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gregori [183]
4 years ago
14

The value of ka for benzoic acid is 6.30×10-5. what is the value of kb, for its conjugate base, c6h5coo-?

Chemistry
1 answer:
Serggg [28]4 years ago
4 0
Benzoic acid release protons in water:
C₆H₅COOH(aq) ⇄ C₆H₅COO⁻(aq) + H⁺(aq).
Benzoic acid conjugate base gain protons in water:
C₆H₅COO⁻(aq) + H⁺(aq) ⇄ C₆H₅COOH(aq).
Ka(C₆H₅COOH) = 6.3·10⁻⁵.
Ka · Kb = 1·10⁻¹⁴.
Kb(C₆H₅COO⁻) = 1·10⁻¹⁴ ÷ 6.3·10⁻⁵.
Kb(C₆H₅COO⁻) = 1.587·10⁻¹⁰.
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2 years ago
What volume of a 6.67 M NaCl solution contains 3.12 mol NaCl? L
harkovskaia [24]

Answer:

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Explanation

Given parameters :

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Volume of NaCl =?

Volume of NaCl = number of moles/Molarity

Volume of NaCl = 3.12mol/6.67M

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4 0
4 years ago
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swat32
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