How many grams of NaF should be added to 2.00 L of 0.100 M HF to yield a buffer with a pH = 3.50?

Question

How many grams of NaF should be added to 2.00 L of 0.100 M HF to yield a buffer with a pH = 3.50?

options given:

12.4 g

5.56 g

2.28 g

7.32 g

i keep getting 17.96 grams but thats not an option

Answer 1

I get an answer of 19.65 g which also is not an option. Is it possible the options given are incorrect? In fact, if you take each of the options given, plug them into the Henderson Hasselbalch equation and solve for pH, in no case do you get a pH of 3.50. Plugging in your answer of 17.96 g, I get a pH = 3.46, and plugging in 19.65 g, I get a pH = 3.50. This slight difference may be related to our use of different values for Ka and pKa. My approach is as follows:

Henderson Hasselbalch equation:

pH = pKa + log [salt] / [acid]

pH = 3.50

pKa = -log Ka and Ka for HF = 7.2x10-4; pKa = 3.13 (looked it up in a table)

[HF] = 0.100 M

[NaF] = ?

So, we will first solve for [NaF] and then determine grams needed:

3.50 = 3.13 + log (NaF/0.100)

0.37 = log (NaF/0.100)

[NaF] / 0.100 = 2.34

[NaF] = 0.234 M

Molar mass NaF = 41.99 g / mol

0.234 mol / L x 2 L x 41.99 g /mol = 19.65 g NaF needed

NOTE: We are assuming no change in volume upon addition of the NaF