Question

For the coagulation of $100ml$ of arsenious sulphide solution,$5ml$ of $1M$ $NaCl$ is required. What is the coagulating power of $NaCl$ ?

Answer 1

For solving this question, first let’s understand the meaning of coagulation power. The productive ions of an electrolyte when brought into coagulation are the one which carry charge opposite to that of the colloidal particles. Such ions are called coagulating ions.

Complete step by step answer:
As we know, the basic meaning of coagulation ions. Now, let’s understand the property of the coagulating power. Greater is that the valency of the coagulating or the flocculating ion; greater is its power to cause the coagulation.
Also, we need to understand the meaning of a colloidal solution to solve the above question.
A colloid is a tiny or small particle that is spread out uniformly all through by any other substance. So, a colloidal solutions or colloidal suspension is a mixture in which the substances are regularly suspended in a fluid. Every colloid has two parts i.e, colloidal particles and dispersing medium. Also, we need to remember that the concentration of a colloidal solution is measured in millimoles per litre.
After understanding let’s move on for the solution of the question.
Firstly, we need to calculate the number of millimoles of an electrolyte that must be added to 1 liter of a solution for the process of coagulation.
As per the question, for preparing 100mL of a solution 5mL of 1M of $NaCl$ is required.
Thus, 1000mL of a solution will require
$\dfrac{5}{{100}} \times 1000 = 50{\text{ }}mL$
Now, the total volume of the system after addition of 50mL of 1M $NaCl$ solution will become (1000 + 50) mL = 1050 mL
For calculating the new molarity of $NaCl$ through dilution the equation will be:
Formula used:
${M_i}{V_i} = {M_f}{V_f}$
Where,
$M_i$=Initial molarity
$V_i$ =Initial volume
$M_f$ =Final molarity
$V_f$ =Final volume
Now, let’s substitute the values in the above formula
$1 \times 50 = {M_f} \times 1050$
${M_f}$ = $\dfrac{{1 \times 50}}{{1050}}$
On simplification we get,
${M_f}$ = $\dfrac{5}{{105}}\;mol/L$
Now, the concentration will be in millimoles per litre
=$\dfrac{5}{{105}} \times 1000{\text{ }} = {\text{ }}47.6{\text{ }}mol/L$
$\therefore$ The Concentration will be in millimoles per litre $47.6{\text{ }}mol/L$

Note: We need to know that the arsenious sulphide is prepared by the hydrolysis of arsenious oxide then hydrogen sulphide gas is passed through the solution. We can write the chemical equation for this chemical reaction as,
$A{s_2}{O_3} + {H_2}O \to 2As{\left( {OH} \right)_3}$
$2As{\left( {OH} \right)_3} + 3{H_2}S \to A{s_2}{S_3} + 6{H_2}O$
We must know that the particle in arsenious sulphide colloidal solution is surrounded by the HS ions which are formed due to the dissociation of ${H_2}S$ . The sulphide ion layer is surrounded by ${H^ + }$ ions.