Answer:
\[{{H}_{2}}O\rightleftharpoons
{{H}^{+}}+O{{H}^{-}}\]
\[{{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]\] \[[{{H}^{+}}]=[O{{H}^{-}}]\]
\[27\times
{{10}^{-14}}={{[{{H}^{+}}]}^{2}}\]
\[[{{H}^{+}}]=\sqrt{2.7\times
{{10}^{-14}}}\]
\[=1.643\times
{{10}^{-7}}=M\]
\[pH=-{{\log
}_{10}}[{{H}^{+}}]\]
\[=-{{\log
}_{10}}[1.643\times {{10}^{-7}}]=6.78\]
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