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The Kw of water varies with temperature. Calculate the pH of water at 30⁰C with a Kw = 1.469 x 10-14. Show all calculations leading to an answer. @girlover @iceicebaby
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plz i am so stuck
Kw=[H+]*[OH-] Kw=x^2 x=[H+]=[OH-]=(1.469*(10^(-14)))^(1/2) pH=-log[H+]
\[Kw=\sqrt{1.469\times10^{-14}}\]
got it?
\[\left[ H ^{+} \right]=\sqrt{1.469 \times 10 ^{-14}}\] \[pH=-\log[H ^{+}]= - \log \sqrt{1.469 \times 10 ^{-14}}= 6.92\]
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i'm not a calculator but i think that manner is right.
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