Question

$25.0mL$ of an unknown diprotic acid is neutralized by titration with $5.00mL$ of $0.100MNaOH$. What is the molar concentration of the diprotic acid?

• A. $0.0050M$
• B. $0.010M$
• C. $0.015M$
• D. $0.020M$

Answer 1

The correct option is B $0.010M$

Let the diprotic acid be $H_{2}A$

$1$ $mole$ of $H_{2}A$ gives $2$ $moles$ of $H^{+}$.

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

$⟹$ Moles of $H^{+}=2\times volume\times molarity$

$⟹$ Moles of $H^{+}=2\times 25\times {10}^{-3}\times Concentration$

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

$⟹$ Moles of $O{H}^{-}=5\times {10}^{-3}\times 0.1$

$⟹$ Moles of $O{H}^{-}=5\times {10}^{-4}$

On neutralization,

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

$\therefore Concentration\times 50\times {10}^{-3}=5\times {10}^{-4}$

$\therefore Concentration={10}^{-2}$ $M=0.01$ $M$