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Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

100 mL of a ...

Question

$100 m L$ of a $0.25 M (B a (O H)_{2}$ solution was used to titrate $12.5 m L$ of a $H N O_{3}$ solution to neutralization. What was the concentration of the acid?

1.0 M

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2.0 M

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3.0 M

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4.0 M

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Solution

The correct option is D $4.0 M$
The acid solution is titration to neutralized.
Hence, at the equivalence point,
Moles of $O H^{-}$ ions $=$ Moles of $H^{+}$ ions. $(1)$
$1$ $m o l e$ of $B a (O H)_{2}$ gives $2$ $m o l e s$ of $O H^{-}$
$∴$ $M o l e s$ of $O H^{-} = 2 \times$ Moles of $B a (O H)_{2}$
$M o l e s$ of $O H^{-} = 2 \times [M o l a r i t y \times V o l u m e o f B a (O H)_{2}]$
$M o l e s$ of $O H^{-} = 0.05 (2)$
Moles of $H^{+}$ ion $= M o l e s$ of $H N O_{3}$
Moles of $H^{+}$ ion $= C o n c e n t r a t i o n \times V o l u m e$ of $H N O_{3}$
Moles of $H^{+}$ ion $= C o n c e n t r a t i o n \times 12.5 \times 10^{- 3}$
From $(1)$ and $(2)$ ,
$0.05 =$ (Concentration of acid) $\times 12.5 \times 10^{- 3}$
$∴$ Concentration of acid $= 4.0$ $M$

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100 mL of a 0.25 M (Ba(OH)2 solution was used to titrate 12.5 mL of a HNO3 solution to neutralization. What was the concentration of the acid?

$100 m L$ of a $0.25 M (B a (O H)_{2}$ solution was used to titrate $12.5 m L$ of a $H N O_{3}$ solution to neutralization. What was the concentration of the acid?