According to the experiment
procedure, given that the NaOH
solution has a concentration of
0.100M, and that the vinegar
(acetic acid) has a concentration of
5.0%m/m (mass percent), what
volume(mL) of titrant would be
required to reach the "equivalence
point", if you place 2.500g of
vinegar in the flask (show
calculations) ?
I know this question was posted before but I don't understand how you go about a question like this can you explain?
Certainly! To calculate the volume of titrant required to reach the equivalence point, we need to first determine the moles of acetic acid in the vinegar and then use stoichiometry to find the volume of NaOH solution needed to neutralize it.
Here's a step-by-step explanation of the calculations:
Step 1: Calculate the moles of acetic acid.
Given that the concentration of acetic acid (vinegar) is 5.0% m/m, we can assume that 100g of vinegar contains 5.0g of acetic acid. We have 2.500g of vinegar, so we can calculate the moles of acetic acid using the molar mass of acetic acid:
Molar mass of acetic acid (CH3COOH) = 12.01g/mol (C) + 1.01g/mol (H) + 12.01g/mol (C) + 16.00g/mol (O) + 1.01g/mol (H) = 60.05g/mol
Moles of acetic acid = Mass of acetic acid / Molar mass of acetic acid
Moles of acetic acid = 2.500g / 60.05g/mol
Step 2: Determine the stoichiometry of the reaction.
The balanced chemical equation for the neutralization reaction between acetic acid and sodium hydroxide can be written as:
CH3COOH + NaOH -> CH3COONa + H2O
From the equation, you can see that the stoichiometry between acetic acid and sodium hydroxide is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.
Step 3: Calculate the volume of titrant required.
Now that we have the moles of acetic acid, we can use the concentration of the NaOH solution to calculate the volume of NaOH solution needed to neutralize the acetic acid.
Moles of NaOH = Concentration of NaOH (in mol/L) * Volume of NaOH (in L)
Since we are looking for the volume of NaOH in mL, we can rearrange the formula:
Volume of NaOH (in mL) = Moles of NaOH * (1000 mL / Concentration of NaOH)
Using the molarity of NaOH, which is 0.100 M, and the moles of acetic acid we calculated earlier:
Moles of NaOH = Moles of acetic acid
Finally, substituting everything into the equation:
Volume of NaOH (in mL) = (Moles of acetic acid) * (1000 mL / Concentration of NaOH)
Plugging in the values, the equation becomes:
Volume of NaOH = (2.500g / 60.05g/mol) * (1000 mL / 0.100 mol/L)
Now you can calculate the final volume of NaOH solution required to reach the equivalence point.