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

pH of the solution is 10.37

Step-by-step explanation:

`pOH=pkb+log([salt])/([base])`

kb =
`4.4 * 10^(-4)`

pkb = -log kb

=
`-log4.4 * 10^(-4)`

= 3.35

salt is methylammonium bromide and methylamine is base

Substitute the values in the above expression as follows:

`pOH=pkb+log([salt])/([base]) \\=3.35+log(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