Question

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

• A. $0.022M$
• B. $0.042M$
• C. $0.084M$
• D. $0.16M$

Answer 1

Answer: (B)

$0.042M$

At the given point i.e., at first equivalence point, all of $H_{2}A$ converts to $H{A}^{-}$.

Moles of $H^{+}=$ Moles of $O{H}^{-}$

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

$⟹$ Moles of $H^{+}=25\times {10}^{-3}\times$ Concentration of acid

Moles of $O{H}^{-}=0.1\times 10\times {10}^{-3}$

$\therefore 25\times {10}^{-3}\times$ Concentration of acid $=0.1\times 10\times {10}^{-3}$

$\therefore$ Concentration of acid $=0.04$ $M$

In this solution, we have taken the volume of $NaOH$ to be equal to $10$ $mL$. But in the graph, the volume is not exactly $10$ $mL$ but slightly greater than $10$ $mL$.

Hence, concentration is slightly greater than $0.04$ $M$.