Question

Calculate the equivalent weight of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in the following reaction ?
${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{ + 3S}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ }} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + C}}{{\text{r}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$

Answer 1

For an acid, the equivalent weight is the ratio of the molecular weight to its basicity. For an oxidizing agent or a reducing agent, the equivalent weight is the ratio of the molecular weight to change in the oxidation number.

Complete answer:
In presence of sulphuric acid, sodium dichromate reacts with sulphur dioxide to form sodium sulphate, chromium sulphate and water. Write a balanced chemical equation for the above reaction.
${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{ + 3S}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ }} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + C}}{{\text{r}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
In the above reaction, the oxidation number of sulphur increases from +4 (in sulphur dioxide gas) to +6 (in chromium(III) sulphate). Increase in the oxidation number is oxidation. Thus, sulphur dioxide is oxidized during this reaction.
The oxidation number of chromium decreases from +6 (in sodium dichromate) to +3 (in chromium(III) sulphate). Decrease in the oxidation number is reduction. Thus, sodium dichromate is reduced during the reaction.
Thus, it is a redox reaction involving simultaneous oxidation and reduction half reactions.
During this reaction, there is no change in the oxidation number of sulphur present in sulphuric acid. This is because the oxidation number of sulphur in sulphuric acid is +6. The oxidation number of sulphur in sodium sulphate is also +6. The oxidation number of sulphur in chromium(III) sulphate is also +6. Thus, during the reaction, sulphuric acid neither undergoes reduction nor it undergoes oxidation. However, sulphuric acid acts as an acid as it loses two protons.
${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ }} \to 2{{\text{H}}^ + }{\text{ + SO}}_4^{2 - }$
Thus, the basicity of sulphuric acid is 2. In other words, sulphuric acid is dibasic acid.
You can calculate the equivalent weight of an acid by dividing its molecular weight with basicity.
${\text{Equivalent weight = }}\dfrac{{{\text{Molecular weight}}}}{{{\text{Basicity}}}}$
The atomic weights (in grams per mole) of hydrogen, sulphur and oxygen are 1, 32 and 16 respectively.
Calculate the molecular weight of sulphuric acid ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$
$2\left( 1 \right) + 32 + 4\left( {16} \right) = 2 + 32 + 64 = 98$
Calculate the equivalent weight of sulphuric acid in the reaction with sodium dichromate and sulphur dioxide.

$${\text{Equivalent weight = }}\dfrac{{{\text{Molecular weight}}}}{{{\text{Basicity}}}} \\\ \Rightarrow {\text{Equivalent weight = }}\dfrac{{98}}{2} \\\ \Rightarrow {\text{Equivalent weight = 49}}$$

**Hence, the equivalent weight of sulphuric acid in the reaction with sodium dichromate and sulphur dioxide is 49.

Note:**
The equivalent weight of a substance is not only the characteristic of that substance but also depends on the reaction. For example, consider following two reactions.

$${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + NaOH}} \to {\text{NaHS}}{{\text{O}}_{\text{4}}}{\text{ + }}{{\text{H}}_2}{\text{O}} \\\ {{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 2NaOH}} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 2}}{{\text{H}}_2}{\text{O}}$$

The equivalent weight of sulphuric acid in the first reaction is 98 and in the second reaction is 49.