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If you titrate 35 drops of vinegar to the equivalence point with 43 drops of 0.600M sodium hydroxide what is the concentration of the vinegar?

User GameZelda
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To determine the concentration of vinegar (acetic acid) based on the titration with sodium hydroxide, we need to use the concept of stoichiometry and the volume and concentration of the sodium hydroxide solution.

Let's assume that each drop of the sodium hydroxide solution is approximately 0.03 mL. Based on this assumption, the volume of sodium hydroxide solution used would be:

Volume of NaOH = 43 drops × 0.03 mL/drop = 1.29 mL

Now, we need to determine the moles of sodium hydroxide used in the titration. Using the volume and concentration of the sodium hydroxide solution:

Moles of NaOH = Volume of NaOH (L) × Concentration of NaOH (mol/L)

= 1.29 mL × (1 L / 1000 mL) × 0.600 mol/L

= 0.000774 mol

Since acetic acid (CH3COOH) and sodium hydroxide (NaOH) react in a 1:1 molar ratio, the moles of acetic acid present in the vinegar can be determined as well:

Moles of CH3COOH = Moles of NaOH = 0.000774 mol

Finally, we need to calculate the concentration of acetic acid (vinegar) in the solution. Assuming the volume of vinegar is 1 L:

Concentration of CH3COOH = Moles of CH3COOH / Volume of CH3COOH

= 0.000774 mol / 1 L

= 0.000774 M

Therefore, the concentration of the vinegar (acetic acid) is approximately 0.000774 M.

User Fabian Sievert
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