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Acetylsalicylic acid (aspirin) behaves as an acid according to the equation shown. Calculate K_b for the $C_{9} H_{7} O_{4}^{-} (a q) ion : (K_{a} = 3.0 \times 10^{- 4})$
${HC}_{9} H_{7} O_{4} (a q) + H_{2} O ⇌ H_{3} O^{+} (a q) + C_{9} H_{7} O_{4}^{-} (a q)$

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3.0 X 10^-17

3.3 X 10^-11

9.0 X 10^-8

3.3 X 10³

answer is B.

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Detailed Solution

$k_{\dot{b}} = \frac{k_{w}}{k_{a}}$

$k_{b} = \frac{10^{- 14}}{3 \times 10^{- 4}} k_{b} = \frac{1}{3} \times 10^{- 10} k_{b} = 3.33 \times 10^{- 11}$

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