An aqueous solution of prop ionic acid, CH3CH2COOH, is prepared in the lab by di

Question

An aqueous solution of prop ionic acid, CH3CH2COOH, is prepared in the lab by dissolving 3.6 g of the acid in a liter of solution. In a titration of this acid with an aqueous solution of NaOH, 25.00mL of the acid required 12.15 mL of a 0.100 M NaOH solution for complete neutralization using phenolphthalein as an indicator.
What is the mole of NaOH required to completely react with the 25.00mL aliquot of the acid? An aqueous solution of prop ionic acid, CH3CH2COOH, is prepared in the lab by dissolving 3.6 g of the acid in a liter of solution. In a titration of this acid with an aqueous solution of NaOH, 25.00mL of the acid required 12.15 mL of a 0.100 M NaOH solution for complete neutralization using phenolphthalein as an indicator.
What is the mole of NaOH required to completely react with the 25.00mL aliquot of the acid?
What is the mole of NaOH required to completely react with the 25.00mL aliquot of the acid?

Explanation / Answer

mol of NaOH = MV = 12.15*0.1 = 1.215 mmol of NaOH

mol of NaOH = 1.215*10^-3 mol of NaOH needed

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