You have to prepare a pH 3.50 buffer and you have .10M HCOOH and .10M HCOONa, how many milliliters of each solution do you neeed for 1 liter of buffer
Use the Henderson-Hasselbalch equation.
You will need to look up Ka for formic acid and convert to pKa.
Calculate the ratio Base/acid.
Then acid + base = 1
and base/acid = from above.
Two equations and two unknowns. Solve for acid and base. I get something like 600 mL acid and 400 mL base but those are just close numbers.
To prepare a pH 3.50 buffer using HCOOH (formic acid) and HCOONa (sodium formate), you need to calculate the required amounts of each solution.
First, let's determine the acid-to-salt ratio (HCOOH to HCOONa) needed to achieve the desired pH. Since formic acid is a weak acid, its conjugate base, sodium formate, will act as a salt. The Henderson-Hasselbalch equation for pH of a buffer is given by:
pH = pKa + log([Salt]/[Acid])
The pKa of formic acid is about 3.75. Since we want a pH of 3.50, we substitute these values into the equation and solve for [Salt]/[Acid]:
3.50 = 3.75 + log([Salt]/[Acid])
-0.25 = log([Salt]/[Acid])
Next, convert this logarithmic equation to exponential form:
10^(-0.25) = [Salt]/[Acid]
0.562 = [Salt]/[Acid]
This means you need a ratio of approximately 0.562 parts salt (HCOONa) for every part acid (HCOOH).
Since we want to make a 1 L buffer, you can choose any total volume as long as the ratio remains constant. For ease of calculation, let's assume a total volume of 1000 mL (or 1 L).
Let x be the volume (in mL) of HCOOH (acid) needed.
Therefore, the volume of HCOONa (salt) needed will be 0.562x mL.
Given that the concentration of both HCOOH and HCOONa is 0.10 M, we can use the equation:
Molarity (M) = moles/volume (L)
Applying this equation to each component:
0.10 M HCOOH = moles HCOOH / (x mL / 1000 mL/L)
0.10 M HCOONa = moles HCOONa / (0.562x mL / 1000 mL/L)
To achieve the required pH, the moles of HCOOH and HCOONa must be equal. Therefore, we equate the two equations:
moles HCOOH / (x mL / 1000 mL/L) = moles HCOONa / (0.562x mL / 1000 mL/L)
Now, substituting the molar mass of HCOOH (46.03 g/mol) and HCOONa (68.01 g/mol):
(0.10 M * x mL / 1000 mL/L) * 46.03 g/mol = (0.10 M * 0.562x mL / 1000 mL/L) * 68.01 g/mol
Simplifying:
(0.04603 x g) = (0.0568017 x g)
0.04603 x = 0.0568017 x
x = 0.0568017 x / 0.04603
x = 1.2367x
Therefore, x = 0.04603 L (or 46.03 mL).
The volume of HCOOH (acid) required for a 1 L buffer is approximately 46.03 mL, while the volume of HCOONa (salt) required is approximately 0.562 * 46.03 mL = 25.88 mL.
Therefore, for 1 L of pH 3.50 buffer, you need approximately 46.03 mL of 0.10 M HCOOH (formic acid) and 25.88 mL of 0.10 M HCOONa (sodium formate).