Question

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

Answer 1

Answer:

$[OH^-]=9.24x10^{-3}M$.

$pH=11.97$.

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:

$Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}$

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:

$3.7x10^{-4}=\frac{x*x}{024-x}$

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.