Question

A buffer with a pH of 3.90 contains 0.15 M of sodium benzoate and 0.30 M of benzoic acid. What is the concentration of [H ] in the solution after the addition of 0.056 mol of HCl to a final volume of 1.4 L? Assume that any contribution of HCl to the volume is negligible. WHAT IS CONCENTRATION OF HCL ADDED TO BUFFER??

Answer 1

Benzoic acid pKa = 4.2

Given that

pH = 3.90

and molarity of sodium benzoate =0.15 M

molarity of benzoic acid =0.30 M

pH = pKa + log (moles of sodium benzoate / moles of benzoic acid)

and here initial volume is 1.4 litre

molarity = number of moles / volume in L

number of moles = molarity * volume in L

Therefore

salt moles = 0.15 x 1.4 = 0.21

acid moles = 0.30 x 1.4 = 0.42

HCl moles = 0.056

pH =pKa + log [salt - HCl moless / acid + HCl moles ]

pH = 4.2 + log [0.21 -0.056 / 0.42 + 0.056 ]

pH = 4.2 + log [0.154 / 0.476 ]

pH = 3.71

[H+] = 10^-pH = 10^-3.71

[H+] =   1.95 x 10^-4 M