pH of sample of `KOH` and another of `NaOH` are `10` and `12` respectively. Their normalities are related as `N_(NaOH) = xN_(KOH)`. What is the value of x?

Options

A. `5//6`
B. `6//5`
C. `10^(2)`
D. `10^(-2)`

Answer

C

Explanation

`pH` of `KOH = 10, pH` of `NaOH = 12`
So, `pOH` of `NaOH = 14 - 10 = 4`
`pOH` of `NaOH = 4-12 = 2`.
`N_(KOH) = [OH^(-)] = 10^(-4)`,
`N_(NaOH) = 10^(-2)` (Because for base, `[OH^(-)]` `=` normality `= pOH`)
As per the question,
`10^(-2) = x xx 10^(-4)`
`rArr x = (10^(-2))/(10^(-4)) = 10^(2)`
`P^(H) = -log (10^(-1)) = 1`