Answer:
a. [tex]2.818 \times 10^{-3} mol\ NaOH[/tex]
b. [tex]1.409 \times 10^{-3} mol\ acid[/tex]
c. [tex]135.98 \frac{g}{mol}[/tex]
Explanation:
[tex]H_2X + 2NaOH \rightarrow Na_2X +2H_2O[/tex]
a. To calculate the number of moles of NaOH we need the following definition of molarity:
[tex]M= \frac{mol}{V}\\mol=M\times V= 0.1025 \frac{mol}{L}\times 27.5mL\times \frac{1L}{1000mL}= 2.818 \times 10^{-3} mol[/tex]
b. As we have a diprotic acid, it means that when the second equivalence point is reached, two moles of NaOH have reacted by one mol of acid, therefore the relation we use to determine the moles of acid is:
[tex]\frac{1\ mol\ acid}{2\ mol NaOH}\\2.818 \times 10^{-3} mol\ NaOH \times \frac{1\ mol\ acid}{2\ mol NaOH}= 1.409 \times 10^{-3} mol\ acid[/tex]
c. Finally, we determine the molar mass as follows:
[tex]molar \ mass=\frac{g\ acid}{mol\ acid}= \frac{0.1916\ g}{1,409\times 10^{-3} mol}= 135.98 \frac{g}{mol}[/tex]
