How much mL of 0.2 M NaOH are needed to titrate vinegar while doing a titration with Phenolphthalein?
To figure out how much mL of 0.2 M NaOH are needed to titrate vinegar using phenolphthalein, you will need to know the concentration of the vinegar (acetic acid) you are titrating.
Here's the step-by-step process:
1. Write the balanced chemical equation for the reaction between acetic acid (vinegar) and sodium hydroxide:
CH3COOH (acetic acid) + NaOH (sodium hydroxide) → CH3COONa (sodium acetate) + H2O (water)
2. Determine the stoichiometry of the reaction. In this case, the stoichiometric ratio between acetic acid and sodium hydroxide is 1:1.
3. Calculate the number of moles of acetic acid present in the vinegar using its concentration and volume. For example, if you have 50 mL of vinegar with a concentration of 0.5 M, the calculation would be:
Moles of acetic acid = concentration × volume
Moles of acetic acid = 0.5 M × 0.050 L = 0.025 moles
4. Since the stoichiometric ratio is 1:1, the number of moles of sodium hydroxide required to completely react with the acetic acid is also 0.025 moles.
5. Now, calculate the volume of 0.2 M NaOH required to reach the equivalence point using the following equation:
Volume of NaOH (mL) = moles of NaOH ÷ concentration of NaOH
Substituting the values, we have:
Volume of NaOH (mL) = 0.025 moles ÷ 0.2 M = 0.125 L (125 mL)
So, 125 mL of 0.2 M NaOH would be needed to titrate 50 mL of vinegar with a concentration of 0.5 M using phenolphthalein.