Calculate the pH of a solution that results from mixing 36 mL of 0.11 M benzoic acid with 13 mL of 0.14 M sodium benzoate.

Question

Calculate the pH of a solution that results from mixing 36 mL of 0.11 M benzoic acid with 13 mL of 0.14 M sodium benzoate. The K_a value for C₆H₅COOH is 6.5 x 10^-5.

the answer is: 3.85

please explain how to find the answer so I can follow along. thank you!

Answer 1

This is a buffer consisting of a weak scid and the salt of that acid. You can use the Henderson Hasselbalch equation

pH = pKa + log [salt]/[acid]

pKa = - log Ka = -log 6.5x10^-5 = 4.19

Now all we need is [salt] and [ acid] in a final volume of 49 ml = 0.049 L

[salt] = 0.013 L x 0.14 mol/L = 0.00182 mol/0.049 L = 0.037 M

[acid] = 0.036 L x 0.11 mol/L = 0.00396 mol/0.049 L = 0.081 M

pH = 4.19 + log (0.037/0.081)

pH = 4.19 + (-.33)

pH = 3.86

Answer 2

To answer this question, you need to use the Henderson Equation:

pH = pKa + log[A-/HA]

Where HA is the concentration of the acid (benzoic acid in this problem)

And A- is the concentration of the conjugate base (sodium benzoate in this problem)

pKa = -logKa = -log(6.5 x 10^-5) = 4.187

To find the concentrations: (volume*molarity)/total volume of solution

Total volume = 36 + 13 = 49 mL

Concentration of benzoic acid

[C₆H₅COOH] = (36 mL*0.11 M)/49 mL = 0.0808 M

Concentration of Sodium Benzoate

[C₆H₅COO^-] = (13 mL*0.14 M)/49 mL = 0.0371 M

Plug in the values in the Henderson Equation and solve.

pH = pKa + log[C₆H₅COO^-]/[C₆H₅COOH]

pH = 4.187 + log(0.0371/0.0808) = 3.85