A student found a 500 cm3 sample of vinegar. Knowing that vinegar is a solution of ethanoic acid (CH3COOH) and water, this student decided to take a 25.0 cm3 sample, dilute it with 25.0 cm3 of water and then react it with sodium hydroxide. The dilute sample required 37.0 cm3 of 0.850 mol dm–3 sodium hydroxide solution to react entirely with it via the following reaction: CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l)Calculate the concentration of ethanoic acid in the original 500 cm3 sample of vinegar.
Question
A student found a 500 cm3 sample of vinegar. Knowing that vinegar is a solution of ethanoic acid (CH3COOH) and water, this student decided to take a 25.0 cm3 sample, dilute it with 25.0 cm3 of water and then react it with sodium hydroxide. The dilute sample required 37.0 cm3 of 0.850 mol dm–3 sodium hydroxide solution to react entirely with it via the following reaction: CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l)Calculate the concentration of ethanoic acid in the original 500 cm3 sample of vinegar.
Solution
To calculate the concentration of ethanoic acid in the original 500 cm3 sample of vinegar, we first need to find the number of moles of sodium hydroxide that reacted with the diluted vinegar sample.
The number of moles of a substance can be calculated using the formula:
Moles = Concentration (mol/dm3) x Volume (dm3)
So, the number of moles of sodium hydroxide is:
Moles of NaOH = 0.850 mol/dm3 x 37.0 cm3 x (1 dm3/1000 cm3) = 0.03145 mol
From the balanced chemical equation, we know that one mole of sodium hydroxide reacts with one mole of ethanoic acid. Therefore, the number of moles of ethanoic acid in the 50.0 cm3 diluted sample is also 0.03145 mol.
Next, we need to find the concentration of ethanoic acid in the diluted sample. The concentration can be calculated using the formula:
Concentration = Moles / Volume
So, the concentration of ethanoic acid in the diluted sample is:
Concentration of CH3COOH = 0.03145 mol / 50.0 cm3 x (1 dm3/1000 cm3) = 0.629 mol/dm3
Finally, we need to find the concentration of ethanoic acid in the original 500 cm3 sample of vinegar. Since the volume of the vinegar sample was doubled when it was diluted, the concentration of ethanoic acid in the original sample is half of the concentration in the diluted sample. Therefore, the concentration of ethanoic acid in the original sample is:
Concentration of CH3COOH = 0.629 mol/dm3 / 2 = 0.315 mol/dm3
Similar Questions
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.
The pH of a 0.001 M aqueous solution of sodium hydroxide will be:
A chemistry student weighs out 0.137g of citric acid H3C6H5O7, a triprotic acid, into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1600M NaOH solution.Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Round your answer to 3 significant digits.
Sodium hydroxide solution (in water) reacts with hydrochloric acidsolution (in water) to produce sodium chloride solution and water
Mass of empty flask (grams)71.39171.42572.095Mass of flask + vinegar (grams)81.00381.37982.014Mass of vinegar (grams)Volume of vinegar (L)0.010000.010000.01000Concentration of NaOH (mol/L)0.20000.20000.2000Initial volume of NaOH (mL)0.600.290.20Final volume of NaOH (mL)41.6541.9441.53Volume of NaOH added (mL)Volume of NaOH added (L)Moles of NaOH used (mol)Moles of acetic acid (mol)Mass of acetic acid (grams)Mass percent of acetic acid
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