At `25^(@)C`, pH of a `10^(-8)`(M) aqueous KOH solution will be-

Options

A. 6
B. 7.02
C. 8.02
D. 9.02

Explanation

In `10^(-8)` of KOH solution, `[HO^-]=10^(-8)(M)+` concentration of `OH^-` ions obtained form ionisation of water`=10^(-8)`(M)+x(M)
In thies solutioni `[H_3O^+]`=concentration of `H_3O^+` ions obtained from ionisation of water=x(M) [on ionisation of water equal number of `H_3O^+` and `OH^-` ions are formed]
`:.[H_3O^+][OH^-]=K_w=10^(-14)`
or, `x(x+10^(-8))=10^(-14)` or `x^2+10^(-8)x-10^(-14)=0`
or, x`=(-10^(-8)+-sqrt(10^(-16)+4xx1xx10^(-8)))/(2xx1)=9.51xx10^(-8)`
So, in `10^(-8)`(M) solution in KOH
`[HO^-]=10^(-8)+9.51xx10^(-8)(M)=1.051xx10^(-7)(M)`
`:.pOH=-log_(10)[OH^-]=-log_(10)(1.051xx10^(-7))=6.98`
and pH=14-pOH=14-6.98=7.02