The `pH` of a solution of weak base at neutralisation with strong acid is `8. K_(b)` for the base is

Options

A. `1.0 xx 10^(-4)`
B. `1.0 xx 10^(-6)`
C. `1.0 xx 10^(-8)`
D. None of these

Answer

B

Explanation

At half neutralisation,
`[B^(o+)] = [BOH] , ["Salt"] = ["base"]`
`pOH = pK_(b) + "log" ([B^(o+)])/([BOH])`
`pOH = pK_(b), (pH = 8, pOH = 6)`
`:. pK_(b) = 6, K_(b) = 1 xx 10^(-6)`