What was the volume of 6.12 M acetic acid HC2H3O2 needed to prepare the 250 mL acetic acid/acetate ion buffer solution required in this part? Show your calculations.
By the use of Henderson Hasselbalch equation;
pH = pKa + log{[acetate ion]/[acetic acid]}
4.5 = 4.75 + log{[0.10 M]/[acetic acid]}
-0.25 = log{[0.10 M]/[acetic acid]}
[Acetic acid] = 0.10 M/ 10-0.25
[Acetic acid] = 0.10 M/0.56
[Acetic acid] = 0.1786 M
Moles of sodium acetate dissolved in 250 mL buffer solution = 0.10 M× (250mL/1000mL) × 1L
= 0.025 mol
Weight (w) of sodium acetate (purity 100%) dissolved to prepare 250 mL of solution with buffer concentration of 0.10 M is calculate as follow;
w100% = 0.025 mol × 82.0343 g/mol = 2.051 g
Weight (w) of sodium acetate (purity 99%) is calculate as follow;
w99% = 2.051 g× (100/99) = 2.072 g
What was the volume of 6.12 M acetic acid HC2H3O2 needed to prepare the 250 mL acetic acid/acetate ion buffer solution required in this part? Show your calculations.
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