InterviewSolution
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InterviewSolution
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\left. \begin{array} { l } { ( 55.4 \times 10 ^ { 14 } ) ^ { - 1 } } \\ { \frac { 1 \times 10 ^ { - 14 } } { 18 } } \end{array} \right. |
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Answer» [H+] = [OH-]= 1 x 10 -7 mol -3 Kw = [H3O+] [OH -] Kw = (1 x 10 -7 mol dm -3) ( 1 x 10 -7 mol dm -3) = 1 x 10 -14 mol2 dm -6 The degree of dissociation of water = 10 -7 mol dm -3 / 55.55 mol dm -3 = 1.8 x 10-9 Since with increase in temperature dissociation of water increases, therefore, value of Kw increases as the temperature is increased. However, at all temperatures [H+] remains equal to [OH-] in pure water. |
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