Find the equivalent weight of Aluminium permanganate in acidic, neutral and basic medium.
Answer
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Hint:We know that equivalent weight means that the quantity of a substance which it exactly reacts with, or which is equal to the combining value of an arbitrarily fixed quantity of another substance in that particular reaction taken in consideration.
Complete step by step answer:
To find the equivalent weight the formula used is as follows –
Equivalent weight = Molecular mass of the compound / n – factor
Aluminium permanganate is $Al{\left( {Mn{O_4}} \right)_3}$.
Now in acidic medium,
$Mn$$O_4^ - $ $ + $ $8{H^ + }$ $ + $ $5{e^ - }$ $ \to $ $M{n^{2 + }}$ $ + $ $4{H_2}O$
We can see that there is a gain of $5{e^ - }$ .
Therefore now, for $Al{\left( {Mn{O_4}} \right)_3}$ we can find out the equivalent weight in acidic medium from the known formula –
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $\dfrac{{383.7}}{5}$
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $76.68$
Now in basic medium,
$Mn$$O_4^ - $ $ + $ ${e^\_}$ $ \to $ $Mn$$O_4^{2 - }$
We can notice that there is a loss of an electron, $1{e^ - }$ .
Therefore now, for $Al{\left( {Mn{O_4}} \right)_3}$ we can find out the equivalent weight in basic medium from the known formula –
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $\dfrac{{383.7}}{1}$
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $383.7$
Now in neutral medium,
$Mn$$O_4^ - $ $ + $ $4{H^ + }$ $ + $ $3{e^ - }$ $ \to $ $Mn{O_2}$ $ + $ $2{H_2}O$
Therefore now, for $Al{\left( {Mn{O_4}} \right)_3}$ we can find out the equivalent weight in neutral medium from the known formula –
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $\dfrac{{383.7}}{3}$
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $127.9$
The equivalent weight of $Al{\left( {Mn{O_4}} \right)_3}$ in acidic medium is $76.68$ and in basic medium $383.7$
And in neutral medium, $127.9$
Note:
Basically in common words, the equivalent weight of a particular element or radical is equal to its own atomic weight or formula weight which is divided by the valence it assumes in the taken compounds. The unit of equivalent weight is known as the atomic mass unit or in short $amu$. We know that the amount of a substance in grams which is numerically equal to the equivalent weight is called the gram equivalent of the substance.
Complete step by step answer:
To find the equivalent weight the formula used is as follows –
Equivalent weight = Molecular mass of the compound / n – factor
Aluminium permanganate is $Al{\left( {Mn{O_4}} \right)_3}$.
Now in acidic medium,
$Mn$$O_4^ - $ $ + $ $8{H^ + }$ $ + $ $5{e^ - }$ $ \to $ $M{n^{2 + }}$ $ + $ $4{H_2}O$
We can see that there is a gain of $5{e^ - }$ .
Therefore now, for $Al{\left( {Mn{O_4}} \right)_3}$ we can find out the equivalent weight in acidic medium from the known formula –
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $\dfrac{{383.7}}{5}$
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $76.68$
Now in basic medium,
$Mn$$O_4^ - $ $ + $ ${e^\_}$ $ \to $ $Mn$$O_4^{2 - }$
We can notice that there is a loss of an electron, $1{e^ - }$ .
Therefore now, for $Al{\left( {Mn{O_4}} \right)_3}$ we can find out the equivalent weight in basic medium from the known formula –
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $\dfrac{{383.7}}{1}$
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $383.7$
Now in neutral medium,
$Mn$$O_4^ - $ $ + $ $4{H^ + }$ $ + $ $3{e^ - }$ $ \to $ $Mn{O_2}$ $ + $ $2{H_2}O$
Therefore now, for $Al{\left( {Mn{O_4}} \right)_3}$ we can find out the equivalent weight in neutral medium from the known formula –
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $\dfrac{{383.7}}{3}$
$Al{\left( {Mn{O_4}} \right)_3}$ $ = $ $127.9$
The equivalent weight of $Al{\left( {Mn{O_4}} \right)_3}$ in acidic medium is $76.68$ and in basic medium $383.7$
And in neutral medium, $127.9$
Note:
Basically in common words, the equivalent weight of a particular element or radical is equal to its own atomic weight or formula weight which is divided by the valence it assumes in the taken compounds. The unit of equivalent weight is known as the atomic mass unit or in short $amu$. We know that the amount of a substance in grams which is numerically equal to the equivalent weight is called the gram equivalent of the substance.
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