To find the pH of a 0.080 M solution of methylamine (CH3NH2), we first need to determine the concentration of hydroxide ions (OH–) produced by the dissociation of the base in water.
The dissociation of methylamine in water can be represented as:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
The dissociation constant Kb for methylamine is given as 4.4 x 10-4.
Using the expression for Kb:
Kb = rac{[CH3NH3+][OH-]}{[CH3NH2]}
Let x be the concentration of OH– produced at equilibrium. Then we can denote the equilibrium concentrations as:
- [CH3NH2] = 0.080 – x
- [CH3NH3+] = x
- [OH–] = x
Substituting these into the Kb expression gives:
4.4 x 10-4 = rac{x imes x}{0.080 - x} ≈ rac{x2}{0.080} (assuming x is small)
This simplifies to:
x2 = 4.4 x 10-4 × 0.080
Calculating that gives:
x2 = 3.52 x 10-5
Therefore:
x = √(3.52 x 10-5) ≈ 5.93 x 10-3 M
The concentration of hydroxide ions [OH–] is about 5.93 x 10-3 M.
Next, we can find the pOH:
pOH = -log[OH-] = -log(5.93 x 10-3) ≈ 2.23
Finally, we convert pOH to pH:
pH + pOH = 14
pH = 14 - pOH = 14 - 2.23 ≈ 11.77
Thus, the pH of a 0.080 M solution of methylamine is approximately 11.77.