Question

Monosodium salt of oxalic acid is titrated with NaOH solution. In the second titration it is titrated with
KMnO4 solution. The ratio of equivalent weights of monosodiumoxalate is
1:1
1:2
2:1
3:2​

Answer 1

Answer:

The equivalent weight of oxalic acid in the titration with NaOH is equal to its molecular weight and in the titration with KMnO4 it is equal to half of its molecular weight. So, the ratio of equivalent weights of monosodium oxalate in both the titrations will be 2.

Explanation:

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Answer 2

The correct answer is option (c). 2:1.

Given:

Given that the monosodium salt of oxalic acid is first titrated against NaOH solution and then against KMnO₄ solution.

To Find:

We have to find the ratio of equivalent weights of monosodium oxalate in NaOH and KMnO₄.

Solution:

The reaction equation of monosodium salt of oxalic acid (NaHC₂O₄) with NaOH is given below.

$NaHC_2O_4 + NaOH$ → $Na_2C_2O_4 + H_2O$.

This reaction involves only one ionizable hydrogen. Therefore, the valency factor for the above reaction is 1.

The equivalent weight ($E_1$) of this reaction is given by,

$E_1 = \frac{Molar mass}{Valency factor} = \frac{M}{1}$.

The reaction equation of monosodium salt of oxalic acid (NaHC₂O₄) with KMnO₄ is as follows.

$NaHC_2O_4 + KMnO_4$ → $2CO_2$.

Here, the carbon in NaHC₂O₄ has $+6$ charge and that in CO₂ ha $+8$ charge. So the valency factor for the above reaction is 2.

The equivalent weight ($E_2$) of this reaction is given by,

$E_2 = \frac{Molar mass}{Valency factor} = \frac{M}{2}$.

∴, $\frac{E_1}{E_2} = \frac{\frac{M}{1} }{\frac{M}{2} } = \frac{2}{1}$.

$i.e., E_1 : E_2 = 2:1.$

Hence, the ratio of equivalent weights of monosodium oxalate in NaOH and KMnO₄ is 2:1.

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