Question

The pKa of formic acid is 3.74, a monoprotic weak acid. A 1.0 L sample of a buffer with a pH of 4.5 is combined with .09 mol of NaOH. What is the new pH of the solution? The ratio of HCOO- to H+ is .86 M to .15 M.

Answer 1

Use the Henderson-Hasselbach equation:

pH = pKa + log [base/acid]

pH = 3.74 + log [0.86/0.15]

pH = 3.74 + 0.758

pH = 4.498

Calculate the number of moles of HCOOH and HCOO^- present in the buffer before the addition of NaOH. To do this multiply molarity by amount of solution.

HCOOH = 0.15 mole/L X 1L = 0.15 moles.

HCOO^- = 0.86 mole/L X 1L = 0.86 moles.

These are the number of moles already present in the buffer.

After addition of NaOH, it's going to react with formic acid.

NaOH + HCOOH -----------> HCOONa + H₂O

Now, the new number of moles are going to be:

HCOOH = 0.15 moles - 0.09 moles of NaOH = 0.06 moles

NaOH = 0.09 moles - 0.09 moles = 0 moles.

HCOO^- = 0.86 moles + 0.09 moles of NaOH = 0.95 moles

Now, naturally the concentration of HCOOH and HCOO^- in the buffer has changed.

HCOOH = 0.06 moles / 1 L = 0.06 M

HCOO^- = 0.95 moles / 1 L = 0.95 M

Use the Henderson-Hasselbach equation again to calculate the new pH:

pH = pKa + log [base/acid]

pH = 3.74 + log [0.95/0.06]

pH = 4.939

As can be seen, after addition of 0.09 moles of NaOH the pH of buffer has increased as expected.