If a 0.775 M solution of a base is found to have a hydroxide concentration of 0.050 M at equilibrium, what is the percent ionization of the base

Question

asked Feb 11, 2021 167k views

1 Answer

Ionizer · Answer 1 · 2021-02-18T07:00:40+0000

Answer:

$\%\ ionization=6.45\%$

Step-by-step explanation:

Hello.

In this case, since the ionization of a hypothetical base MOH is:

$MOH(aq)\rightleftharpoons M^+(aq)+OH^-(aq)$

The equilibrium expression is:

Kb=([M^+][OH^-])/([MOH])

And the percent ionization is:

$\%\ ionization=([OH^-])/([MOH]) *100\%$

For the equilibrium concentration of hydroxide, it turns out:

$\%\ ionization=(0.050M)/(0.775M)*100\%\\\\\%\ ionization=6.45\%$

Best regards!