Using the determined equivalence point from question 2 and the balanced reaction of acetic acid and sodium hydroxide, calculate the molarity of the acetic acid in your hot sauce packet. Equivalance point is 3.0 mL NaOH, 0.6 g hot sauce, 0.1 M NaoH.

Molarity of the packet is 0.5M

Explanation:

In the reaction of acetic acid with NaOH:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

1 mole of acetic acid reacts with 1 mole of NaOH.

When you are titrating the acid with NaOH, you reach equivalence point when moles of acid = moles of NaOH.

Moles of NaOH are:

3.0mL = 3.0x10⁻³L ₓ (0.1 mol / L) = 3.0x10⁻⁴ moles of NaOH = moles of CH₃COOH.

Now, you find the moles of acetic acid in the hot sauce packet. But molarity is the ratio between moles of the acid and liters of solution.

As you don't know the volume of your packet, you can assume its density as 1g/mL. Thus, volume of 0.6g of hot sauce is 0.6mL = 6x10⁻⁴L.

And molarity of the packet is:

3.0x10⁻⁴ moles acetic acid / 6x10⁻⁴L =

the answer is definitely c.

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