HC2H3O2 + NaOH -> NaC2H3O2 + H2O
At the equivalence point, all of the HC2H3O2 will have reacted with NaOH to form NaC2H3O2. Therefore, the number of moles of NaOH used at the equivalence point will be equal to the number of moles of HC2H3O2 initially present.
The number of moles of HC2H3O2 initially present is:
$$0.10 \text{ M} \times 0.040 \text{ L} = 4.0 \times 10^{-3} \text{ mol}$$
Therefore, the number of moles of NaOH used at the equivalence point is also 4.0 x 10-3 mol.
The volume of NaOH used at the equivalence point can be calculated using the following formula:
$$V = \frac{n}{C}$$
where:
* V is the volume in liters
* n is the number of moles
* C is the concentration in moles per liter
Substituting the values we know into the formula, we get:
$$V = \frac{4.0 \times 10^{-3} \text{ mol}}{0.15 \text{ M}}$$
= 0.0267 L
Therefore, the volume of NaOH used to reach the equivalence point is 0.0267 L or 26.7 mL.
At the equivalence point, the concentration of C2H3O2- can be calculated using the following formula:
$$[C2H3O2-] = \frac{n}{V}$$
where:
* [C2H3O2-] is the concentration of C2H3O2- in moles per liter
* n is the number of moles of C2H3O2-
* V is the volume in liters
At the equivalence point, the number of moles of C2H3O2- is equal to the number of moles of HC2H3O2 initially present, which is 4.0 x 10-3 mol. The volume at the equivalence point is the sum of the volumes of HC2H3O2 and NaOH used, which is 40.00 mL + 26.7 mL = 66.7 mL or 0.0667 L.
Substituting these values into the formula, we get:
$$[C2H3O2-] = \frac{4.0 \times 10^{-3} \text{ mol}}{0.0667 \text{ L}}$$
= 0.0600 M
Therefore, the concentration of C2H3O2- at the equivalence point is 0.0600 M.
