Question

Calculate the ph of a 0.40 m solution of sodium benzoate (nac6h5coo) given that the ka of benzoic acid (c6h5cooh) is 6.50 x 10-5.
a. 2.29
b. 9.81
c. 5.11
d. 8.89
e. 11.71

Answer 1

Hello!

The dissociation reaction for Benzoic Acid is the following:

C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺

The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:

$Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X} (assume: 0,40-X\approx0,40)\\ \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\ \\ pH=-log([H_3O^{+}]=2,29$

So, the pH of this Benzoic Acid solution is 2,29

Have a nice day!