Experimental Chemistry

1. Label all pernent components (i.e. cathode, anode, direcon of electron ﬂow, etc.) of this representaon of a voltaic cell given that: - The reacon Cu 2+ (aq) + 2e -  Cu (s) is occurring at one electrode, which is submerged in a sulfate soluon. - The overall E 0 for the cell is 0.575 V. Idenfy and label the second metal, also submerged in a sulfate soluon. a) Write the full, balanced redox reacon that is occurring between the two electrodes, including physical states of all species. b) Calculate the value of the reacon quoent when E = 0.615 V c) If the concentraon of CuSO 4 is increased, what eﬀect should this have on the cell potenal? (30 points) Labels: Cu(s), cathode; Ni(s), anode; voltmeter; salt bridge; electron ﬂow TO cathode, sulfate soluons a) Cu 2+ (aq) + 2e -  Cu(s) has E 0 red = 0.339 V E 0 = 0.575 V = E 0 red + E 0 ox = 0.339 V + E 0 ox E 0 ox = 0.236 V Nickel has E 0 red = -0.236 V but since Cu is being reduced, the other metal must be oxidized, so E 0 ox =+0.236 V for Ni Cu 2+ (aq) + Ni (s)  Cu(s) + Ni 2+ (aq) (2 electrons transferred) b) E = E 0 – 0.05916/n log Q 0.615 V = 0.575 V – 0.05916/2 log Q 0.04 = -0.02958 log Q log Q = -1.35 Q = 10 -1.35 = 0.044 Name___________________________________________________ CH2640 Exam I c) Since copper is the cathode and the reducon reacon is taking place, copper ions are being removed from the soluon and reduced as solid copper on the electrode. More copper ions in soluon will drive the equilibrium forward, favoring the reacon and increasing the number of mes that the reacon can occur. Mathemacally, since [Cu 2+ ] is in the denominator of the reacon quoent, increasing this number will make Q and log Q smaller. Since the cell potenal reacon subtracts log Q from the standard state potenal, this results in an overall larger (or less negave) value of E. 2. Given E probe = C + (0.05916/n)log[H + ], a) Explain why two buﬀer soluons are needed for the proper calibraon of a pH-meter electrode. b) If the meter recognizes pH 4 and 7 buﬀer soluons and a student uses contaminated pH 4.2 and 7.2 calibraon standards, explain which type of error this will introduce into subsequent soluon measurements. c) Calculate the percent error associated with the measurement of the pH of pure water (NOT from the GH306 DI tap…) using this improperly calibrated meter. (15 points) a) Each calibraon soluon used gives one set of (E probe , [H + ] values). Since there are essenally two constants (C and n) in the operang equaon for the meter, we need two points to solve for two unknowns. Also can be thought of as needing two points to ﬁnd the equaon of a line. b) A constant error that will cause all future measurements to be oﬀset by a parcular value (higher or lower) is a systemac error. c) pH of pure water = 7 (actual/correct value) Meter will read pH 0.2 units too low = 6.8 (experimental value) % error = (7-6.8)/7 x 100 = 2.86% 3. Describe the fundamental diﬀerence between determining water hardness through EDTA traon and with a conductance measurement. Which is more accurate at ﬁnding total hardness? (15 points) EDTA will only react with divalent caons (typically Mg 2+ and Ca 2+ in water samples). While these two ions are the most signiﬁcant players in terms of contribuons to water hardness, they are not the only ones. Conductance measurements will report a value based on contribuons of all ions in a water sample, not just divalent caons. This means that it includes other hardness-contribung ions, but also other ions that don’t necessarily contribute to water hardness. Arguments can be made for either method being more accurate at ﬁnding total hardness.

CHEM | 2640 | Experimental Chemistry | homework | 1. Label all pernent componen

Answer Practice Quiz 3 Copper-Silver Phase Diagram 1. At 700 ֯ C, what is maximum wt% solubility of Cu in Ag? Show line on above graph. If I draw a horizontal line at 700 ֯C (left red line going from 100 wt% Ag), and it meets the curve line at roughly 94 wt% Ag, hence solubility of Cu in Ag is 6 wt% Cu. 2. At 800 ֯ C, what is the maximum at% solubility of Ag in Cu? Show line on the graph. Draw horizontal line at 800 ֯C from 100 wt% Cu (green line), roughly 9 wt% Ag is soluble in Cu and 5 at% Ag is soluble in Cu. 3. Is Cu-Ag system isomorphous? Explain your answer Cu-Ag is not isomorphous because at room temperature there are two phases present (α and β), whereas isomorphous system has 100% solubility of components into each other at room temperature or presence of only one phase at room temperature. 4. How much wt% of Cu is present in Cu-Ag system at eutectic point? At eutectic point, there is 71.9 wt% of Ag and 28.1 wt% of Cu. Please draw the line in the exam. 5. At eutectic point, how much of α-phase is present if I have 5 kg of Cu-Ag system? Using the level rule at eutectic point to calculate α-phase W α = 91.2 − 71.9 91.2 − 8.0 W α = 0.2319 There is 5 Kg in the system, hence there is 0.2319 * 5 = 1.16 kg of α phase 6. At what wt% of Cu, there is eutectoid reaction? Show line on graph Eutectoid reaction is point or temperature at which a solid phase transforms into two solid phase. In above phase diagram, δ solid phase transforms into two solid phases γ and ε as shown by red line. This transformation is at 75 wt% Zn and 25 wt% Cu. 7. At what temperature solidification start for 60 wt% Cu in Cu-Zn system? Show on the graph Solidification starts at 900 ֯ C as shown by black line in the graph. 8. At what temperature solidification ends for 60 wt% Cu in Cu-Zn system? Show on the graph Solidification ends roughly at 875 C as shown by black line in the graph. 9. Calculate wt% of Zn in α-phase at 600 ֯C and 40 wt% Zn in Cu-Zn system? 37 wt% Zn is present in α-phase at 600 C with 40 wt% Zn in the system as shown with blue line. 10. Write one thing that you would like to see change in this course?

CHEM | 2640 | Experimental Chemistry | assessment | Answer Practice Quiz 3 Copper-

CH2640_Problem_set_1_Answer_key.docx.docx-CH2640 Problem set I (Answers) 1.

CH2640 Problem set II Answers Due at the start of class on Monday, September 18 th . Show all work for full credit! 1) Express a 30.50 mass% aqueous solution of ethanol in units of molarity and molality, assuming a density of 0.900 g/mL for the solution . 30.5 mass % = 30.5 g ethanol/100 g solution 0.305 g / g solution x (1 mol / 46.07 g C 2 H 6 O) x 0.900 g solution/ mL x 1000 mL/L = 5.96 M 30.5 g ethanol/(30.5 g ethanol + 69.5 g water)  30.5 g ethanol/69.5 g water 30.5 g ethanol/69.5 g water x (1 mol / 46.07 g C 2 H 6 O) x 1000 g/kg = 9.53 m 2) How many milliliters of 4.65 M SnCl 4 are needed to prepare 6.00 L of a 750 mM solution ? M 1 V 1 = M 2 V 2 (4.65 M) V 1 = (0.750 M)(6000 mL) V 1 = 968 mL 3) #4.4, p. 99 in the textbook (5 th ed), #4.3 p. 102 (4 th ed) a) Mean = 1.52767 b) s = 0.00126, relative s = 0.00126/1.52767 x 100 = 0.0825% c) t parameter is for n – 1 or 6 degrees of freedom, in the 99% confidence column = 3.707 ts/sqrt(n) = 3.707(0.00007)/sqrt(7) = 0.00010 Confidence interval = 1.52793 +/- 0.00010 4) A solution is made from 0.844 g Ca(OH) 2 dissolved into 815 mL water. If 3.30x10 2 mL of 0.0310 M hydroiodic acid are added, what is the pH of the resulting solution? Write the balanced reaction first, and assume that the product salt is a solid and does not hydrolyze (will not affect pH). Ca(OH) 2 (aq) + 2HI (aq)  CaI 2 (s) + 2H 2 O (l) mol Ca(OH) 2 = 0.844 g/(74.1 g/mol) = 0.0114 mol mol OH - = 2(0.0114) = 0.0228 mol mol H + = mol HI = 0.0310 M x 0.330 L = 0.01023 mol mol OH - > mol H + , so the hydroxide will control the pH of the solution After reaction: excess mol OH - = 0.0228 mol – 0.01023 mol = 0.01257 mol OH - [OH - ] = 0.01257 mol/1.145 L = 0.01098 M pOH = 1.959 and pH = 12.041 5) Lactic acid has one acidic hydrogen atom in its structure. A 35 mM solution of lactic acid has a pH of 2.66. Calculate the K a of lactic acid. K a = [H + ][lactate ion]/[Lactic acid] = (10 -2.66 ) 2 /(0.035 – 10 -2.66 ) = 1.46 x 10 -4 6) #9.8, p. 204 in the textbook (p. 210 4 th ed)…AND, calculate the degree of deprotonation of methylamine at each quotient given in the problem. pH = pK a + log([base]/[acid]) pH = 10.632 + log( [CH 3 NH 2 ]/[CH 3 NH 3 + ] ) Degree of deprotonation,  = ([CH 3 NH 2 ]/[CH 3 NH 3 + ] )/(1+ [CH 3 NH 2 ]/[CH 3 NH 3 + ] See handout! a) [CH 3 NH 2 ]/[CH 3 NH 3 + ] = 10 (4.00 – 10.632) = 2.33 x 10 -7  = 2.33 x 10 -7 /(1+2.33x10 -7 ) = 2.33x10 -7 = 2.33x10 -5 % deprotonated b) [CH 3 NH 2 ]/[CH 3 NH 3 + ] = 10 (10.632 – 10.632) = 1  = 1/(1+1) = 0.5 = 50% deprotonated c) [CH 3 NH 2 ]/[CH 3 NH 3 + ] = 10 (12.00 – 10.632) = 23.3  = 23.3/(1+23.3) = 0.9588 = 95.88% deprotonated 7) #9.14, p.204 in the textbook (p. 210 4 th ed) How many mL of 0.626 M KOH should be added to a solution containing 5.00 g of HEPES to give a pH of 7.40? mol H + HEPES = 5.00 g/238.30 g/mol = 0.0210 mol acid pH = pKa + log(mol KOH/(mol HEPES-mol KOH)) 7.40 = 7.56 + log(x/(0.0210 – x)) 0.692 = x/(0.0210 – x) 0.0145 – 0.692x = x x = 0.00859 mol KOH V KOH = 0.00859 mol/0.626 mol/L x 1000 mL/L = 13.7 mL 8) After performing an EDTA titration experiment on 125.00 mL of an unknown water sample you find that it takes 12.70 mL of a 10.00 mM EDTA solution to sequester all Mg 2+ and Ca 2+ in solution. What is the concentration of [Mg 2+ + Ca 2+ ] in the water in parts per million (ppm) and in French degrees? M EDTA V EDTA = M ions V water (10.00 mM)(12.70 mL) = M ions (125.00 mL) M ions = 1.016 mM = 1.016 x 10 -3 M Mg 2+ and Ca 2+ [Mg 2+ + Ca 2+ ] = 1.106 x 10 -3 M x (32.2 g/mol avg. atomic weight) x 1000 mg/g =

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