What is the pH of a 0.085 M solution of a weak base that is 1.6% ionized?

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Homework Statement

A 0.085 M aqueous solution of a weak base is known to be 1.6 percent ionized. Calculate the pH of the solution.

Homework Equations

%Ionization= ([OH-] derived from base/ initial [base]) x 100%
pOH=-log[OH-}

The Attempt at a Solution

%Ionization= ([OH-] derived from base/ initial [base]) x 100%
1.6%=[OH]/0.085M
[OH]=0.136M

pOH=-log[OH-]
pOH=-log(o.136M)
pOH=0.866
pH+pOH=14, 14-0.866=13.134...

I normally do these without a problem, so I'm assuming it's a stupid mistake. The problem clearly says weak base...Where am I going wrong?

Regards,
Fragment

 

[Borek]

%Ionization= ([OH-] derived from base/ initial [base]) x 100%
1.6%=[OH]/0.085M
[OH]=0.136M

Check your math. You are right - it is a stupid mistake. Not that uncommon though.

 

[Fragment]

Hahaha, I even double checked it the other night. How lame. Thanks