CHM 215 Final Key, Fall 2007

Student Name: Notre Dame University Faculty of Natural and Applied Sciences Department of Sciences Quantitative Analysis / CHM 215 Final Exam — Duration: 120 minutes Fall 2008 Multiple-choice questions 14- Calculate the pH of the solution that results when 20.0 ml of 0.20 M formic acid are diluted to 45.0 ml with distilled water. (Ka for formic acid = 1.8 x 10°) a) b) c) d) e) i 47 1.5 3.8 2.4 2- Calculate the percent ionization of 0.0010 M CH3COOH. (Ka = 1.75x10°) a) b) c) ) e) 6.71 x 10° 4.83 x 10° 2.56 x 10° 1.32 x 107 3.82 x 10° 3- In the Fajans method for signaling the end point in argentometric titration: a) b) c) d) e) The indicator used is an adsorption indicator The indicator used is fluoroscein The end point is indicated by back titration Two of the above All of the above 4- Precipitation titration a) b) c) qd) e) Is based on weighing the mass of the precipitate obtained during precipitation Is based on reactions that yield soluble complexes Is based on reactions that yield ionic compounds of limited solubility Includes EDTA titrations None of the above NDU — CHM 215 — Final examination — Fall 2008 ID#: CGxl-.Wo Moe Koy, 5— Select the false statement about carbonic acid (H2COs). a) Itis a diprotic acid b) It is a weak acid c) Its neutralization includes 2 stages with 1 buffering regions d) Its half titration yield an amphiprotic species 6- Which solution will have the lowest pH value? a) 0.1 MHCN (Ka = 4.9 x 107°) b) 0.1 M HNO; c) 0.1M NaCl d) 0.1 MH2CO3 e) 0.1 MNaOH 7- Which one of the following statements on buffers is false? a) The more the concentration of buffer components is large, the more the buffer capacity is important. b) A buffer system is always composed from a weak acid and its conjugate weak Base c) A buffer resists change in pH d) The buffer capacity is the best when the concentration of the weak acid is equal to the concentration of its conjugate base e) All of the above are true 8— What does the ionization constant tell us about the strength of an acid? aS Pp The larger the ionization constant, the stronger the acid =< by The smaller the ionization constant, the weaker the acid c) The ionization constant does not reflect the strength of an acid d) The strength of an acid is inversely proportional to the ionization constant 9- An EDTA solution is added to a solution containing 0.1 mol/L of Ag’, Cu?", Ca", and Zn”. Referring to the formation constants for EDTA complexes, which cation is complexed first? K (Ca¥) = 5x10'°, K (CuY*) =6 x 10", K(AgY”) =2x 10’, K (ZnY”) =6 x 10" a) Ag’ b) Cu** ¢)Cae Gyan NDU - CHM 215 — Final examination — Fall 2008 op Questions 10 trough 12: During an argentometric titration where a 0.1 M AgNO; solution is used to titrate 50 mL of an NaBr solution: 10- Select the incorrect statement: a) pAg decreases as titration proceeds b) pAg increases as titration proceeds c) Atthe equivalence point pAg = pKgp/2 d) The equivalence point is characterized by a major change in pAg 11- If the equivalence point is obtained upon the addition of 100 mL of AgNOs, this means that the initial concentration of NaBr is: a) 0.1 ™M, since Ag+ react with sodium ion and with bromine ion with a 1:1 ration b) 0.2M c) 0.05M d) Itis impossible to answer this question, since the [NaBr] is not given 12- Before the equivalence point, the Ag+ concentration can be computed from: Ksp Ksp and initial NaBr Ksp and excess NaBr Excess Agt+ Excess NaBr 22eee NDU — CHM 215 — Final examination — Fall 2008 Problems (7 problems) 1- Calculate Ni2* concentration in a solution that is buffered to a pH of 3 and prepared by mixing 50.0 ml of 0.025 M Ni** with: a) 25 mL ofa0.05 M EDTA solution (6 points) K(NiY*) = 4.2 x 10° og = 2.51 x 10") Answer: 1.26 x 10°M b) 50 ml of a 1.86% (m/v) EDTA (372.24 g/mol) solution. (6 points) Answer: 9.52 x 10° 2- Calculate the equilibrium constant for the reaction (10-points) 2MnO, + 3Mn** 2H,0 < 5MnO,(s) + 4H" E° MnO,4/ MnO, = +1.695 V and E° MnO,/ Mn*" = +1.23 V Hint: multiply equations by appropriate integers so that the numbers of electrons are equal. NDU — CHM 215 — Final examination — Fall 2008 4 Answer: 1.44 x 10” 3- A solution containing 4.48 ppm KMnO, (158.04 g/mol) has a transmittance of 0.309 ina 1.0 cm cell at 520 nm. An unknown KMnO4 solution has a transmittance of 0.502 in the same cell at 520 nm. What is the concentration in ppm of the unknown KMnO, solution? NDU — CHM 215 - Final examination — Fall 2008 5 4- Calculate the hydronium ion concentration in a solution that is 0.040 M in (Given Data on H3sPO,: Ka; = 7.11 x 10°; Kaz = 6.32 x 10°; Kas = 4.5 x 10°) a) Na3PO, (6 points) Answer: 1.18 x 10" b) Na,HPO, (6 points) Answer: 2.4 x 10°” NDU — CHM 215 - Final examination — Fall 2008 6 5- Calculate the pH of a solution made by mixing 50.0 ml 0.2 M NH,CI (aq) and 75 mL of 0.1 M NaOH(aq). Kb (NH3) = 1.8 x 10° at 25°C. (10 points) Answer: 9.72 NDU - CHM 215 - Final examination — Fall 2008 a 6- A 100 mL sample of spring water was treated to convert any iron present to Fe". Addition of 25.00 mL of 0.002107 M K2Cr20; to the previously treated sample resulted in the following reaction: 6Fe™ + Cr.07;* + 14H’ > 6Fe* + 2Cr* + 7H20 The excess K,Cr207 was back titrated with 7.47 mL of a 0.00979 M standard Fe** solution. Calculate the parts per million (mg/L) of iron in the sample. (10 points) Answer: 135.4 ppm 7- BONUS: Calculate the potential of a platinum electrode immersed in a solution that is 0.0183 M in K3Fe(CN)g. and 0.00566 M in K,Fe(CN)g. (6 points) E° Fe(CN),*/Fe(CN)," = 0.36 V 0.329 V NDU —- CHM 215 — Final examination — Fall 2008 8