Correct Answer - 3
pH of HCl solution `=5`
`[H^(+)]_("initial")=10^(-5)`
`[H^(+)]_("initial")xxV_(1)==[H^(+)]_("initial")xxV_(2)`
`[H^(+)]_("initial")=10^(-8)`
Let the `H^(+)` from water be x.
`{:(H_(2)O,hArr,H^(+),+,OH^(-)),(,,x+10^(-8),,x):}`
`x(x+10^(-8))=10^(-14)`
`x^(2)+10^(-8)-10^(-14)=0`.
`x=(-10+-sqrt(10^(-16)+4xx10^(-14)))/(2)rArr" "x=(-10^(-8)+10^(-8)sqrt(401))/(2) =19xx10^(-8)=9.5xx10^(-8)`
`[H^(+)]_("Total")=[H^(+)]_(HCl)+[H^(+)]_(Water)`
`=10^(-8)+10^(-8)xx9.5`
`10.5xx10^(-4)`
`pH=8-log 10.5`
Slightly less then 7
Hence `: (3)`