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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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<h3>Answer:</h3>

             7.57 × 10⁻²² g of F

<h3>Solution:</h3>

Data Given:

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Putting value,

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Step 2: Calculate Mass of BF₃:

                   Moles  =  Mass ÷ M.Mass

Solving for Mass,

                   Mass  =  Moles × M.Mass

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                   Mass  =  1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹

                   Mass  =  9.0 × 10⁻²² g

Step 3: Calculate Mass of Fluorine Atoms:

As,

                         67.82 g BF₃ contains  =  57 g of F

So,

                    9.0 × 10⁻²² g will contain  =  X g of F

Solving for X,

                       X =  (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g

                        X  =  7.57 × 10⁻²² g of F

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