Question

The flocculation value of ${\text{HCl}}$ for arsenic sulphide sol. is ${\text{30mmol}}{{\text{L}}^{{\text{ - 1}}}}$ . If ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is used for the flocculation of arsenic sulphide, the amount, in grams, of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in 250 ml required for the above purpose is______. ( molecular mass of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ = 98g/mol}}$ )

Answer 1

Hint :(1) The flocculation value of an electrolyte or the precipitation value of an electrolyte is the minimum amount of the electrolyte in millimoles which must be added to one litre of a colloidal solution so as to bring about complete coagulation or precipitation.
(2) Using the normality and volume relation, the normality/ molarity of the sulphuric acid can be determined from the normality/ molarity of hydrochloric acid. This, in turn, will give the amount of sulphuric acid in grams.

Complete Step By Step Answer:
Given, the flocculation value of ${\text{HCl}}$ for arsenic sulphide sol ${\text{ = 30mmol}}{{\text{L}}^{{\text{ - 1}}}}$ . Also, the molecular mass of sulphuric acid is given to be 98g/mol.
We need to find the amount of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in grams in 250 ml for the flocculation process of the same sol.
From the given data, the number of millimoles of ${\text{HCl}}$ required for 1 litre $= 30$
Now, we need the number of millimoles of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in 1 litre.
We know, normality and molarity are units of concentration in chemistry. Molarity is the number of moles of solute present in one litre of solution and normality is the number of equivalents present in one litre of solution.
Also, basicity of an acid refers to the difficulty for an acid to react with bases which is determined by the number of replaceable hydrogen atoms in the acid.
The relation between normality and volume of a solution is:
${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}$
where ${{\text{N}}_{\text{1}}}$ is the normality of solution 1, ${{\text{N}}_{\text{2}}}$ is the normality of solution 2 and ${{\text{V}}_{\text{1}}}$ and ${{\text{V}}_2}$ are the volume of solutions 1 and 2 respectively.
Again, the relation between normality and molarity for acidic solutions:
$Normality= Basicity\times Molarity$
If ${{\text{M}}_{\text{1}}}$ and ${{\text{M}}_2}$ are the molarities of solutions 1 and 2 respectively, and ${{\text{n}}_{\text{1}}}$ and ${{\text{n}}_2}$ are the basicities of acids 1 and 2 respectively, then the overall relation between molarity and volume becomes:
$n_{1}\times M_{1}\times V_{1}= n_{2}\times M_{2}\times V_{2}$
For the given question, let solution 1 be ${\text{HCl}}$ and solution 2 be ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ . Also, ${{\text{V}}_{\text{1}}}{\text{ = }}{{\text{V}}_{\text{2}}}$ and the basicity of ${\text{HCl}}$ (solution1) is 1 and that of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ (solution2) is 2.
Therefore, we get:
$1\times 30\times 1= 2\times M_{2}\times 1 \Rightarrow M_{2}= \dfrac{30}{2} \Rightarrow M_{2}= 15$
Thus, the number of millimoles of sulphuric acid in 1 litre $= 15$ .
Therefore, number of millimoles of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in 250 ml
$= \dfrac{15\times 250}{1000} = \dfrac{15}{4} = 3.75$
Now, 1 mole of a molecule is equal to its molecular mass.
Therefore, ${\text{98g}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ = 1mole = 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{millimole}}$ .
Thus, $10^{^{-3}}millimoleH_{2}SO_{4}= 98g \Rightarrow 3.75millimoleH_{2}SO_{4}= \dfrac{98\times 3.75}{1000}g \Rightarrow 3.75millimoleH_{2}SO_{4}= 0.3675g$
∴ The amount of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in grams in 250 ml required for flocculation of arsenic sulphide is 0.3675g.

Note :
The method used for calculation of flocculation value involves the calculation of the amount of electrolyte in millimoles which is required to bring about complete coagulation of 1 litre of the colloidal solution. The effective ions of the electrolyte which brings about coagulation contain charge opposite to that of the colloidal particles. The coagulating power is inversely proportional to the flocculation value of the electrolyte and directly proportional to the valency of the flocculating ion.