Question

8. Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution. a. 0.18 M CH,NH b. 0.18 M CH,NH CI c. a mixture that is 0.18 M in CH NH, and 0.18M in CH NH CI and

Answer 1

CHINH2 H0 » CH,NH, + OH A) 1(м) 0.18 0 0 с (М): E (M) x 0.18-x Baseionizati on constant of CHNH2, K = 4.47x10 [CH,NH, OH [CH,NH2 Baseionization constant, 4.47x10 (0.18-x) x 9.0x103 M From the equilibrium table, OH=x = 9.0x103 M 1.00x10- Thus, [H,OOH9.0x10 = 1.1x1012 M pH = -log [H,0] -1log (1.1x10-2) = 11.96 B)Ionization of CHNH2C1 CHNH2CI (aq) >CH,NH (aq) Since lm ol of CH,NH,CI forms lm ol of CH,NH. CH,NH CH,NH, (a) =0.18 M ICE table CHNH HO CH3NHH Initi al concentrati on (M): 0.18 0 0 Change (M) Equilibrium concentration (M) 0.18-x x [CH,NH, HO] CH,NH Acidionization constant,K, = 1.0x10-14 К, 4.47x10 0.18-x х 2.2x1011 0.18-x x= 2.0x10 M From the equilibrium table, x = H20*2.0x10 M pH-log [H,0]log (20x104) -5.70 C) concentrati on of CH,NH2CHNH,] = 0.18 M concentrati on of CH,NH;, [CH,NH [CH,NHC] 0.18 M p, of CH,NH, log K -log(4.47x10)= 3.35 pK of CHNH =14.00-pK, = 14.00 3.35 10.65 [base] conjugate acid [CH,NH, According to Henderson equation, pH pKog pH pKog CH,NH 0.18 = 10.65 =10.65+1og 0.18