Correct Answer - 1
`(1)" " 10^(-pH)xx10^(-pOH)=10^(-Kw)`
`(2)" "`as `[H^(+)]` increases, pH decreases,
`(3)` since ionization of water is endothermic in nature hence increase in temperature increases `K_(w)`.
`(4)" "K_(a)` of `H_(2)O=([H^(+)][OH^(-)])/([H_(2)O])=(10^(-14))/(55.55)=1.8xx10^(-16)`
`10^(-pH)xx10^(-pOH)=10^(-Kw)`