My question is #14, how do you find the [oh-] of the problem?

Hey! I can help you with this question!! 

So you first need to understand that [OH-][H+]=1.0 × 10⁻¹⁴. Remember that 1.0 x 10⁻¹⁴ is equal to Kw just for future reference. So now that you know this you can plug and chug.

(3.4 x 10⁻⁶)×[OH-] = 1.0 × 10⁻¹⁴

[OH-]= (1.0 × 10⁻¹⁴) ÷ (3.4 × 10⁻⁶)
[OH-]=2.941167 × 10⁻⁹

Rounded for sig figs= 2.9 × 10⁻⁹

Had a test on this today! Hope it helps!