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Question

Design Layout References Mailings Review View Foxit PDF Design Layout anbcdai AaBbc Aarbcel Aab A six x, x' la-?·A-| | 15-la-B+ || 1 Normal!! No Spac Heading 1 Heading 2 Title . Subtitle Subtle Ern." Font Paragraph Styles 2. What is the equivalence point? Use your graph to determine this. 13 ml 3. Using the determined equivalence point from question 2 and the balanced reaction of acetic acid and sodium hydroxide, calculate the molarity of the acetic acid in your hot sauce packet eScience Labs, LLC 2015 Acid-Base Titrations

Explanation / Answer

Amount of NaOH used = 1 M, 13 mL

1 M = 1 mol / 1 L = 1 mol / 1000 mL

No. of moles of NaOH in 1M, 13 mL solution = 1 M x 13 mL/ 1000 mL = 0.013 mol.

The balance equation for the reaction

CH3COOH + NaOH ---> CH3COONa + H2O

1 mole of CH3COOH reacts with 1 mole of NaOH at equivalence point.

So 0.013 mol of NaOH reacts with 0.013 mol of CH3COOH

M.W. of CH3COOH = ?60.052 g/mol

Mass of CH3COOH of 0.013 mol of CH3COOH = 60.052 g/mol x 0.013 mol = 0.78 g

The mass of CH3COOH in 1 g of sauce = 0.78 g

Since Volume of sauce is not given molarity cannot be calculated, if the volume (V mL) is known the molarity can be calculated by the equation = (V /1000)x (0.78 g/ 60.052 g/mol)

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