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
To calculate the molarity of acetic acid in a vinegar sample, you need to use the information provided about the titration.
First, write the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH → CH3COONa + H2O
From the equation, you can see that the ratio between acetic acid and sodium hydroxide is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.
Next, calculate the number of moles of NaOH used in the titration:
moles of NaOH = volume of NaOH (in L) × molarity of NaOH
Since the volume of NaOH used is given as 43.50 ml, you need to convert it to liters by dividing by 1000:
volume of NaOH = 43.50 ml / 1000 ml/L = 0.04350 L
Now you can calculate the moles of NaOH:
moles of NaOH = 0.04350 L × 0.105 M = 0.00457 moles
Since the stoichiometry of the reaction is 1:1 between acetic acid and NaOH, the moles of acetic acid in the vinegar sample would also be 0.00457 moles.
Finally, calculate the molarity of acetic acid in the vinegar sample:
molarity of acetic acid = moles of acetic acid / volume of vinegar sample (in L)
The volume of vinegar sample used in the titration is given as 5.00 ml, so you need to convert it to liters:
volume of vinegar sample = 5.00 ml / 1000 ml/L = 0.00500 L
Now you can calculate the molarity of acetic acid:
molarity of acetic acid = 0.00457 moles / 0.00500 L = 0.914 M
Therefore, the molarity of acetic acid in the vinegar sample is 0.914 M.