methylamine is a weak base. calculate the pH of a .025 M solution of methylamine.
..........CH3NH2 + HOH --> CH3NH3^+ + OH^-
I..........0.025............0..........0
C...........-x..............x..........x
E........0.025-x............x..........x
You need the Kb for CH3NH2. You can find that in your text/notes/web.
Substitute the E line into Kb expression and solve for x = (OH^-) and convert that to pH. You may need to solve for quadratic depending upon how close your prof wants the answer.
To calculate the pH of a solution of a weak base like methylamine (CH3NH2), you need to consider its equilibrium reaction with water. The dissociation of methylamine in water can be represented as follows:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
Methylamine accepts a proton (H+) from water and forms the methylammonium ion (CH3NH3+) and hydroxide ion (OH-).
The equilibrium expression for this reaction can be written as:
Kb = [CH3NH3+][OH-] / [CH3NH2]
The Kb value for methylamine is 4.4 × 10^-4 (at 25°C).
Since the initial concentration of methylamine is 0.025 M and it is a weak base, we can assume that most of it will remain undissociated. Therefore, the concentration of methylamine ([CH3NH2]) can be approximated as 0.025 M.
Let x be the concentration of OH- formed and CH3NH3+ formed.
Using the equilibrium expression, we can write:
4.4 × 10^-4 = (x)(x) / 0.025
Simplifying the equation, we get:
x^2 = (4.4 × 10^-4)(0.025)
x^2 = 1.1 × 10^-5
Taking the square root of both sides, we have:
x ≈ 1.05 × 10^-3
Since hydroxide ion concentration is roughly equal to the concentration of methylammonium ions formed, [OH-] = [CH3NH3+] ≈ 1.05 × 10^-3 M.
Now, to calculate the pOH, we can use the equation:
pOH = -log10[OH-]
pOH = -log10(1.05 × 10^-3)
pOH ≈ 2.98
Finally, to find the pH, we can use the equation:
pH = 14 - pOH
pH ≈ 14 - 2.98
pH ≈ 11.02
Therefore, the pH of the 0.025 M solution of methylamine is approximately 11.02.