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

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.

To solve this equilibrium problem using an ICE (Initial-Change-Equilibrium) table, we need to understand the equilibrium reaction and the corresponding equilibrium expression.

The given equilibrium reaction is:
CH3NH2 (aq) + H2O (l) ⇌ CH3NH3+ (aq) + OH- (aq)

The equilibrium expression for this reaction is:
Kb = [CH3NH3+][OH-] / [CH3NH2]

Now, let's set up the ICE table:

Initial:
[CH3NH2] = 0.266 M
[CH3NH3+] = 0 M
[OH-] = 0 M

Change:
[CH3NH2] decreases by x
[CH3NH3+] increases by x
[OH-] increases by x

Equilibrium:
[CH3NH2] = 0.266 - x
[CH3NH3+] = x
[OH-] = x

Now, use the Kb expression to write the equilibrium equation:
Kb = [CH3NH3+][OH-] / [CH3NH2]
4.4e-4 = x * x / (0.266 - x)

Since the Kb value is very small, we can assume that x is much smaller than 0.266. Therefore, we can approximate 0.266 - x to be approximately 0.266.

Now, solve the equation for x:
4.4e-4 = x * x / 0.266
x^2 = 4.4e-4 * 0.266
x^2 = 1.1664e-4
x ≈ 0.0108

Now, calculate the concentration of OH-:
[OH-] = x ≈ 0.0108 M

Since the solution is basic, the concentration of H+ (or H3O+) can be obtained using the equation:
[H+] * [OH-] = Kw = 1.0e-14
[H+] = Kw / [OH-]
[H+] = 1.0e-14 / 0.0108
[H+] ≈ 9.26e-13 M

Lastly, calculate the pH of the solution using the equation:
pH = -log[H+]
pH = -log(9.26e-13)
pH ≈ 12.03

Therefore, the pH of the solution is approximately 12.03.