Question

0.095 g of an unknown diprotic acid is titrated with 0.095 M NaOH. The first equivalence point occurred in the titration at a volume of 6.70 mL of NaOH added; the second equivalence point occurred at a volume of 13.40 mL of NaOH added. How many moles of NaOH were used to reach the first equivalence point in this diprotic acid titration

Answer 1

To determine the moles of NaOH used to reach the first equivalence point in the titration of a diprotic acid, multiply the molarity of the NaOH solution (0.095 M) by the volume of NaOH added at the first equivalence point (6.70 mL or 0.00670 L), yielding 0.0006365 mol of NaOH.

Step-by-step explanation:

The number of moles of NaOH used to reach the first equivalence point in a diprotic acid titration can be calculated using the concentration of the NaOH solution and the volume of NaOH added. In a diprotic acid titration, two moles of NaOH are required to neutralize one mole of the diprotic acid.

To find the number of moles of NaOH at the first equivalence point, you can use the formula:

moles of NaOH = molarity of NaOH × volume of NaOH in liters

Given that the concentration of NaOH is 0.095 M and the volume at the first equivalence point is 6.70 mL, the equation becomes:

moles of NaOH = 0.095 M × 0.00670 L

This calculation yields:

moles of NaOH = 0.0006365 mol

Answer 2

Step-by-step explanation:

The first equivalence point occurred in the titration at a volume of 6.70 mL of NaOH added . The molarity of NaOH is .095 M .

6.70 mL of NaOH = .00067 L of NaOH .

.00067 L of .095 M NaOH will contain .00067 x .095 moles of NaOH .

= 6.365 x 10⁻⁵ moles .

Moles of NaOH used to reach the first equivalence point = 6.365 x 10⁻⁵ moles.