Calculate the amount (in mL) of 1.520M NaOH that is required to add the following acetic acid solution to prepare a buffer with the corresponding pH:
30.00mL of a 5.00% (w/v%) acetic acid; the resulting acetate buffer has a pH of 5.75.
pKa of acetic acid = 4.74
I would do this.
5% w/v means 5g HAc/100 mL solution which is (5/60) mols/100 and that is 5/60/0.1L or 0.0833 M.
Then millimols HAc is 30 x 0.0833 = 25.
So acid + base = 25
pH = pKa + log(base)/(acid)
5.75 = 4.74 + log b/a
b/a = about 10.3 but you need a better answer than that so go through it yourself.That's equation 1.
equation 2 is a + b = 25
Solve the two equation simultaneously for a and b. I get something like 2 for acid and 23 for base.
Then knowing that M = millimols/mL.
1.52 = 23/mL; solve for mL = about 15 mL of the 1.52 M NaOH. Of course that is about 10 for the acid that is left.
To check all of this I would plug the acid you find and the base you find into the HH equation and see if the final pH is 5.75.
1. What is 5/60?
2. How did you get 10.3 for b/a?
Hi DrBob222, can you explain what you mean by "Solve the two equation simultaneously for a and b. I get something like 2 for acid and 23 for base." I don't know how you got that and what you started with.
Thanks!
To calculate the amount of NaOH required to prepare the buffer solution, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where:
- pH is the desired pH of the buffer solution
- pKa is the dissociation constant of the weak acid, which in this case is acetic acid (CH3COOH)
- [A-] is the concentration of the conjugate base of the weak acid (acetate ion, CH3COO-)
- [HA] is the concentration of the weak acid (acetic acid, CH3COOH)
First, let's calculate the concentrations of [A-] and [HA] using the given information:
Concentration of acetic acid (CH3COOH):
5.00% (w/v%) means 5.00 g acetic acid per 100.00 mL solution.
Convert grams to moles:
(5.00 g / 60.05 g/mol) = 0.0832 mol
Concentration of acetate (CH3COO-) in the buffer:
Since acetic acid is a weak acid, it partially dissociates into its conjugate base (acetate).
Use the equation: [A-] = [HA]
Next, calculate the pH difference from the pKa:
pH difference = pH - pKa
pH difference = 5.75 - 4.74
pH difference = 1.01
Now, calculate the concentration of acetate ([A-]):
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
5.75 = 4.74 + log([A-]/0.0832)
1.01 = log([A-]/0.0832)
Taking the antilog (10 raised to the power of both sides):
[A-]/0.0832 = 10^1.01
[A-]/0.0832 = 10^1.01
Now, calculate the concentration of acetate ([A-]):
[A-] = 10^1.01 * 0.0832
[A-] = 10.652 * 0.0832
[A-] = 0.886 M
Since the concentrations of [A-] and [HA] are the same, we can conclude that the total Moles of acetic acid (CH3COOH) and acetate (CH3COO-) present in the final solution is twice the Moles of acetic acid.
Total Moles of acetic acid (CH3COOH) in the final solution:
2 * 0.0832 mol = 0.1664 mol
Now, let's calculate the volume of 1.520 M NaOH required to react with the acetic acid:
NaOH reacts with acetic acid based on the stoichiometry of the balanced equation:
CH3COOH + NaOH -> CH3COONa + H2O
The balanced equation shows that for every 1 mol of acetic acid (CH3COOH), 1 mol of NaOH is required.
Therefore, the volume of 1.520 M NaOH can be calculated using the following equation:
volume (L) = moles / concentration (M)
volume (L) = 0.1664 mol / 1.520 M
Convert liters to milliliters:
volume (mL) = 1000 * 0.1664 L
volume (mL) = 166.4 mL
Therefore, 166.4 mL of 1.520 M NaOH is required to prepare the buffer solution.