Question

In the standardization of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$using ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$. The equivalent weight of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$is:

Answer 1

Step 1: Finding the valence factor in the molecule of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$:

The ionic equation between ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ and ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is:

$4{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+3{\mathrm{S}}_2{\mathrm{O}}_3^{2-}+26{\mathrm{H}}^{+}\to 6{\mathrm{SO}}_4^{2-}+8{\mathrm{Cr}}^{3+}+13{\mathrm{H}}_2\mathrm{O}$

So, the oxidation number of $\mathrm{Cr}$ in ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ =$2\mathrm{x}+7(-2)=-2$

So, $\mathrm{x}=+6$

The oxidation number of $\mathrm{Cr}$ in ${\mathrm{Cr}}^{+3}$ $=+3$

The net change in oxidation number : 3

$$\begin{gathered} \mathrm{Valence}\mathrm{factor}=\mathrm{number}\mathrm{of}\mathrm{Cr}\mathrm{atom}\times \mathrm{Change}\mathrm{in}\mathrm{oxidation}\mathrm{state} \\ =2\times 3 \\ =6 \end{gathered}$$

Step 2: Finding the equivalent weight of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$:

$$\begin{gathered} \mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{valence}\mathrm{factor}} \\ \Rightarrow \mathrm{Equivalent}\mathrm{weight}\mathrm{of}{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7=\frac{\mathrm{Molecular}\mathrm{weight}}{6} \\ \Rightarrow \mathrm{Equivalent}\mathrm{weight}\mathrm{of}{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7=\frac{2\left(39\right)+2\left(52\right)+7\left(16\right)}{6} \\ \Rightarrow \mathrm{Equivalent}\mathrm{weight}\mathrm{of}{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7=\frac{294}{6} \\ \Rightarrow \mathrm{Equivalent}\mathrm{weight}\mathrm{of}{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7=49 \end{gathered}$$

Therefore, the equivalent weight of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is 49.