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Get the equation of ionic product (K_w) of water. |
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Answer» SOLUTION :Water are UNIQUE in their ability of acting both as an acid and a base. In presence of an acid, HA it accepts a proton and acts as the base while in the presence of a base, `B^-` it acts as an acid by donating a proton. In pure water, one `H_2O` molecule donates proton and acts as an acid and another water molecules accepts a proton and acts as a base at the same time. The following equilibrium exists: `{:(H_2O_((l))+, H_2O_((l)) hArr, H_3O_((aq))^(+)+, OH_((aq))^(-)),("Acid","Base" , "Conjugate acid","Conjugate base"):}` `therefore` The dissociation constant is represented by, `K=([H_3O^+][OH^-])/([H_2O][H_2O])=([H_3O^+][OH])/([H_2O]^2)`....(Eq.-i) The concentration `[H_2O]^2 = 1` is omitted from the denominator of water concentration REMAIN constant `[H_2O]^2`is incorporated within the equilibrium constant to give a new constant. `K[H_2O]^2=[H_3O^+][OH^-]`=K (constant) `=[H_3O^+][OH]` `therefore K_w=[H_3O^+][OH^-]` where, `K_w=K` (constant) ....(Eq.-ii) `K_w` is ionic PRODUCT of water. The concentration has been found experimentally as `1.0xx10^(-7)` M at 298 K. Dissociation of water produces EQUAL number of `H^+` and `OH^-` `therefore [OH^-] = [H^+]= 1.0xx10^(-7) M = [H_3O^+]` `K_w=(1.0xx10^(-7))(1.0xx10^(-7))` `therefore K_w=1.0xx10^(-14) M^2` ...(Eq.-ii) The very small value `10^(-14)` of `K_w` is indicate the much less self ionization of water. `K_w` is equilibrium constant so the value of `K_w` temperature dependent. |
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