question_answer 1)

Ionic product of water at 310 K is\[2.7\times {{10}^{-14}}\]. What is the pH of neutral water at this temperature?

Answer:

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