calculate the molarity of acetic acid in a vinegar sample,knowing that 5.00ml of vinegar requires 43.50ml of 0.105 M NaOH to just reach the phenolphthalein endpoint in a titration
As from the equation the ratio is 1:1
Use M1V1=M2V2 (i.e. working in mmoles)
thus
43.50 ml x 0.105 M (NaOH) = 5.00 ml x X
where X is the molarity of the vinegar and taek care with the sig figs.
0.05
To calculate the molarity of acetic acid in the vinegar sample, we'll use the equation and the volume of NaOH required in the titration.
The balanced chemical equation for the reaction is:
CH3COOH (acetic acid) + NaOH (sodium hydroxide) → CH3COONa (sodium acetate) + H2O (water)
In the given titration, the volume of NaOH required to reach the phenolphthalein endpoint is 43.50 mL. Since the reaction has a 1:1 stoichiometric ratio between acetic acid and sodium hydroxide, we can say that the number of moles of acetic acid is equal to the number of moles of NaOH used in the titration.
First, we convert the volume of NaOH used from milliliters to liters:
43.50 mL = 43.50 / 1000 = 0.04350 L
Next, the number of moles of NaOH can be calculated using the molarity and volume of NaOH:
moles of NaOH = Molarity × Volume
moles of NaOH = 0.105 M × 0.04350 L = 0.0045675 moles NaOH
Since the reaction has a 1:1 stoichiometric ratio between acetic acid and NaOH, the number of moles of acetic acid is also 0.0045675.
Now, we need to calculate the molarity of acetic acid in the vinegar sample. This can be determined using the following formula:
Molarity = Moles of solute / Volume of solution (in liters)
The volume of the vinegar sample is given as 5.00 mL. To convert it to liters:
5.00 mL = 5.00 / 1000 = 0.00500 L
Finally, we can calculate the molarity of acetic acid:
Molarity = 0.0045675 moles / 0.00500 L
Molarity ≈ 0.9135 M
Therefore, the molarity of acetic acid in the vinegar sample is approximately 0.9135 M.
To calculate the molarity of acetic acid in the vinegar sample, we can use the concept of stoichiometry in a titration.
Step 1: Write the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH --> CH3COONa + H2O
Step 2: Determine the molar ratio between acetic acid (CH3COOH) and sodium hydroxide (NaOH) based on the balanced equation. In this case, the ratio is 1:1, as both reactants have a coefficient of 1.
Step 3: Calculate the number of moles of NaOH used in the reaction. We can use the formula:
moles of NaOH = molarity of NaOH * volume of NaOH used
moles of NaOH = (0.105 M) * (43.50 mL) / 1000 mL/L
moles of NaOH = 0.0045675 mol
Step 4: Using the molar ratio from step 2, we can determine the number of moles of acetic acid that reacted. Since the ratio is 1:1, the moles of acetic acid are also 0.0045675 mol.
Step 5: Calculate the molarity of acetic acid in the vinegar sample.
Molarity = moles of acetic acid / volume of vinegar sample (L)
Since the volume of vinegar sample is given as 5.00 mL, we need to convert it to liters:
Volume = 5.00 mL / 1000 mL/L = 0.005 L
Molarity = 0.0045675 mol / 0.005 L
Molarity = 0.9135 M
Therefore, the molarity of acetic acid in the vinegar sample is 0.9135 M.