Question
Question: Out of \[{\text{BaC}}{{\text{l}}_{\text{2}}}\] and \[{\text{ KCl}}\] , which one is more effective i...
Out of BaCl2 and KCl , which one is more effective in causing coagulation of a negatively charged colloidal sol? Give reason.
Solution
Use the Hardy-Schulze law. Coagulation of a negatively charged colloidal sol depends on the valency of positively charged ions. The greater the valency of cations greater is the power of coagulation of a negatively charged colloidal sol.
Complete answer:
The process of precipitating a colloidal solution by electrolyte is known as coagulation. The coagulation power of different electrolytes is different and depends on the valency of the effective ions. This phenomenon is known as Hardy-Schulze law.
Hardy-Schulze law states that the ion which is effective in causing a coagulation of a sol is the one whose charge is of the opposite sign of that of the colloidal particle. It also states that the precipitating effect increases with an increase in valency of the ion.
Here colloidal particles are negatively charged so positively charged ions are effective in causing coagulation.
The two salts given to us are BaCl2 and KCl. The valency of barium ion is +2 and the valency of potassium ion is +1. As Ba2 + ion has greater valency than K + ion so according to Hardy-Schulze law BaCl2 is more effective in causing a coagulation of a negatively charged colloidal sol than KCl.
Thus, BaCl2 is more effective in causing a coagulation of a negatively charged colloidal sol.
Note: Coagulation power of similar charged ions is the same and varies with a charge on the ion. The greater the valency of active ions, the greater is the precipitation action. Barium is IIA group metal so has valency +2. Potassium is an IA group metal so has valency +1. So, the coagulation power of BaCl2> KCl.