chemistry 1046
posted by Anonymous .
Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution.
a solution that is 0.266 M in CH3NH2 and 0.134 M in CH3NH3Br.
Kb = 4.4e4

It's easier to solve this one with the HendersonHasselbalch equation BUT if you must use the ICE chart, here it is.
......CH3NH2 + HOH ==> CH3NH3^+ + OH^
I.....0.266.............0..........0
C........x.............x..........x
E.....0.266x............x.........x
.......CH3NH3Br ==> CH3NH3^+ + Br^
I.......0.134.........0..........0
C.......0.134.......0.134....0.134
E..........0.........0.134.....0.134
Kb = 4.4E4 = (CH3NH3^+)(OH^)/(CH3NH2)
Substitute as follows:
(CH3NH3^+) = 0.134 +x = 0.134 from salt and x from base.
(OH^) = x
(CH3NH2) = 0.266x from salt
Solve for x and convert to pH.
If using the HH equation first convert Kb to pKb = log 4.4E4 = 3.36, then
pH = pKa + log(base)/acid)
pH = 10.64 + log(0.134/0.266) = ?
Same answer either way.