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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» Correct Answer - 1 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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