Question

Be sure to answer all parts. What is the [H3O+] and the pH of a benzoic acid-benzoate buffer that consists of 0.24 M C6H5COOH and 0.35 M C6H5COONa? (Ka of benzoic acid = 6.3 × 10−5) Be sure to report your answer to the correct number of significant figures.

Answer 1

Answer:

[H3O+] = 4.32 * 10^-5 M

pH = 4.36

Explanation:

Step 1: Data given

Molarity of C6H5COOH = 0.24 M

Molarity of C6H5COONa = 0.35 M

(Ka of benzoic acid = 6.3 * 10^−5

Step 2: Calculate [H3O+]

Benzoic acid = weak acid which, in aqueous solution dissociates as follows:

C6H5COOH (aq)  <=>  C6H5COO- (aq)  +  H+ (aq)

Ka = [C6H5COO-][H+]  /  [C6H5COOH]     = 6.3 * 10^−5

Because benzoic acid  is a  weak acid and so only slightly dissociated in aqueous solution, we can assume that all of the C6H5COOH comes from the benzoic acid, and all of the C6H5COO- comes from the ionic sodium benzoate in the buffer mixture.

Ka =6.3 * 10^−5  =  0.35 x [H+(aq)]  / 0.24

[H+]  = [H3O+] = 6.3 * 10^−5 *  0.24  / 0.35

[H3O+] = 4.32 * 10^-5 M

Step 3: Calculate pH

pH = -log [H3O+]  = - log( 4.32 * 10^-5)  =  4.36

Answer 2

The pH of a benzoic acid-benzoate buffer that consists of 0.24 M C₆H₅COOH and 0.35 M C₆H₅COONa is 5.7 and the concentration of H₃O⁺ is 2.0 × 10⁻⁶ M.

We have a buffer system formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ from C₆H₅COONa).

We can calculate the pH using Henderson-Hasselbach's equation.

$pH = pKa + log\frac{[C_6H_5COO^{} ]}{[C_6H_5COOH]} \\\\pH = -log (6.3 \times 10^{-5} )+ log \frac{0.35}{0.24} = 5.7$

Given the pH is 5.7, we can calculate the concentration of H₃O⁺ using the following expression.

$pH = -log [H_3O^{+} ]\\[H_3O^{+} ] = antilog-pH = antilog -5.7 = 2.0 \times 10^{-6} M$

The pH of a benzoic acid-benzoate buffer that consists of 0.24 M C₆H₅COOH and 0.35 M C₆H₅COONa is 5.7 and the concentration of H₃O⁺ is 2.0 × 10⁻⁶ M.

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