3) Formic acid, HFor, has a Ka value equal to about 1.8x10-4. A student is asked

Question

3) Formic acid, HFor, has a Ka value equal to about 1.8x10-4. A student is asked to prepare a buffer having a pH of 3.40 from a .10M HFor and a .10M NaFor solution. How many milliliters of the NaFor solution should she add to 20mL of the .10M HFor to make the buffer?

answer: 9.043 mL .....correct???? or incorrect????

4) When 5 drops of .10 M NaOH were added to 20mL of the buffer in problem 3, the pH went from 3.40 to 3.43. Write a net ionic equation to explain why the pH didn't go up to about 10, as it would have it that amount of NaOH were added to distilled water or to 20mL .00040 M HCl, which also would have a pH of 3.40.

Explanation / Answer

3) According to Henderson - Hasselbach equation for a weak acid (HFor) and its salt with a strong base (NaFor), pH = pKa + log10[salt]/[acid]

Now, pKa = - log Ka = - log (1.8 x 10-4) = 3.745

log10[salt]/[acid] = pH - pKa = 3.40 - 3.745 = - .345

[salt]/[acid] = antilog (-.345) = 0.45185594437

Now, we have [salt]/[acid] ratio is 0.45185594437, i.e, for every 1 volume of the acid, we have to add 0.45185594437 volumes of same concentration of salt (0.10 M).

Therefore, for 20 mL acid, we need 20 x 0.45185594437 = 9.037 mL. So accounting for rounding errors, yes, your answer is CORRECT!

4) Formic acid is a weak acid and is not completely dissociated when in aqueous solution, and there is an equilibrium:

HCOOH + H2O <---> HCOO- + H3O+ ---- (1)

Sodium formate is a salt of strong base (NaOH), so when placed in water it will undergo dissociation almost to full extent. Hence this reaction goes to completion:
NaHCOO --> Na+ + HCOO- ------- (2)

Thus, a solution of a weak acid and its salt with a strong acid can act as a buffer solution and resist large changes in pH from happening. If you add a strong base, it completely dissociates and produces OH-, which can consume H+ and thus usually increases the pH of the solution.

NaOH --> Na+ + OH-

OH- + H3O+ <--> 2H2O

However, in the buffer, the OH- is consumed by the Na+ which was formed by the dissociation of sodium formate, thus not allowing it to deplete the H3O+. Hence the pH remains same.

If you added a strong acid, usually the pH decreases because of the increase in H+ ions from dissociation of acid. In the buffer, these H+ are consumed by the formate ions to drive the reaction in (1) in reverse direction to produce HCOOH. And hence pH would not change much.

But in case of a strong acid being added to a strong like HCl in the example, there is a complete neutralization of the H+ ions of HCl (or H3O+ considering they are hydrolysed in aqueous form) by the OH- from NaOH, and hence pH increases very quickly.

Hope this helps. Good luck!

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