Calculate the amounts and pH values for preparing the following acetic acid/acetate buffer solution.
1.) In the first step, If you plan to prepare a 2.00 M sodium acetate in 200.0 mL distilled water, how many grams of sodium acetate should be added in water?
2.) B. In the 2nd step, if 1.00M acetic acid will be added in the above solution, what will the pH value of the acetic acid/sodium buffer solution be? Assuming the change in volume will not be significant when all the chemicals are added, and the dissociation value (Ka) of acetic acid is 1.78x10-5
3.) C. If you plan to use the above stock solution from the 2nd step to prepare a new buffer solution in pH=4.50, how many additional moles of acetic acid should be added in this 200.0 mL solution. Assuming the change in volume will not be significant?
I think something is missing here. I don't get the connection between step 1 and step 2 and the final product.
To prepare 200 mL of 2.00 M sodium acetate, you will need 0.2 x 2 = 0.4 mol sodium acetate and that x molar mass = about 33 grams (but you need to go through and do it more accurately).
2) There is no indication of the amount of acetic acid added.
1.
200.0 mL = 0.20 L
Molarity = moles of solute/liter of solution
2.00M = moles of solute/0.20 L
Moles of solute = 2.00M x 0.20 L = 0.40 moles sodium acetate
Grams of solute = moles of solute x molar mass
Grams of solute = 0.40 moles x 82.03 g/mol = 32.8 grams sodium acetate
1.) To calculate the amount of sodium acetate needed, you first need to convert the given volume of solution (200.0 mL) into liters. This can be done by dividing 200.0 mL by 1000, which gives you 0.2 L.
Next, you can use the formula:
Molarity (M) = moles of solute / volume of solution (L)
Since you want to prepare a 2.00 M sodium acetate solution, you can rearrange the formula to solve for moles of solute:
moles of solute = Molarity (M) * volume of solution (L)
moles of solute = 2.00 M * 0.2 L
moles of solute = 0.4 moles
To convert moles of sodium acetate to grams, you need to use the molar mass of sodium acetate. The molar mass of sodium acetate is 82.03 g/mol. Multiply the moles of sodium acetate by the molar mass to obtain the grams of sodium acetate:
grams of sodium acetate = 0.4 moles * 82.03 g/mol
grams of sodium acetate = 32.81 grams
Therefore, you should add 32.81 grams of sodium acetate to 200.0 mL of water to prepare a 2.00 M sodium acetate solution.
2.) To calculate the pH of the acetic acid/sodium acetate buffer solution, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([Salt]/[Acid])
In this case, the salt is sodium acetate (CH3COONa) and the acid is acetic acid (CH3COOH).
The pKa value of acetic acid is given as 1.78x10^-5, so you can substitute that into the equation.
The concentration ratio [Salt]/[Acid] can be calculated by dividing the moles of sodium acetate by the moles of acetic acid. Since we don't have the moles of acetic acid yet, let's calculate it.
Using the given 1.00 M acetic acid solution, you can determine the moles of acetic acid by multiplying the concentration (M) by the volume (L). Since the volume is not provided, we'll assume it's the same as the 200.0 mL from the previous step, which is 0.2 L.
moles of acetic acid = 1.00 M * 0.2 L
moles of acetic acid = 0.2 moles
Now, you can substitute the values into the Henderson-Hasselbalch equation:
pH = -log(1.78x10^-5) + log(0.4/0.2)
pH = 4.73
Therefore, the pH value of the acetic acid/sodium acetate buffer solution is approximately 4.73.
3.) To prepare a new buffer solution at pH 4.50 using the stock solution from the second step, you need to determine the additional moles of acetic acid required.
Since we want to maintain the same volume (200.0 mL) and assuming the change in volume is not significant, the mole ratio of acetic acid to sodium acetate in the final buffer solution should be the same as in the stock solution.
From the second step, we already have 0.2 moles of acetic acid in the 200.0 mL solution.
To find the additional moles needed, you can use the Henderson-Hasselbalch equation again, but this time rearrange it to solve for the concentration ratio ([Salt]/[Acid]):
[Salt]/[Acid] = 10^(pH - pKa)
Substituting the values:
[Salt]/[Acid] = 10^(4.50 - (1.78x10^-5))
[Salt]/[Acid] = 382818.5
Since we have 0.2 moles of acetic acid, we can find the moles of sodium acetate required:
Moles of sodium acetate = [Salt]/[Acid] * moles of acetic acid
Moles of sodium acetate = 382818.5 * 0.2
Moles of sodium acetate = 76563.7
To convert moles of sodium acetate to grams, use the molar mass of sodium acetate:
grams of sodium acetate = 76563.7 moles * 82.03 g/mol
grams of sodium acetate = 6.29 x 10^6 g
Therefore, you should add an additional 6.29 x 10^6 grams of acetic acid to the 200.0 mL solution to obtain a new buffer solution with a pH of 4.50.