Answer:
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
I hope it helps!
