ] AgI [Ag + ][I - ] K sp /[I - ] A. AgCl Ag 2 CrO 4 AgBr AgI B. AgI AgBr Ag 2 CrO 4 AgCl C. Ag 2 CrO 4 AgCl AgBr AgI D. Ag 2 CrO 4 AgI AgBr AgCl E. AgI AgBr AgCl Ag 2 CrO 4 page 8

9 29. (4 pts) Barium sulfite (s) and barium fluoride (s) are in equilibrium with a solution containing 8.70 x 10-3 M ammonium fluoride (completely soluble). Calculate the concentration of sulfite ion present in this solution. K sp (BaSO 3 ) = 6.0 x 10-7, K sp (BaF 2 ) = 1.7 x A M [F - ] is used to calculate [Ba 2+ ], which is then B M used to compute [SO 2-3 ]: C. 2.7 x 10-5 M [Ba 2+ ] = (1.7 x 10-6 )/(8.70 x 10-3 ) 2 = M D. 3.1 x 10-3 M [SO 2-3 ] = (6.0 x 10-7 )/(0.0225) = 2.67 x 10-5 M E. 6.0 x 10-7 M 30. A solution contains 0.10 M potassium sulfide and 0.10 M sodium carbonate. Solid nickel acetate is added slowly to this mixture. What substance precipitates first? A. K 2 CO 3 Soluble B. NiS K sp = [Ni 2+ ][S 2- ] = 3.0 x (from exam front page) C. CH 3 CO 2 K Soluble D. NiCO 3 K sp = [Ni 2+ ][CO 2-3 ] = 6.6 x 10-9 (from exam front page) E. No precipitate will form. The solubility product for NiS is much smaller, NiS will precipitate first, since S 2- and CO 3 2- are present in equivalent concentrations. 31. The ionic solids AgNO 3 and KI are dissolved in water and mixed together and in various proportions to form solutions with the Ag + and I - concentrations listed below: [Ag + ] [I - ] [Ag + ][I - ] Precipitate? (M) (M) Yes No x Yes Which of these solutions will form a precipitate? (K sp = 1.5 x ) A. 1 B. 1 and 3 C. 1 and 2 D. all three E. None of the solutions form precipitate. page 9

10 32. (4pts) A solution contains 6.8 x 10-4 M KOH. Solid iron(iii) nitrate is added slowly to this mixture. What is the concentration of iron(iii) ion when precipitation first begins? A. 6.8 x 10-4 M Take the K sp for (Fe(OH) 3 and solve for [Fe 3+ ]. B. 1.2 x M K sp = 6.3 x (from the exam front page) C. 1.4 x M [Fe 3+ ] = K sp /[OH - ] 3 = 6.3 x /(6.8 x 10-4 ) 3 D. 2.0 x M [Fe 3+ ] = 2.0 x E. 1.7 x 10-8 M EXTRA CREDIT 33. (3 pts) At 25 C, only 1.04 x 10-3 g Cu(NO 3 ) 2 will dissolve per liter of a solution that is buffered at ph What is the value of K sp for Cu(OH) 2? The molar mass of Cu(NO 3 ) 2 is g/mol. A B C D E poh = = 7.2 [OH - ] = moles Cu(NO 3 ) 2 dissolved = (1.04 x 10-3 g)/(187.6 g/mol) = 5.54 x 10-6 Cu 2+ moles dissolved/l K sp = [Cu 2+ ][OH - ] 2 = (5.54 x 10-6 )( ) 2 = 2.21 x (3 pts) What is the concentration of PO 4 3- in a solution created by mixing equal volumes of 0.1 M NaH 2 PO 4 and 0.2 M Na 2 HPO 4? (The acid dissociation constants for phosphoric acid are K a1 = , K a2 = , and K a3 = ) A M B. 0.2 M C D E. zero [H 2 PO 4 - ] = 0.1 M; [HPO 4 2- ] = 0.2 M Find [H 3 O + ] with K a2, then use K a3 to find [PO 4 3- ] K a2 = [H 3 O + ][HPO 4 2- ]/[H 2 PO 4 - ] [H 3 O + ] = (K a2 [H 2 PO 4 - ])/[HPO 4 2- ] = (6.2 x 10-8 )(0.05)/(0.1) = (3.1 x 10-8 ) K a3 = [H 3 O + ][PO 4 3- ]/[HPO 4 2- ] [PO 4 3- ] = K a3 [HPO 4 2- ]/[H 3 O + ] = (3.6 x )(0.1)/(3.1 x 10-8 ) = 1.16 x 10-6 M The closest correct answer, C, was computed incorrectly, 0.2 M was used mistakenly for [HPO 4 2- ] to compute [PO 4 3- ]. page 10