So I already posted this question earlier today and got a reply but now that Im working through it, its not working.
What mass of solid sodium formate (of MW
68.01) must be added to 130 mL of 0.63 mol/L
formic acid (HCOOH) to make a buffer solution
having a pH of 3.68? Ka = 0.00018 for
HCOOH.
Answer in units of g.
Using pH = pKa + log [salt]/[acid] formula.
To find pKa, I did -log(.00018) = 3.745
so...
3.68 = 3.745 + log [salt]/[.63]
-.0647 = log [salt]/[.63]
.86153 = [salt]/[.63]
[salt] = .54276
Therefore mass should be [.54276] x 68.01 = 36.91
But Dr Bob said it should be around 5 so im not sure what ive done wrong
3 answers
You didn't finish. You're right with 36.91 grams but that is for 1,000 mL of solution (remember M = mols/L). You want grams in 130 mL and 36.91 x 130/1000 is very close to 5g but you can work out the exact number.
By the way, I usually follow up at the original post in case a student has a follow up question. Had you showed your work there I would have caught it faster than looking at a new post. However, I'm glad you worked it out. For another thought, if you had used mols and not mols/L for the (acid) you would not have needed that last step I showed of 36.91 x (130/1000). To be honest, however, most profs prefer that you use M or mols/L. Technically, it is correct, but it makes the problem longer. The answer is the same no matter how you do it.
Good to know, thank you so much!
3 answers