Question

for the coagulation of 10 ml of ferric hydroxide 2 ml of 1 molar potassium bromide is required what is the coagulation value of potassium bromide

Answer 1

Thanks for the question!

It is definitely a very interesting question to solve and do some brainstorming.

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Coagulation is the process which is exhibited by the precipitation of dispersed or colloidal particles which is the main cause for the neutralisation of the charges present on the either of the sides when the electrolytes are added to the provided solution. The colloids are some heterogeneous mixtures having dispersed particulate matters or phases of dispersion or fine particles into a colloidal media. After the addition of electrolytes the solution is thoroughly boiled (after entrapment of impurities present) this removes the electrolytes completely and the solution eventually loses the stability and performs the Coagulation say settling down at a bottom surface.

The Coagulation value is regarded through the presentation of millimoles of a specified electrolyte solution with must get added to form a 1 litre colloidal or dispersed particulate solution for a completely performed Coagulation.

Henceforth, we can put up, 2 ml of 1 molar potassium bromide or KBr =

$\frac{1}{1000} \times 2 = \frac{2}{1000} = 0.002 \: or \: 2 \: m \: moles \: or \: 2 \: milli \: moles.$

Now, 10 ml of a colloidal solution will require for the completion of ferric hydroxide which will be equal to;

Again for this case we have to apply the Coagulation or flocculation formula.

The Coagulation value or the flocculation value will be directly proportional to;

$\frac{1}{coagulation \: g \: power}$

So we can get, by this property, and implying the 1 litre of total colloidal solution;

that is, $\frac{1000}{10}$ × 2 millimoles

= 200 millimoles.

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Hope it helps and solves your query!!