Question

I am struggling with this.

Using the average volume of NaOH, calculate the molar concentration of your vinegar solution.

The average volume of 0.10 NaOH is 7.524ml

There is 1ml of vinegar

10ml of distilled water

Answer 1

Answer 2

To calculate the molar concentration of the vinegar solution, we first need to determine the number of moles of NaOH used in the reaction with the vinegar.

The balanced chemical equation between NaOH and acetic acid (the main component of vinegar) is:

CH3COOH + NaOH → CH3COONa + H2O

From the equation, we can see that 1 mole of NaOH reacts with 1 mole of acetic acid (CH3COOH).

Given that the average volume of 0.10 M NaOH used is 7.524 mL, we can calculate the number of moles of NaOH as follows:

Number of moles of NaOH = Volume of NaOH (in L) * Molar concentration of NaOH Number of moles of NaOH = 7.524 mL * (1 L / 1000 mL) * 0.10 mol/L Number of moles of NaOH ≈ 0.0007524 mol

Since the stoichiometry between NaOH and acetic acid is 1:1, the number of moles of acetic acid (vinegar) is also approximately 0.0007524 mol.

Now, let's find the total volume of the vinegar solution:

Total volume of vinegar solution = Volume of vinegar + Volume of distilled water Total volume of vinegar solution = 1 mL + 10 mL = 11 mL

Finally, we can calculate the molar concentration of the vinegar solution (acetic acid) as follows:

Molar concentration of vinegar = Number of moles of vinegar / Total volume of vinegar solution Molar concentration of vinegar ≈ 0.0007524 mol / (11 mL * (1 L / 1000 mL)) Molar concentration of vinegar ≈ 0.0684 mol/L

So, the molar concentration of the vinegar solution is approximately 0.0684 mol/L.