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At 25^@C the degree of ionization of water was found to be 1.8 xx 10^(–9). Calculate the ionization constant and ionic product of water at this temperature |
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Answer» Solution : If x is the degree of ionization of water, then `{:(H_2O, hArr , H^(+), + , OH^(-)),(c(1-ALPHA), , CALPHA , , calpha):}` `c=[H_2O]=1000/18`=55.56 M `K_(eq)=([H^+][OH^+])/([H_2O])=((calpha)^2)/(c(1-alpha))=calpha^2` (since `alpha` is very muchless than 1) `K_(eq)=55.56xx(1.8xx10^(-9))^2=1.8xx10^(-16)` M `K_W=[H^+][OH^-]=(c alpha )^2 =(55.56xx1.8xx10^(-9))^2` `K_W=1.0xx10^(-14) M^2` |
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