A buffer with a pH of 413 contains 027 M of sodium benzoate

A buffer with a pH of 4.13 contains 0.27 M of sodium benzoate and 0.32 M of benzoic acid. What is the concentration of [H⁺] in the solution after the addition of 0.050 mol of HCl to a final volume of 1.6 L? Assume that any contribution of HCl to the volume is negligible.

Solution

Answer

[H+] = 9.33×10^-5M

Explanation

Henderson - Hasselbalch equation is

pH = pKa + log([A-] /[HA])

pKa of Benzoic acid = 4.20

pH = 4.20 + log([C6H5COO-] /[C6H5COOH])

Initial mole of C6H5COO- = (0.27mol/1L)×1.6L = 0.432

Initial mole of C6H5COOH = (0.32mol/1L)×1.6L =0.512

HCl react with conjucate base C6H5COO-

After addition of HCl

No of moles of C6H5COO- = 0.432 - 0.05 = 0.382

No of moles of C6H5COOH = 0.512 + 0.05 = 0.562

[C6H5COO-]= (0.382mol/1.6L)×1L =0.2388M

[C6H5COO-]= (0.562mol/1.6L)×1L = 0.3513M

Therefore, new pH is

   pH = 4.20 + log(0.2388M/0.3513M)

   = 4.20 - 0.17

   = 4.03

pH = - log[H+]

-log[H+] = 4.03

   [H+] = 9.33×10^-5M