Define ionic product of water. The degree of dissociation of pure water at `18^(@)C` is found to be `1.8times10^(-9)` Find the final ionic product of water and its dissociation constant at `18^(@)C`

Explanation

Cocnetration of water `=(1000)/(18)=55.56" moles/litre" ("supposing density of water =1g/mL")`.
Thus,
`overset(55.56)(H_(2)O) Leftrightarrow overset(0)(H^+)+overset(0)(OH^(-))` Initial concentration
55.56 `(1-1.8 xx 10^(-9)) (55.56 xx 1.8 xx 10^(-9)) (55.56 xx 1.8 xx 10^(-9)) `
`K_(w) =[H^+] [OH^+] =(55.56 xx 1.8 xx 10^(-9)) (55.56 xx 1.8 xx 10^(-9))`
`=1 xx 10^(-14) ("mole/litre")^2.`
Further
`K=([H^+] [OH^(-)])/([H_(2)O])` (K=dissociation const. of `H_(2)O)`
`or K=(K_w)/([H_(2)O])=(1 xx 10^(-14))/(55.56) =1.8 xx 10^(-16)"mole/litre"`