Question

A student creates 150ml of a buffer that is .018M in formic acid and .016M in sodium formate. What would the pH of the buffer be after the addition 15ml of .070M HCl solution to the buffer?

a) 3.69

b) 3.85

c) 4.18

d) 1.15

e) 3.30

Answer 1

To solve this kind of problem you need first calculated the quantity of mol that you have in solution for the formic acid and for sodium formate:

Formic acid: mol = 0,018 M x 0,150 L = 0,0027 mol formic acid

Sodium formate: mol = 0,016 M * 0,150 L = 0,0024 mol sodium formate

The mol of acid that are added are: mol HCl = 0,070 M * 0015 L = 0,0015 mol HCl

The reaction that follows the process is:

HCOOH <=> H⁺ + COO^-

If we added a little bit of HCl , the H⁺ is going to be produce a little bit of formic acid, and less quantity of sodium formate is going to be present in the solution. In the process, the 0.0024 moles of COO^- is reduced:

0,0024 mol initial sodium formate - 0,00105 mol H⁺ = 0,00135 mol sodium formate

For the formic acid we have:

0,0027 mol initial formic acid + 0,00105 mol H⁺ = 0,00375 mol formic acid

Plugging these new values into Henderson-Hasselbalch gives:

pH = pK_a + log (base/acid) = 3.75 + log (0.0135 moles COO^-/0.0375 moles HCOOH) = 4.18

The answer is c)