Question

Consider a 1.0 L buffer containing 0.083 mol L^-1 HCOOH and 0.095 mol L^-1 HCOO^-. What is the pH of the solution after adding 6.0 x 10^-3 mol of NaOH? Express your answer to 2 decimal places.

i'm repeatedly getting 3.87 as my answer....would really appreciate some help!

Answer 1

Formic acid and formate ion mixture forms acidic buffer.

To this buffer, when base is added, concentration of acid decreases and concentration of salt of acid increases.

Ka of formic acid: 1.77 × 10^-4

pKa = -logKa = -log(1.77×10^-4 )

= 4 - log(1.77)

= 4 - 0.248 = 3.752

Number of moles of Formic acid in 1.0 L of 0.083 M formic acid = Molarity* Volume in L = 0.083 M * 1.0 L = 0.083 moles

Number of moles of formate ion in 1.0 L of 0.095 M formate ion = 0.095 M * 1.0 L = 0.095 moles

When 0.006 moles of NaOH is added to 1 L of buffer, 0.006 moles of Hydroxide ions react with 0.006 moles of Hydronium ion. Hence number of moles of formic acid decreases by 0.006 moles and number of moles of formate ion increases by 0.006 moles.

Final number of moles of formic acid = 0.083 - 0.006

= 0.077 moles

Final number of moles of formate ion = 0.095 + 0.006 = 0.101 moles

Molarity is the number of moles of solute present in one L of the solution.

M = Number of moles of solute/ Volume of solution in Litres

Concentration of formic acid = 0.077 moles /1.0 L = 0.077 M

Similarly [Formate ] = 0.101 M.

pH of buffer is given by

pH = 3.752 + 0.118 = 3.87