What volume of 0.200M of aqueous solution of formic acid, a weak monoprotic acid (KA =

Question

1.78x10-4) and 0.200M aqueous solution of NaOH would you mix to prepare a 500mL of a
buffer solution of pH = 4.0.

Answer 1

I am assuming that the problem does not mean that mL acid + mL base = 500 mL. With that assumption, we see that a starting quantity for neither formic acid nor NaOH is given; therefore, an infinite number of answers are possible depending upon how much formic acid we take initially. So let's choose 25 mL of the 0.200 M HCOOH.
25 mL x 0.200 = 5.00 millimols HCOOH.
..........HCOOH + NaOH ==> HCOONa + H2O
initial...5.0......0.........0.......0
add................x..................
change.....-x......-x........x.......x
equil......5-x.....0..........x

pH = pKa + log (b/a)
4.0 = 3.75 + log (x/5-x)
solve for x and that gives me
3.2 mmols NaOH
M = mmols/mL and 0.2 = 3.2/mL
mL = 16
I like to check these things to make sure the final numbers give the right product.
25 mL HCOOH = 25*0.200 = 5.00 mmols
16 mL NaOH = 16*0.200 = 3.20 mmols

........HCOOH + NaOH ==> HCOONa + H2O
I.......5.0.......0........0........0
C.......-3.2...............3.2
E........1.8................3.2

pH = 3.75 + log(3.2/1.8) = ?
You would need to add water to the 500 mL mark.

Having said all of that if the problem actually means that volume formic acid + volume NaOH = 500 mL, we can do this. 500 x 0.2 = 100 mmols and we rework with that number.
4.0 = 3.75 x (x/100-x)
Repeat everything.
x = 64 mmols or 320 mL of 0.2M NaOH
100-x = 36 mmols or 180 mL of 0.2M HCOOH.