It's easier to solve this one with the Henderson-Hasselbalch 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.....0.266-x............x.........x
.......CH3NH3Br ==> CH3NH3^+ + Br^-
I.......0.134.........0..........0
C.......-0.134.......0.134....0.134
E..........0.........0.134.....0.134
Kb = 4.4E-4 = (CH3NH3^+)(OH^-)/(CH3NH2)
Substitute as follows:
(CH3NH3^+) = 0.134 +x = 0.134 from salt and x from base.
(OH^-) = x
(CH3NH2) = 0.266-x from salt
Solve for x and convert to pH.
If using the HH equation first convert Kb to pKb = -log 4.4E-4 = 3.36, then
pH = pKa + log(base)/acid)
pH = 10.64 + log(0.134/0.266) = ?
Same answer either way.
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 = 4.4e-4
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