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Calculate the percent error in the `[H_(3)O^(o+)]` made by neglecting the ionisation of water in `10^(-6)M NaOH` solution. |
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Answer» a. Neglecting the ionisation of `H_(@)O, [oversert(Θ)OH]` from `10^(-6)M NaOh`, and `[H_(3)O^(o+)] = 10^(-8)M`. ltbRgt b. Including the ionisation of water `x = [H_(3)O^(o+)]` and `(x+10^(-6)) = [overset(Θ)OH]` `:. (x+10^(-6)) (x) = 10^(-14)` or `x^(2) +10^(-6) x- 10^(-14) = 0` `:.x=(-10^(-6)+sqrt(10^(-12)+4xx10^(-14)))/(2)=9.9xx10^(-9)` % error `= ((10^(-8))-(9.9 xx 10^(-9)))/(9.9xx10^(-9))xx100` `= ((10xx10^(-9))-(9.9xx10^(-9)))/(9.9xx10^(-9))xx100% =1%` |
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