Question

$0.98\text{ }gm$ of the metal sulphate was dissolved in water and an excess of barium chloride was added. The precipitated barium sulphate weighed $0.95\text{ }gm$. Calculate the equivalent weight of the metal.

Answer 1

Hint: We can calculate the equivalent weight of the metal by using the law of equivalent proportions according to which the ratio of the mass of the metal and barium sulphate will be equal to the ratio of equivalent weight of the dissociated ions in the solution.

Complete Solution :
Given that,
$0.98\text{ }gm$ of the metal sulphate was dissolved in water and an excess of barium chloride was added. And barium reacts with the metal sulphate and forms precipitate of barium sulphate.
As excess barium chloride was added, we can say that all the sulphate of the metal was used up in the formation of the precipitate. Therefore, according to the law of equivalent proportions we can write that:
$\dfrac{{{E}_{metal}}+{{E}_{S{{O}_{4}}^{2-}}}}{{{E}_{B{{a}^{2+}}}}+{{E}_{C{{l}^{-}}}}}=\dfrac{\text{Mass of metal sulphate}}{\text{Mass of barium sulphate}}$
Here, E represents the equivalent weight of the metal, barium and the respective ions in the solution.
We know that, equivalent weight is the molar mass of the substance divided by the number of protons lost or gained i.e. the charge.

The molar mass of sulphur is $32$ and the molar mass of oxygen is $16$. So, the molar mass of sulphate which will be $96$. And thus, the equivalent weight of the sulphate ion will be:
${{E}_{S{{O}_{4}}^{2-}}}=\dfrac{\text{molar mass of sulphate}}{2}=\dfrac{96}{2}=48\text{ g/eq}$

Similarly, ${{E}_{C{{l}^{-}}}}=\dfrac{\text{molar mass of chlorine}}{1}=35.5\text{ g/eq}$

${{E}_{B{{a}^{2+}}}}=\dfrac{\text{molar mass of barium}}{1}=\dfrac{137.327}{2}\text{=68}\text{.66 g/eq}$

Now, putting these values we can get the equivalent weight of the metal as:
$\dfrac{{{E}_{Metal}}+48}{68.66 + 35.5}=\dfrac{0.98}{0.95}$
Then, ${{E}_{metal}}=59.4\text{ g/eq}$
Hence, the equivalent weight of the metal is $59.4\text{ gram/equivalent}$.

Note: Different formulas can be used to find out the equivalent weight of the required compound depending upon the reaction. If the reaction is a redox reaction, the equivalent weight will be the formula weight divided by the change in oxidation number, whereas for acid and base it is the molecular weight divided by the basicity of acid and the acidity of the base respectively.