make 300.0 mL of a pH 5.30 buffer from 0.500 M acetic acid and 0.500 M sodium acetate. Calculate the volume of 0.500 M acetic acid and the volume of 0.500 M sodium acetate required to make 300.0 mL of a buffer solution with pH = 5.30.
(The Ka for acetic acid at 298 K is 2.00 x 10-5)
pH = pKa + log (base)/(acid)
5.30 = 4.699 + log(b/a)
b/a = 3.99 for one equation.
Second equation is
mLbase x 0.5M = mLacid x 0.5M
The 0.5M divides out. Substitute base 3.99*acid in the above and
mLbase =3.99(300-mLbase)
solve for mL base and subtract from 300 to obtain mL acid. You should get something close to 120 and 31 but those aren't exact.
To calculate the volume of 0.500 M acetic acid and 0.500 M sodium acetate required to make a buffer solution with pH = 5.30, we need to use the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is: pH = pKa + log([A-]/[HA])
In this case, acetic acid (CH3COOH) acts as the acid (HA) and sodium acetate (CH3COONa) acts as the conjugate base (A-).
Step 1: Calculate the ratio of [A-]/[HA]
pH = pKa + log([A-]/[HA])
5.30 = -log10(2.00 x 10^-5) + log([A-]/[HA])
Step 2: Solve for [A-]/[HA]
5.30 + log10(2.00 x 10^-5) = log([A-]/[HA])
Take the antilog of both sides:
10^(5.30 + log10(2.00 x 10^-5)) = [A-]/[HA]
10^5.30 × 2.00 x 10^-5 = [A-]/[HA]
2.00 x 10^-5 × 10^5.30 = [A-]/[HA]
2.00 × 10^(5.30 - 5) = [A-]/[HA]
2.00 × 10^0.30 = [A-]/[HA]
2.00 × 1.9952623149688795 = [A-]/[HA]
3.990524629937759 = [A-]/[HA]
Step 3: Calculate the moles of acetic acid and sodium acetate required.
Let's assume the volume of acetic acid (HA) required is V1 mL, and the volume of sodium acetate (A-) required is V2 mL.
The initial moles of acetic acid (HA) = 0.500 M × V1 mL = 0.500 V1 mmol
The initial moles of sodium acetate (A-) = 0.500 M × V2 mL = 0.500 V2 mmol
Using the ratio from Step 2, the final moles of acetic acid (HA) and sodium acetate (A-) will be:
Final moles of acetic acid (HA) = (3.990524629937759 / (1 + 3.990524629937759)) × 0.500 V1 mmol
Final moles of sodium acetate (A-) = (1 / (1 + 3.990524629937759)) × 0.500 V2 mmol
Step 4: Convert moles to volume using the molarity (M) of each solution:
Final volume of acetic acid (HA) = (Final moles of acetic acid (HA) / 0.500 M)
Final volume of sodium acetate (A-) = (Final moles of sodium acetate (A-) / 0.500 M)
Substituting the values:
Final volume of acetic acid (HA) = ((3.990524629937759 / (1 + 3.990524629937759)) × 0.500 V1 mmol) / 0.500 M
Final volume of sodium acetate (A-) = ((1 / (1 + 3.990524629937759)) × 0.500 V2 mmol) / 0.500 M
Simplifying the equations gives:
Final volume of acetic acid (HA) = (0.798105539)./M × V1 mL
Final volume of sodium acetate (A-) = (0.2018944606)./M × V2 mL
Therefore, to make a 300.0 mL buffer solution with pH = 5.30, you would need to measure approximately (0.798105539)./M × V1 mL of 0.500 M acetic acid and (0.2018944606)./M × V2 mL of 0.500 M sodium acetate.