Question

A solution is prepared at 25°C that is initially 0.18 M in methylamine (CH3NH2), a weak base with Kb = 4.4 x 10^-4, and 0.35 M in methylammonium bromide (CH3NH3Br).
1. Calculate the pH of the solution. Round your answer to 2 decimal places.

Answer 1

Answer:

pH of the solution is 10.37

Explanation:

$pOH=pkb+log\frac{[salt]}{[base]}$

kb = $4.4 \times 10^{-4}$

pkb = -log kb

       = $-log4.4 \times 10^{-4}$

       = 3.35

salt is methylammonium bromide and methylamine is base

Substitute the values in the above expression as follows:

$pOH=pkb+log\frac{[salt]}{[base]} \\=3.35+log\frac{0.35}{0.18} \\=3.35+0.28\\=3.63$

pH = 14 - pOH

       = 14 - 3.63

       = 10.37

pH of the solution is 10.37