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At `90^(@)C` , pure water has `[H_(3)O^(+)]` as `10^(-6)` mol `L^(-1)`. What is the value of `K_(w)` at `90^(@)C` ?A. `10^(-14)`B. `10^(-6)`C. `10^(-12)`D. `10^(-8)` |
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Answer» Correct Answer - C `K_(w)=C_(H_(3O^(+))C_(OH^(-))` for pure water (neutral), `C_(H_(3)O^(+))=C_(OH^(-))`. Thus, `K_(w)=(10^(-6))(10^(-6))=10^(-12)` |
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