If the [H+] of a solution is 5.7 x 10-14 , what is the [OH-]? Round to four decimal places.
Kw=[H+][OH−]Kw=[H+][OH-]Kw=[H+][OH−]
WhereKw=1×10−14Where Kw=1\times 10^{-14}WhereKw=1×10−14
Therefore:
OH−=KwH+OH- =\frac{Kw}{H+}OH−=H+Kw
OH−=1×10−145.7×10−14=0.1754OH- = \frac{1\times 10^{-14}}{5.7\times10^{-14}}=0.1754OH−=5.7×10−141×10−14=0.1754
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