Describe the laboratory preparation of z1Litre of 1.3M formic acid buffer of pH 3.9 from solid sodium formats (molecular weight =68) and formic acid liquid 30% pure (Density of formic acid = 1.029g/ml, pKa = 3.75)
Use HH equation.
3.9 = 3.75 + log (base)/(acid)
Solve for base/acid = ? which is equation 1.
equation 2 is base + acid = 1.3 M
Solve the two equations simultaneously for (base) and (acid).
Convert to grams each needed and correct fo the 30% purity of HCOOH.
Please explain it how I will understand it especially the last part. Solve the two equations and do the conversions for me. Thanks
To prepare 1 liter of a 1.3M formic acid buffer of pH 3.9, we can follow these steps:
Step 1: Calculate the number of moles of formic acid required.
To do this, we need to determine the amount of formic acid needed to prepare the desired final concentration. The molarity (M) is defined as the number of moles of solute per liter of solution. In this case, we are looking to prepare a 1.3M solution in a final volume of 1 liter.
The number of moles (n) can be calculated using the formula:
n = M * V
where M is the desired molarity and V is the volume in liters.
So, n = 1.3M * 1L = 1.3 moles of formic acid.
Step 2: Determine the amount of solid sodium formate needed.
The molecular weight of solid sodium formate is given as 68g/mol. To convert the number of moles (1.3 moles) to grams, we can use the formula:
mass = moles * molecular weight
mass = 1.3 moles * 68g/mol = 88.4 grams
So, we will need 88.4 grams of solid sodium formate.
Step 3: Calculate the volume of formic acid (30% pure) required.
To prepare the formic acid buffer, we need a 30% pure formic acid solution. The density of formic acid is given as 1.029 g/ml. Since we are preparing 1 liter of solution, we need to calculate the volume (V) of formic acid required in milliliters.
First, we need to determine the total mass of formic acid in the 30% solution. Since the solution is 30% pure, we can find this by multiplying the total volume of the solution (1 L) by the density of formic acid (1.029 g/ml) and the desired percentage (30% or 0.3):
mass of formic acid = density * volume * purity
mass of formic acid = 1.029 g/ml * 1 L * 0.30 = 0.3087 g
Next, we need to convert the mass of formic acid to moles using the molecular weight (46.03 g/mol). This can be calculated as follows:
moles of formic acid = mass / molecular weight
moles of formic acid = 0.3087 g / 46.03 g/mol = 0.0067 moles
Finally, we can determine the volume using the equation:
volume = moles / concentration = 0.0067 moles / 1.3 M = 0.00515 L = 5.15 mL
So, we will need 5.15 mL of 30% pure formic acid.
Step 4: Preparation of the buffer solution.
Now that we have both the solid sodium formate (88.4 grams) and the liquid formic acid (5.15 mL), we can proceed to prepare the buffer solution.
1. Add the calculated mass of solid sodium formate (88.4 grams) to a suitable container.
2. Carefully measure out 5.15 mL of 30% pure formic acid (ensure accurate measurement using a suitable measuring device) and add it to the container containing the sodium formate.
3. Slowly add distilled water to the container while stirring until the final volume reaches 1 liter.
4. Continue stirring until the solid sodium formate dissolves completely.
5. Measure the pH of the buffer solution using a pH meter or indicator paper. Adjust the pH, if necessary, using either more solid sodium formate or more formic acid until the desired pH of 3.9 is achieved.
And there you have it! You have prepared 1 liter of a 1.3M formic acid buffer with a pH of 3.9 using solid sodium formate and 30% pure formic acid.