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Sergio039 [100]
3 years ago
13

In an exactly.200M solution of benzoic acid, a monoprotic acid, the [H+] = 3.55 × 10–3M. What is the Ka for benzoic acid?

Chemistry
2 answers:
SashulF [63]3 years ago
7 0
6.3 *10^-5 I hope this helps
Law Incorporation [45]3 years ago
4 0

Answer : The value of k_a for benzoic acid is, 6.4\times 10^{-5}

Solution :

The balanced equilibrium reaction will be,

                             C_6H_5COOH\rightleftharpoons H^++C_6H_5COO^-

initial conc.          0.2 M                  0                0

at eqm.  (0.2-0.00355)M       0.00355M 0.00355M

The expression for dissociation constant for a benzoic acid will be,

k_a=\frac{[H^+]\times [C_6H_5COO^-]}{[C_6H_5COOH]}

Now put all the given values in this formula, we get the value of k_a

k_a=\frac{(3.55\times 10^{-3})\times (3.55\times 10^{-3})}{(0.2-3.55\times 10^{-3})}=6.4\times 10^{-5}

Therefore, the value of k_a for benzoic acid is, 6.4\times 10^{-5}

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