Diebolt Summer 010 CHM 15 EXAM II Key Part 1. multiple choice 56 Part II. Problems 69 Total Pts 15 Part One: Multiple choice. There are 14 m.c. questions worth 4 points each (56 points) 1. The concentration of H for a solution is 1.010 9 M. The poh is and the solution would be. = log 1.010 9 = 9 so solution is basic; poh = 14 9 = 5 A. 9, basic B. 9, acidic C. 5, basic D. 5, acidic E. none of these. Select the conjugate acid for (CH ) NH. Conjugate acid: add 1 H = (CH ) NH A. H O B. (CH ) N C. OH D. (CH 4 ) NH E. (CH ) NH. Select the solution with the highest. Highest = most basic; SB with largest [OH = 0.10 M Ca(OH) since [OH = 0.10 M = 0.0 M A. 0.10 M Ca(OH) B. 0.10 M NH C. 0.10 M H O D. 0.10 M HBr E. 0.10 M KOH 4. Equilibria lies on the reactant side for the following reaction: CHO (aq) HCN(aq) HCHO (aq) CN (aq) Select the weakest acid and weakest base for this reaction. Eq lies on side with weakest acid and base so acid and base on left are weakest. Acid Base A. HCN CN B. HCN CHO C. HCHO CN D. HCHO CHO 5. Consider the K a values for the following acids: 5.6 10 10 HC H O 1.8 10 5 HCHO 1.8 10 4 HOC 6 H 5 1.6 10 10 smallest weakest acid has strongest conj base HF 7. 10 4 NH 4 Which of the following is the strongest conjugate base? A. NH B. C H O C. CHO D. OC 6 H 5 E. F 6. One molar solutions of KNO, AlBr, Ba(ClO 4 ) will result in solutions that are, respectively, N B, A N, N N A. neutral, acidic, and neutral B. basic, acidic, and neutral C. basic, neutral, and acidic D. neutral, acidic, and basic E. acidic, neutral, and acidic CHM15 Sum 10 Diebolt Eam 1
7. When comparing two different 0.10 M acid solutions, which statement is False? A. The stronger acid has a higher percent ionization. T, ionizes more B. The stronger acid has a lower reading. T, has lower = more acidic C. The stronger acid has a smaller [H O. F, has more H O D. The stronger acid has a higher K a value. T, has more ions so larger K a E. Both acids require the same volume of NaOH(aq) to be neutralized. T, both rns have same # moles of acid and both rns go to completion so V NaOH is same. 8. Which combination cannot function as a buffer solution? Can t have or SB A. HCN and NaCN WA/WB B. NH and NH 4 Cl WB/WA C. HF and NaF WA/WB D. NaOH and NaNO SB/NSalt E. HNO and NaNO WA/WB 9. According to the Lewis theory, an acid is A. an electron pair acceptor B. an electron pair donor C. a proton acceptor D. a proton donor E. a hydroide ion donor Most metal cations are acidic because they can accept electrons! 10. When NH (aq) is titrated by a solution of HNO (aq), the at the equivalence point is NH (WB) titrated by HNO (), so solution is acidic = below 7 A. 7 B. less than 7 C. greater than 7 D. can t tell 11. Which plot shows the titration curve epected when HCO H(aq) is titrated by KOH(aq)? A. SB B. SB C. SB = KOH D. WB WA = HCO H 1. Consider a buffer solution prepared from HOCl and NaOCl. Which is the net ionic equation for the reaction that occurs when NaOH is added to this buffer? A. OH OCl HOCl O B. Na HOCl NaCl OH C. H OCl HOCl D. NaOH HOCl H O NaCl E. OH HOCl H O OCl OH from SB reacts with WA = HOCl 1. What change will be caused by the addition of a small amount of NaOH(aq) to a buffer solution containing NH 4 NO (aq) and NH (aq)? Add OH (SB), WA and WB a. [H will significantly increase CHM15 Sum 10 Diebolt Eam
b. c. [OH will significantly increase [NH 4 and [NH will both increase d. [NH 4 will decrease and [NH will increase WA = NH 4, WB = NH e. [NH 4 will increase and [NH will decrease 14. How will the of a HCN(aq) solution change after KCN(aq) is added to it? Adding conj base so soln is more basic and goes up! Also H O (common ion effect) A. will not change B. will decrease C. will increase D. need more information to tell Part Three. Numerical Problems. (69 pts) 1. Fluoroacetic acid, HCO CH F, is a weak acid found in the poisonous gifblaar plant. a) A 0.0861 M solution of fluoroacetic acid has a of 1.86. Calculate the K a value for HCO CH F. (9 pts) HCO CH F H O H O CO CH F 0.0861 0 0 0.0861 = 1.86 = [H O = [CO CH F [H O = 10 [H O = 10 1.86 = 1.4 10 ( sig figs since after dec pt for value) [HCO CH F = 0.0861 0.014 = 0.07 ( after decimal point from sig fig subtraction rule) [H O [CO CH F [HCO CH F (1.410 ) 0.07.710 b) Calculate the %ionization for the fluoroacetic acid solution given above. ( pts) 1. 410 %ionization = 100% %ionization = 100% [HCO CH F 0. 0861. A NaCN solution is prepared in the laboratory. 0 a) Write the hydrolysis reaction that occurs for the NaCN solution. (4 pts) CN H O HCN OH = 16 % b) Calculate the of a solution that contains 4.8 grams of NaCN in 775 ml of solution. K a for HCN is 4.90 10 10. (9 pts) CN H O HCN OH [CN = [NaCN = mol NaCN/L 0.65 0 0 0.65 4.8gNaCN 1000mL 1molNaCN CN 0.65M 775mL 1L 49.01gNaCN a 14 Kw 1 10 Kb.0410 5 [HCN[OH K 10 b = K 4.90 10 [CN 5 =.04 10 0.65 =.65 10 = [OH =.65 10.0410 5 = 0.65 poh = log.65 10 =.44 = 14.44 = 11.56 = 11.56 CHM15 Sum 10 Diebolt Eam
. Calculate the, poh, and [OH for a 4.510 M HClO 4 (aq) solution. (7 pts) Strong acid, so [H = [HClO 4 = 4.5 10 ( sig figs for given concentration) = log 4.5 10 =.6 ( after dec pt for ) poh = 14.6 poh = 11.68 ( after dec pt for poh) [OH = 10 11.68 [OH =.010 1 M ( sig figs for concentration) 4. You have 50.0 ml of a buffer solution containing 0.75 M HC H O and 0.10 M KC H O. What is the after 50.0 ml of 0.600 M HCl is added to this buffer solution? For HC H O, 5.5 10 5. (1 pts) strong acid: HCl H O H O Cl ; [H O = [HCl [H O = 0.600M 50.0 = 0.100 M 00.0 [HC H O = 0.75M 50 = 0.1 M [C H O = 0.10M 50 (00) (00) = 0.58 M H O will react with the conjugate base, C H O : [C H O [HC H O Reestablish Equilibria: [HO [CH O [HC H O AB neut. Rn: H O C H O HC H O H O Initial 0.100 0.58 0.1 Change 0.100 0.100 0.100 final 0 0.158 0.41 WA rn: HC H O H O H O C H O I 0.41 0 0.158 C E 0.41 0.158 5.510 5 = CHM15 Sum 10 Diebolt Eam 4 0. 158 0. 41 = [H O = 1.4410 4 = log [H O = log 1.4410 4 =.84 =.84 using moles (or mmoles) and HH eq: moles H O = 0.0500 L 0.600 M = 0.000 moles H O (= 0 mmoles) moles HC H O = 0.50 L 0.75 M = 0.098 moles HA (= 9.8 mmoles) moles C H O = 0.50 L 0.10 M = 0.0775 moles A (=77.5 mmoles) AB neut. Rn: H O C H O HC H O H O Initial 0.000 0.0775 0.098 Change 0.000 0.000 0.000 final 0 0.0475 0.18 moles A = 0.0475 mol (= 47.5 mmoles) mol HA = 0.18 mol ( = 1.8 mmoles) = pk a log [A /[HA = log 5.510 5 log (0.0475/0.18) = 4.6 0.416 p H =.84 I love it when the is the same both ways!
5. A solution of nitric acid, HNO, is being titrated with KOH. Calculate the after.5 ml of 0.175 M KOH is added to 10.5 ml of 0.50 M HNO. (10 pts) SB neutralization: HNO KOH KNO H O (Net rn: H OH H O) moles HNO = moles H = 1L 0.50 moles HNO ml 1000mL L 10.5.6510 moles H moles KOH = moles OH = 1L 0.175moles KOH ml 1000mL L.5.9810 moles OH moles ecess OH =.9810 moles.6510 moles = 1.110 moles divide moles by total V to calc M: [OH = 1.110 moles 0.00L.9810 M OH poh = log.9810 = 1.400 = 14 1.400 = 1.600 6. A sample of butyric acid, HC 4 H 7 O, is titrated with NaOH solution. What is the after 1.5 ml of 0.00 M NaOH has been added to 5.0 ml of 0.150 M HC 4 H 7 O? For HC 4 H 7 O, 1.5 10 5 (14 pt) moles HC 4 H 7 O = moles NaOH = 1L 0.150molesHC H7O 5.0 1000mL L 1L 0.00molesNaOH 1000mL L 4.7510 moles HC 4 H 7 O 1.5.7510 moles NaOH AB Neut rn: HC 4 H 7 O OH C 4 H 7 O H O Initial.7510.7510 0 Change.7510.7510.7510 Final 0 0.7510 At eq pt, so have WB hydrolysis rn: C 4 H 7 O 14 110 Find K b since WB rn: K b for C 4 H 7 O = 5 1.510 H O HC 4 H 7 O OH = 6.6710 10.75 10 moles Divide moles by total V to calc M: [C 4 H 7 O = 0.100 M 0.075 L K b = WB rn: C 4 H 7 O H O HC 4 H 7 O OH I 0.100 0 0 C E 0.100 [HC H O [C H O [OH 4 7 4 7 = [OH 10 = 0.1006.67 10 6.6710 10 = 0.100 = 8.17 10 6 poh = log 8.1710 6 = 5.09 = 14 5.09 = 8.91 = 8.91 at equivalence point CHM15 Sum 10 Diebolt Eam 5