1.

Calculate the per cent error in hydronium ion concentration made by neglecting the ionisation of water in 1.0 xx 10^(-6) M NaOH

Answer»


Solution :Because `[H^(+)] = 1.0 XX 10^(-8) =10 xx 10^(-9)`
Which is very SMALL and so NEGLIGIBLE if ionisation of `H_(2)O` is not neglected, then
`{:(H_(2)O,hArr,H^(+)+,H^(-)),(,,a,(10^(-6)+a)):}`
`:. a xx (10^(-6) + a) = 10^(-14) "" a = 9.9 xx 10^(-9)`
`:.` % error `= (10 xx 10^(-9) - 9.9 xx 10^(-9))/(9.9 xx 10^(-9)) xx 100 = 1%`


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