JEE MAIN - Chemistry (2016 - 10th April Morning Slot - No. 20)
The volume of 0.1N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of OH− in aqueous solution is :
200 mL
400 mL
600 mL
800 mL
Explanation
According to law of equivalence,
Equivalence of acid = Equivalence of base.
Equivalence of acid = Normality x volume = 0.1 $$ \times $$ v
As we know base produce OH$$-$$ ion, so moles of base is same as moles of OH$$-$$ ion = 0.04
Another formula of equivalence = n factor $$ \times $$ number of moles
$$\therefore\,\,\,$$ Equivalance of base = n factor of OH$$-$$ $$ \times $$ moles of OH$$-$$ = 1 $$ \times $$ 0.04
As for any ion, the charge of that ion is the n factor of that ion. Here OH$$-$$ has 1 negative charge so it's n factor = 1
$$\therefore\,\,\,$$ 0.1 $$ \times $$ v = 1 $$ \times $$ 0.04
$$ \Rightarrow $$$$\,\,\,$$ v = 0.4 L
= 0.4 $$ \times $$ 1000
= 400 ml.
Equivalence of acid = Equivalence of base.
Equivalence of acid = Normality x volume = 0.1 $$ \times $$ v
As we know base produce OH$$-$$ ion, so moles of base is same as moles of OH$$-$$ ion = 0.04
Another formula of equivalence = n factor $$ \times $$ number of moles
$$\therefore\,\,\,$$ Equivalance of base = n factor of OH$$-$$ $$ \times $$ moles of OH$$-$$ = 1 $$ \times $$ 0.04
As for any ion, the charge of that ion is the n factor of that ion. Here OH$$-$$ has 1 negative charge so it's n factor = 1
$$\therefore\,\,\,$$ 0.1 $$ \times $$ v = 1 $$ \times $$ 0.04
$$ \Rightarrow $$$$\,\,\,$$ v = 0.4 L
= 0.4 $$ \times $$ 1000
= 400 ml.
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