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For a basic buffer made of a weak base B and its salt BHCl with a strong acid, the Henderson equation is

A $pOH = p K_{b} + lo g \frac{[ salt ]}{[ base ]}$
B $pOH = p K_{b} + lo g \frac{[ base ]}{[ salt ]}$
C $pH = p K_{b} + lo g \frac{[ salt ]}{[ base ]}$
D $pOH = p K_{a} + lo g \frac{[ salt ]}{[ base ]}$

Answer: A

From

K_{b} = [OH X^{-}] [BH X^{+}] / [B]

, taking negative logarithm gives

pOH = p K_{b} + lo g ([salt] / [base])

.

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