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.