What volume of 0.200M of aqueous solution of formic acid, a weak monoprotic acid (KA =
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.
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.
To answer this question, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa of the acid and the ratio of the concentration of the acid to its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = desired pH of the buffer solution (4.0 in this case)
pKa = -log(Ka) = -log(1.78x10^-4) = 3.75 (approximated)
[A-] = concentration of the conjugate base (formate ion)
[HA] = concentration of the acid (formic acid)
We want to prepare a buffer solution, so we need to determine the concentrations of formic acid and sodium formate required.
Let's assume that x moles of formic acid (HA) were mixed with x moles of sodium formate (A-) to prepare a 500 mL buffer solution.
To get the concentration of formic acid, we can use the formula:
[HA] = n/V
Where:
n = number of moles of formic acid
V = volume of the buffer solution in liters (500 mL = 0.5 L)
So, [HA] = x/V = x/0.5
Similarly, the concentration of sodium formate is:
[A-] = x/V = x/0.5
Now, we can substitute these values into the Henderson-Hasselbalch equation:
4.0 = 3.75 + log(([A-]/[HA]))
Let's solve for x:
4.0 - 3.75 = log(([A-]/[HA]))
0.25 = log([A-]/[HA])
Now, using the properties of logarithms, we can rewrite this equation as:
10^0.25 = [A-]/[HA]
10^0.25 is approximately 1.778, so:
1.778 = [A-]/[HA]
Since both [A-] and [HA] are x/0.5, we can substitute these values:
1.778 = (x/0.5)/(x/0.5)
Simplifying the expression:
1.778 = 1
This means that any value of x will satisfy the equation, as long as the concentrations of formic acid and sodium formate are the same.
Therefore, the volume of formic acid and sodium hydroxide solutions needed to prepare a 500 mL buffer solution of pH 4.0 is not uniquely determined. As long as the concentrations of the acid and its conjugate base are the same, the exact volumes of the solutions can vary as long as the sum is 0.5 L.