Question

The following titration curve when $25.0mL$ of a diprotic acid was titrated with $0.100MNaOH$. What is the concentation of the acid at the point indicated by the red arrow?

• A. $0.021M$
• B. $0.041M$
• C. $0.082M$
• D. $0.16M$

Answer 1

The correct option is C $0.082M$

At second equivalence point, all of $H_2{A}^{-}$ converts to $A^{2-}$

Moles of $H^{+}=$ Moles of $O{H}^{-}$ At this point $e{q}^n\left(1\right)$

Total moles of $H^{+}=1\times$ Moles of $H_{2}A$

$⟹$ Total moles of $H^{+}=1\times 25\times {10}^{-3}\times$ (Concentration of $H_{2}A$)

Total moles of $O{H}^{-}=Concentration\times Volume$

$⟹$ Total moles of $O{H}^{-}=0.1\times 20\times {10}^{-3}$ [Volume of $NaOH$ obtained from graph]

From equation $\left(1\right)$,

$1\times 25\times {10}^{-3}\times$ (Concentration of $H_{2}A$) $=0.1\times 20\times {10}^{-3}$

$\therefore$ Concentration of acid $=\frac{2}{25}=0.08$ $M$

Since, the volume is nearly $20$ $mL$ and not exactly $20$ $mL$. The concentration is nearly $0.08$ $M$.