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The value of K_(w) is 9.55xx10^(-14) at a certain temperature. Calculate the pH of water at this temperature . |
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Answer» Solution :Here, we are given : `K_(w)= 9.55xx10^(-14)` Now, as for WATER, `[H_(3)O^(+)]=[OH^(-)] :. K_(w)=[H_(3)O^(+)][OH^(-)]=[H_(3)O^(+)][H_(3)O^(+)]=[H_(3)O^(+)]^(2)`, i.e., `[H_(3)O^(+)]^(2) = 9.55xx10^(-14) or [H_(3)O^(+)]=sqrt(9.55xx10^(-14))=3.09xx10^(-7)M` `:. PH = - log [H_(3)O^(+)]=-log (3.09xx10^(-7))=-[log 3.09+ log 10^(-7)]=- [ 0.49-7]=6.51`. |
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