Jawabanmu

2014-04-02T18:23:16+07:00
8 + log  \sqrt{2}
[OH⁻] =  \sqrt }4 x 10^{3} \frac{10^{-14}}{2.10^{-5}  }
[OH⁻]  \sqrt{2 . 10^{-12} }
[OH⁻] =  \sqrt{2} x 10 ⁻⁶
pOH = 6- log  \sqrt{2}
pH = 14 -( 6- log  \sqrt{2} )
pH= 8 + log  \sqrt{2}