Question

a volume of 25 ml of .1 m c6h5co2h (aq) is titrated with .1 M naoh (aq) what is the ph after the addition of 12.5 ml of naoh (ka of benzoic acid = 6.3E-5

Answer 1

C₆H₅COOH is a weak acid, in order to calculate pH we first have to determine the [H⁺] in the solution.

Ka of C₆H₅COOH = 6.3 x 10^-5.  C₆H₅COOH dissociates as follows:

C₆H₅COOH C₆H₅COO^- + H⁺

Ka = ([H⁺] [C₆H₅COO^-]) / [C₆H₅COOH]

6.3 x 10^-5 = x²/ 0.1

  x² = 6.3 x 10^-6

  x = 2.5 x 10^-3

[H⁺] = x = 2.5 x 10^-3

pH = -log[H⁺] = - log (2.5 x 10^-3) = 2.6

After addition of 12.5 mL of NaOH

When NaOH is added to the C₆H₅COOH solution you form C₆H₅COONa. Now, we have a solution of a weak acid and a salt of this acid. This is a buffer solution. The pH of the buffer solution is calculated using the Henderson - Hasselbalch equation.

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

Some preliminary calculations :
pKa = -log Ka = - log (6.3 x 10^-5) = 4.2

Moles of C₆H₅COOH in 25 mL of 0.1 M solution = 0.025 mL x 0.1 M = 0.0025 moles
Moles of NaOH in 12.5 mL of 0.1 M solution = 0.0125 mL x 0.1 M = 0.00125 moles

NaOH reacts with the C₆H₅COOH in 1:1 ratio;
Therefore we produce 0.00125 mol C₆H₅COONa
and remaining unreacted is 0.00125 mol C₆H₅COOH

Now, total volume 25 mL + 12.5 mL = 37.5 mL = 0.0375 L

Molarity of C₆H₅COONa = 0.00125 / 0.0375 L = 0.033 M
Molarity of C₆H₅COOH = 0.00125 / 0.0375 L = 0.033 M

Now use the Henderson - Hasselbalch equation:
pH = pKa + log ([salt] /[acid])
pH = 4.2 + log ( (0.033 / 0.033)
pH = 4.2 + log 1
pH = 4.2 + 0
pH = 4.2