
The equivalent weight of \[NaH{C_2}{O_4}\] in reaction with $NaOH$ is _______________
A. 112
B. 56
C. 224
D. 84
Answer
570.3k+ views
Hint:For the calculation of equivalent weight of a substance we first need to understand the concepts of equivalent weight. The equivalent weight of any chemical compound is equal to the ratio of its gram atomic weight to its valence.
Complete step by step answer:
First, we will write down the formula of equivalent weight of an acid and a base.
Equivalent weight of an acid = $\dfrac{{Molecular\;weight\;of\;acid}}{{Valency}}$
Equivalent weight of a base = $\dfrac{{Molecular\;weight\;of\;base}}{{Valency}}$
As it is mentioned in the question that \[NaH{C_2}{O_4}\] reacts with $NaOH$ then the chemical reaction will be:
$NaH{C_2}{O_4} + NaOH \to N{a_2}{C_2}{O_4} + {H_2}O$
From the reaction, we know that Sodium oxalate and water is formed as the product of the following reaction.
Therefore, the acidity or basicity of Sodium hydrogen oxalate is equal to one. So, to derive the equivalent weight of Sodium hydrogen oxalate we will use the formula:
Equivalent weight of Sodium hydrogen oxalate = $\dfrac{{molecular\;weight\;of\;NaH{C_2}{O_4}}}{{number\;of\;hydrogens\;transferred}}$
As molecular weight of Sodium hydrogen oxalate = $112$
And the number of hydrogen transferred in the chemical reaction = $1$
So, we will substitute the values in the above formula.
Equivalent weight of Sodium hydrogen oxalate = $\dfrac{{molecular\;weight\;of\;NaH{C_2}{O_4}}}{{number\;of\;hydrogens\;transferred}}$
Equivalent weight of Sodium hydrogen oxalate = $\dfrac{{112}}{1}$
Therefore, Equivalent weight of Sodium hydrogen oxalate =$112$.
Therefore, the option A is correct.
Note:
We must remember that the Sodium oxalate formed in the reaction acts as a reducing agent, and it may be used as a primary standard for standardizing potassium permanganate solutions. Also, remember that sodium oxalate is soluble in formic acid and insoluble in alcohol and ether.
Complete step by step answer:
First, we will write down the formula of equivalent weight of an acid and a base.
Equivalent weight of an acid = $\dfrac{{Molecular\;weight\;of\;acid}}{{Valency}}$
Equivalent weight of a base = $\dfrac{{Molecular\;weight\;of\;base}}{{Valency}}$
As it is mentioned in the question that \[NaH{C_2}{O_4}\] reacts with $NaOH$ then the chemical reaction will be:
$NaH{C_2}{O_4} + NaOH \to N{a_2}{C_2}{O_4} + {H_2}O$
From the reaction, we know that Sodium oxalate and water is formed as the product of the following reaction.
Therefore, the acidity or basicity of Sodium hydrogen oxalate is equal to one. So, to derive the equivalent weight of Sodium hydrogen oxalate we will use the formula:
Equivalent weight of Sodium hydrogen oxalate = $\dfrac{{molecular\;weight\;of\;NaH{C_2}{O_4}}}{{number\;of\;hydrogens\;transferred}}$
As molecular weight of Sodium hydrogen oxalate = $112$
And the number of hydrogen transferred in the chemical reaction = $1$
So, we will substitute the values in the above formula.
Equivalent weight of Sodium hydrogen oxalate = $\dfrac{{molecular\;weight\;of\;NaH{C_2}{O_4}}}{{number\;of\;hydrogens\;transferred}}$
Equivalent weight of Sodium hydrogen oxalate = $\dfrac{{112}}{1}$
Therefore, Equivalent weight of Sodium hydrogen oxalate =$112$.
Therefore, the option A is correct.
Note:
We must remember that the Sodium oxalate formed in the reaction acts as a reducing agent, and it may be used as a primary standard for standardizing potassium permanganate solutions. Also, remember that sodium oxalate is soluble in formic acid and insoluble in alcohol and ether.
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