Question

Given,0.98g of the metal sulphate was dissolved in water and excess of barium chloride was added. The precipitated barium sulphate weighted 0.95g. Calculate the equivalent weight of the metal.

Answer 1

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

Complete answer:
In the question, it is given that 0.98g of metal sulphate was dissolved in water and excess barium chloride was added. We know that 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-
     \[\dfrac{{{E}_{Metal}}+{{E}_{S{{O}_{4}}^{2-}}}}{{{E}_{B{{a}^{2+}}}}+{{E}_{C{{l}^{-}}}}}=\dfrac{\text{Mass }of\text{ }metal\text{ }sulphate}{\text{Mass }of\text{ }barium\text{ }sulphate}\]
Here, we have used E to denote 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.
We know that, Molar mass of sulphur is 32 and molar mass of oxygen is 16 so, we can calculate the molar mass of sulphate which will come out to be 96.
Therefore, we can write the equivalent weight of sulphate ion as-\[{{E}_{S{{O}_{4}}^{2-}}}=\dfrac{molar\text{ mass of sulphate}}{2}=\dfrac{96}{2}=48g/eq\]
Similarly,\[{{E}_{C{{l}^{-}}}}=\dfrac{molar\text{ }mass\text{ }of\text{ }chlorine}{1}=35.5g/eq\]
And\[{{E}_{B{{a}^{2+}}}}=\dfrac{molar\text{ }mass\text{ }of\text{ }barium}{2}=\dfrac{137.327}{2}=68.66g/eq\]
Now, we will put these calculated values in the above equation to find the equivalent weight of the metal
     \[\dfrac{{{E}_{Metal}}+48}{68.66+35.5}=\dfrac{0.98}{0.95}\]
We can solve the above equation and find out the equivalent weight of the metal which will come out to be 59.4g/eq.
Therefore, the equivalent weight of the metal is 59.4 gram/equivalent.

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