Investigation 4 Quiz Name: Show all calculations and report all final values wit

Question

Investigation 4 Quiz Name: Show all calculations and report all final values with the correct significant figures. Given the data below. Determine the molarity of acetic acid in the vinegar solution. Molarity of Base (NaOH): Volume of Acetic Acid used:10.00 mL Complete the Balanced Equation for the Neutralization Reaction (Eqn. 4.3, pg.61) 0.4232 M CH3COOH (aq) + NaOH (aq) Titration Data Trial 1 Show Set-up and Calculations 21.54 Final Buret Reading (mL) Initial Buret Reading (mL) Volume of Base Used (mL) 1.23 Moles of Base Used Moles of Acid in Sample Molarity of Acid (M) Shade of pink Faint, perfect Using the Molarity (mole/L), determine the m/v % (g/mL * 100) of acetic acid in the vinegar souti

Explanation / Answer

Answer

MV of NaOH = MV of CH3COOH = 0.4232 (molarity of NaOH)*vol of burette(20.31 ml NaOH) =molarity of acetic acid *vol of acetic acid

That is equal to 0.860 mole/L of acetic acid = 51.6 g/L = 5.16 g/ml percent.

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