Analytical Chem istry - DePauw University

Chapter 6 Equilibrium Chemistry2752+ 3+5o RT [ Mn ][ Fe ]E = E − ln− + +nF [ MnO ][ Fe ] [ H ]42 5 850.05916 ( 0. 015)( 010 . )E = 074 . − log5 ( 0. 025)( 05 . 0)( 1× 10− )5 7 850.05916 ( 0. 015)( 010 . )E = 074 . − log−5 ( 0. 025)( 05 . 0)( 1×10 )Click here to return to the chapter.5 7 8= 012 . VpH2–Practice Exercise 6.5From Appendix 11, the pK a values for H 2 CO 3 are 6.352 and 10.329. Theladder diagram for H 2 CO 3 is shown to the side. The predominate format a pH of 7.00 is HCO 3 – .Click here to return to the chapter.Practice Exercise 6.6The ladder diagram in Figure 6.5 indicates that the reaction between aceticacid and p-nitrophenolate is favorable. Because p-nitrophenolate is inexcess, we assume that the reaction of acetic acid to acetate is complete. Atequilibrium essentially no acetic acid remains and there are 0.040 molesof acetate. Converting acetic acid to acetate consumes 0.040 moles ofp-nitrophenolate; thusmolesp -nitrophenolate = 0. 090− 0. 040 = 0. 050 molCO 3--pK a2= 10.329HCO 3pK a1= 6.352H 2CO 3molesp -nitrophenol= 0.040 molAccording to the ladder diagram for this system, the pH is 7.15 whenthere are equal concentrations of p-nitrophenol and p-nitrophenolate. Becausewe have slightly more p-nitrophenolate than we have p-nitrophenol,the pH is slightly greater than 7.15.Click here to return to the chapter.Practice Exercise 6.7As Hg 2 Cl 2 dissolves, two Cl – are produced for each ion of Hg 2 2+ . If weassume that x is the change in the molar concentration of Hg 2 2+ , then thechange in the molar concentration of Cl – is 2x. The following table helpsus keep track of our solution to this problem.Concentrations Hg 2 Cl 2 (s) Hg 22+ (aq) + 2Cl – (aq)Initial solid 0 0Change solid +x +2x