In this experiment, you will take a 25.00 mL aliquot of vinegar and dilute it to 250.0 mL. You will then take a 25.00 mL aliquot from this diluted vinegar solution and titrate it against the standardised sodium hydroxide. The titration between acetic acid and sodium hydroxide is a 1:1 stoichiometry. If your standardised sodium hydroxide solution was determined to be 0.051 M, and it required an average titre (titration volume) of 25.3 mL, what is the concentration (in M) of the undiluted vinegar sample (the initial vinegar sample)? Provide your answer to 2 significant figures.
Question
In this experiment, you will take a 25.00 mL aliquot of vinegar and dilute it to 250.0 mL. You will then take a 25.00 mL aliquot from this diluted vinegar solution and titrate it against the standardised sodium hydroxide. The titration between acetic acid and sodium hydroxide is a 1:1 stoichiometry. If your standardised sodium hydroxide solution was determined to be 0.051 M, and it required an average titre (titration volume) of 25.3 mL, what is the concentration (in M) of the undiluted vinegar sample (the initial vinegar sample)? Provide your answer to 2 significant figures.
Solution
To solve this problem, we need to first calculate the moles of sodium hydroxide used in the titration.
- Calculate moles of NaOH: Molarity (M) = moles/volume(L) So, moles = Molarity * volume = 0.051 M * 25.3 mL (converted to liters by dividing by 1000) = 0.051 M * 0.0253 L = 0.00129 moles of NaOH
Since the stoichiometry of the reaction between acetic acid and sodium hydroxide is 1:1, this means that the moles of acetic acid in the 25.00 mL aliquot of the diluted vinegar solution is also 0.00129 moles.
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Calculate concentration of acetic acid in the diluted vinegar solution: Molarity = moles/volume = 0.00129 moles / 0.025 L = 0.0516 M
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Calculate concentration of acetic acid in the undiluted vinegar: The 25.00 mL aliquot of vinegar was diluted to 250.0 mL, which is a dilution factor of 10. Therefore, the concentration of the acetic acid in the undiluted vinegar is 10 times the concentration in the diluted vinegar. = 0.0516 M * 10 = 0.516 M
So, the concentration of the undiluted vinegar sample is 0.52 M (rounded to 2 significant figures).
Similar Questions
You complete your titrations and find that the concentration of acetic acid in the diluted vinegar solution is 0.023 M. If your dilution step involved taking 25.00 mL of the orginal vinegar solution and diluting this to 250.00 mL with water to create your diluted solution for titration, calculate the concentration of your original vinegar solution.Question 4Answera.0.57b.2.30c.0.00023d.0.23e.0.0023
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