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Explain concentration of pure water : Equilibrium of pure water is on left side. |
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Answer» Solution :Calculation of concentration of pure water :Density of pure water = 1.0 g mL = 1000 g `L^(-1)` Concentration of water =Water of mol / litre `=("Weight of 1 L water"/"Molecular mass")=1/"1 Litre"` `=(1000g)/(18 g "mol"^(-1) XX L^(-1))` `= 55.55 "mol L"^(-1)` ...(Eq.-iv) The equilibrium DISSOCIATION of water is towards the reactants. The ions of dissociated water is `[H^+][OH^-]=1.0xx10^(-7)` M `therefore` Concentration of pure water = 55.55 M `therefore` The ratio of dissociated water and undissociated water is, `="[Concentration of ions]"/"[Concentration of water]"` `=(1xx10^(-7)M)/(55.55 M)` `=1.8xx10^(-9) = ~ 2xx10^(-9)` ....(Eq. -v) Thus, for this very SMALL value the equilibrium lies mainly towards (left side) undissociated water. |
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