The pH of a solution of weak base at neutralisation with strong acid is 8. $K_{b}$ for the base is

1.0 \times 10^{- 4}

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1.0 \times 10^{- 6}

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1.0 \times 10^{- 8}

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None of these

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Solution

The correct option is B $1.0 \times 10^{- 6}$
$p H = - l o g [H^{+}] = 8$

$p O H = 14 - p H = 6$

$[O H^{-}] = 10^{- 6} = K_{b}$ as base is weak so, $[B^{-}] = [B O H]$

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