The pH of `10^(-8) M` NaOH will be

Explanation

If we use the relation, pOH=-log `[OH^-]`, we get pOH equal to 8 and pH becomes 6 (14 - 8). But this is not correct because a basic solution cannot have pH less than 7. Therefore, for very dilute basic solutions, when `OH^-` concentrations from base and water are comparable, then concentration of `OH^-` from water cannot neglected. Therefore,
`[OH^-]_("total")=[OH^-]_"base"+[OH^-]_"water"`
The concentration of `OH^-` from dissociation of water is equal to the concentration of `H^+` from water,
`[OH^-]_(H_2O) =[H^+]_(H_2O)` =x say
`[OH^-]_("total")=1.0xx10^(-8)+x`
But `[H^+] [OH^-]=1.0xx10^(-14)`
`x(1.0xx10^(-8)+x)=1.0xx10^(-14)`
`1.0xx10^(-8)x+x^2=1.0xx10^(-14)`
`x^2+1.0xx10^(-8)x-1.0xx10^(-14)=0`
Solving for x , we get x = `9.5xx10^(-8)`
`[OH^-]=1.0xx10^(-8)+9.5xx10^(-8)`
`=1.05xx10^(-7)`
`pOH=-log (1.05xx10^(-7))`
=6.98
pH=14-pOH
=14-6.98 =7.02