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- why is the final answer 9.3 millimole instead of 9.27- what sig fig rule reduces the # of sig figs. thanksBy 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.fischer esterification: 5.4mL of isopentyl alcohol, molar mass 88.15g/mol: 8.5mL of acetic acid, molar mass 60.05 g/mol: Experimental final product(Isopentyl acetate) I got: 3.10g. 1.Calculate for theretical yield. 2.Percentage yield 3. % recovery.