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

`\%\ 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!