0.105 g of an unknown diprotic acid is titrated with 0.120 m naoh. the first equivalence point occurred at a volume of 7.00 ml of naoh added; the second equivalence point occurred at a volume of 14.07 ml of naoh added. how many moles of naoh were used to reach the second equivalence point in this diprotic acid titration?

2,53x10⁻³ moles of NaOH

Explanation:

The reactions of a diprotic acid with NaOH are:

H₂X + NaOH → HX⁻ + H₂O + Na⁺

HX⁻ + NaOH → X²⁻ + H₂O + Na⁺

Where the complete first reaction gives the first equivalence point and the complete second reaction gives the second equivalence point.

The total volume spent of NaOH to reach the second equivalence point is:

7,00mL + 14,07mL = 21,07 mL = 0,02107L

As molar concentration of NaOH is 0,120M, the moles used to reach the second equivalence point are:

0,02107L×(0,120mol/L) = 2,53x10⁻³ moles of NaOH

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