Question

What is the pH of a 0.56 M solution of methylammonium chloride, CH3NH3Cl? What is the concentration of methylamine in the solution? Kb (methylamine) = 4.4×10-4 pH = ___________ [MeNH2] =________M

Answer 1

Methlyamine has a Kb of 4.4 * 10^-4.
Kb = Products / Reactants

CH3NH3+ <=> CH3NH2 + H+
1:1 ratio. We start out with 0.56M methlyamine, 0M of each product. A certain amount will be added to each product to get to equilibrium, and that same amount must be taken from the reactants. then the amount x.

Kb = ( 0 + x )( 0 + x ) / ( 0.56 - x )
4.4 * 10^-4 = x² / ( 0.56 - x )
x = 0.0159
Because we had to use Kb, OH- is a product and the concentration is x, 0.0159.

pOH = -log( OH-)
pOH = -log( 0.0159 )
pOH = 1.79
pOH + pH = 14
pH = 14 - 1.79
pH = 12.21

Set up an ICE table.

.................H2O .... +.... CH3NH2..... ----> CH3NH3+ ......... + ......... OH-
Initial........... -- ..............0.56M.................... -- ............................... --
Change........--.................. -x ................... +x .............................. +x
Equilibrium . --................ 0.56-x .................. x ................................ x

Write the equilibrium expression for the equation. Remember products over reactants.

[[OH-][CH3NH3+]] / [CH3NH2] = Kb

Plug in the variables from the ice table into the equilibrium expression.

[(x)(x)]/(0.56-x) = Kb = 4.4*10^-4

Simplify

x^2 / (0.56-x) = 4.4*10^-4
x=0.0159M