Question

Calculate the [OH-] and the pH for a solution of 0.24M methylamine, CH3NH2. Kb = 3.7 X 10-4.

Answer 1

`[OH^-]=9.24x10^(-3)M`.

`pH=11.97`.

Step-by-step explanation:

Hello,

In this case, since the ionization of methylamine is:

`CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)`

The equilibrium expression is:

`Kb=([CH_3NH_3^+][OH^-])/([CH_3NH_2])`

And in terms of the reaction extent
`x` which is equal to the concentration of OH⁻ as well as that of CH₃NH₃⁺ via ice procedure we can write:

`3.7x10^(-4)=(x*x)/(024-x)`

Whose solution for
`x` via quadratic equation is 9.24x10⁻³ M since the other solution is negative so it is avoided. Therefore, the concentration of OH⁻ is:

`[OH^-]=x=9.24x10^(-3)M`

With which we can compute the pOH at first:

`pOH=-log([OH^-])=-log(9.24x10^(-3))=2.034`

Then, since pH and pOH are related via:

`pH+pOH=14`

The pH turns out:

`pH=14-pOH=14-2.034\\\\pH=11.97`

Best regards.