CHM 1220 In-Class Activity: Aqueous Equilibria, Complex Ions, and Electrochemistry Fall 2017 ANSWERS 1. A 25.0 mL solution of quinine was titrated with 1.00 M hydrochloric acid, HCl. It was found that the solution initially contained 0.125 moles of quinine. What was the pH of the solution after 75.00 mL of the HCl were added? Quinine is monobasic with pKb = 5.10. 2. Calculate Ecell for an electrochemical cell based on the two half-reactions given below when the reaction occurs in an acidic aqueous solution in which the pH is buffered at 7.000 and [BrO3] = 0.880 M, [Cl] = 0.144 M, and = 0.260 atm. (Use the smallest whole-number coefficients possible when balancing the overall reaction. BrO3(aq) + 6 H(aq) + 5 e  ½ Br2 ( ) + 3 H2O( ) E° = 1.52 V Cl2(g) + 2 e  2 Cl(aq) E° = 1.36 V 2 BrO3(aq) + 12 H(aq) + 10 Cl(aq )  Br2 ( ) + 6 H2O( ) + 5 Cl2(g) Need to determine where in the titration we are by comparing moles of analyte and titrant: Need concentrations (or mol since volume identical) for Q and QH+: Initially: mol analyte does not equal mol titrant: NOT at equivalence point (or start), can use Henderson-Hasselbalch equation: Plug-in values, solve for pH: quinine(aq) + H2O(l) ⇌ quinine+-H (aq) + OH-(aq) (Q) (QH+) (oxidation, at anode) Net redox reaction: n = 10 mol e- Non-standard conditions, need Nernst equation to calculate cell potential: Standard cell potential: Plug-in values: Number of electrons transferred: (reduction, at cathode)    72.8176.090.8mol 075.0mol 0.05log10.514QHQlogppH aKadded HCl mol 075.0HCl soln. mL 1000HCl mol 1.00soln. HCl mL 75.00 added HCl mol 05.0quinine mol 125.0   QHQlogppHaK QH mol 0.075formed QH moladded HCl molmol 0.075consumed Q molformed QH mol Q mol 0.05 consumed Q mol 075.0initially Q mol 0.125remaining Q mol     V 37.0530.016.0144.0101880.0260.0log100592.0V 16.0cell1012725cellEE V 16.0V 1.36V 52.1ocellocelloanodeocathodeocellEEEEE      1012235ocellocellcellClHBrOlog0592.0log0592.02ClPnEQnEECHM 1220 In-Class Activity: Aqueous Equilibria, Complex Ions, and Electrochemistry Fall 2017 3. What mass of Ca(NO3)2 must be added to 1.0 L of a 1.0 M HF solution to begin precipitation of CaF2(s)? For CaF2, Ksp = 4.0 × 10-11 and Ka for = 7.2 × 10-4. Assume no volume change on addition of Ca(NO3)2(s). 4. EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt, Na2H2EDTA, are used to treat heavy metal poisoning. The equilibrium constant for the reaction below is 1.0 × 1023. Calculate [Cr3+] at equilibrium in a solution originally 0.0010 M in Cr3+ and 0.050 M in H2EDTA2- and buffered at pH = 6.00. Cr3(aq) + H2EDTA2-(aq) ⇌ CrEDTA-(aq) + 2H+(aq ) [Cr3+] [H2EDTA2-] [CrEDTA-] [H+] Initial Concentration (M) 0 0.049 0.001 1×10-6 Change in Concentration (M) +x +x -x -- Equilibrium Concentration (M) x 0.049 + x 0.001 - x 1×10-6 Dissolution of Ca(NO3)2: Ca(NO3)2(s) → Ca2+(aq) + 2 NO3-(aq) Dissolution of CaF2: Should be “complete” all nitrates are soluble CaF2(s) ⇌ Ca2+(aq) + 2 F-(aq) Solubility low, depends on Ca+ and F- Acid dissociation of HF: HF(aq) + H2O(l) ⇌ F-(aq) + H3O+(aq) Will provide “common ion”, effects CaF2 solubility First: figure out how much F- is in solution:    43102.7HFOH FaKAcid dissociation equilibrium expression   4102.70.1 xxxSet up an ICE table or jump directly here: since Ka is small, x << 1.0 M 42102.70.1x Mx 0268.0F Second: figure out how the F- effect the solubility of CaF2: Equilibrium of solubility for CaF2   112100.4F CaspKCaF2(s) ⇌ Ca2+(aq) + 2 F-(aq) y y 0.0268 + 2y solubility:   112100.420268.0 yysince Ksp is small, y << 0.0268 M   112100.40268.0y My821056.5CaFinally: figure out how much Ca(NO3)2 is needed to provide the Ca+ 2362322328)Ca(NO g101.9)Ca(NO mol 1g 088.164Ca mol 1)Ca(NO mol 1LCa mol1056.5 soln. L 0.1anything greater than 23)Ca(NO μg 1.9This is going to be an ICE table type problem. We need to make some necessary assumptions: 1. Kf is large so initially all the Cr3+ will be complexed 2. the pH does not change as it is “buffered” Notice: a small amount of the EDTA is consumed to complex the Cr3+     232232101EDTAH CrH CrEDTAfKcomplex-ion formation equilibrium expression Substitute from the table     2326-1010.049101 0.001xxxsince Kf is huge, x << any of these concentrations     2326-1010.049101 0.001x Mx3731004.2CrliterionsCr1023.1mol 1ions 1002.6Cr 1004.231323337 Messentially no Cr3+ at