Calculate the percent error in the `[H_(3)O^(o+)]` made by neglecting the ionisation of water in `10^(-6)M NaOH` solution.

Options

A. `1%`
B. `2%`
C. `3%`
D. `4%`

Answer

A

Explanation

Neglecting ionisation water in mixture of `NaOH+H_(2)O`
`[H^(+)]=1.0xx10^(-8)= 10xx10^(-9)`
Also if ionisation of `H_(2)O` is not neglected, then
`{:(H_(2)OhArr,H^(+)+,OH^(-)),(, a,(10^(-6)+a)):}`
`:. K_(w)=10^(-14)=axx(a+10^(-6))`
`:. a=9.9xx10^(-9)`
`:. % "error" = (10xx10^(-9)-9.9xx10^(-9))/(9.9xx10^(-9))xx100`
`=1%`