Question

Enter your answer in the provided box. Calculate the pH of a 0.51 M CH3COOLi solution. (K, for acetic acid = 1.8 x 10-5.)

Answer 1

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CH3 Cooli 13 sait So, it solution elissociales completely quocu in CH3C00li (99) after dissociation CH3C00 -(99 U.SIM Lit 1921 U.SIM CH3COO- underpo hydrolysis in water ICE Table concentration (MI CH3COO-1991 + H2olll CH3COOH (aq) + OH-laq) 0.SI +x -X 0.51-6 ka for CH3COOH = kb for CHaroo-= Kw 118810-5 = 10-14 ka 18x10.5 = 5.56 X 10 -10 in comparison Kb = [CH3COOH] [OH- 5.50X10-10 = x xoc [CH3 100 -] 0.51-30 AS value of lo is very small, so we can neplect x to 0.51 5.56 x 10-10 = x2 > x= 1.684 X105M 0:51 [OH-] = x= 1.684 X 10 -SM pot - - 10g (OH) - - 10g (1.684 X10-5) = 4.77 pH = 14 - bol = 14-4,77 - 9,23 IPH = 9.23