Question

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

Answer 1

Answer:

$\%\ ionization=6.45\%$

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:

$Kb=\frac{[M^+][OH^-]}{[MOH]}$

And the percent ionization is:

$\%\ ionization=\frac{[OH^-]}{[MOH]} *100\%$

For the equilibrium concentration of hydroxide, it turns out:

$\%\ ionization=\frac{0.050M}{0.775M}*100\%\\\\\%\ ionization=6.45\%$

Best regards!