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The ionization constant of benzoic acid is 6.46 × 1 0 - 5 and K s p for silver benzoic is 2.5 × 1 0 - 13 . How many times is silver benzoate more soluble in a buffer of p H = 3.19 compared to its solubility in pure water?

Options

  1. A4
  2. B3.32
  3. C3.01
  4. D2.5

Correct answer

B. 3.32

Step-by-step solution

Suppose S is the molar solubility of silver benzoate in water, then C 6 H 5 C O O A g s ⇌ C 6 H 5 C O O a q - + A g a q + K s p = S 2 So S = 2.5 × 1 0 - 13 = 5.0 × 1 0 - 7 M If the solubility of salt of weak acid of ionization constant K a is S, then K s p , K a and S' are related to each other at p H = 3.19 . So [ H + ] = 6.46 × 1 0 - 4 M ∵ pH = 3.19 K s p = S ′ 2 K a K a + H + S ′ = 2.5 × 1 0 - 13 6.46 × 1 0 - 5 6.46 × 1 0 - 5 + 6.46 × 1 0 - 4 1 2 S ′ = 2.5 × 1 0 - 13 × 7.106 × 1 0 - 4 6.46 × 1 0 - 5 1 2 = 2.75 ×

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