Hardy-Schulze rule explains the effect of electrolytes on the coagulation of colloidal solution. According to this rule, coagulation power of cations follow the order

Question

Hardy-Schulze rule explains the effect of electrolytes on the coagulation of colloidal solution. According to this rule, coagulation power of cations follow the order (A) Ba+2 > Na+ > Al+3 (B) Al+3 > Na+ > Ba+2 (C) Al+3 > Ba+2 > Na+ (D) Ba+2 > Al+3 > Na+

Tardigrade Admin · Accepted Answer

According to this law the coagulating effect of an ion on dispersed phase of opposite charge increases with the valency of the ion. The precipitating power of Al3+ , Ba++, Na+ ions is in order Al3+ > Ba2+ > Na+.

Q. Hardy-Schulze rule explains the effect of electrolytes on the coagulation of colloidal solution. According to this rule, coagulation power of cations follow the order

$B a^{+ 2} > N a^{+} > A l^{+ 3}$

$A l^{+ 3} > N a^{+} > B a^{+ 2}$

$A l^{+ 3} > B a^{+ 2} > N a^{+}$

$B a_{2}^{+} > A l_{3}^{+} > N a^{+}$

According to this law the coagulating effect of an ion on dispersed phase of opposite charge increases with the valency of the ion. The precipitating power of $A l^{3 +}, B a^{++}, N a^{+}$ ions is in order $A l^{3 +} > B a^{2 +} > N a^{+}$ .

Q. Hardy-Schulze rule explains the effect of electrolytes on the coagulation of colloidal solution. According to this rule, coagulation power of cations follow the order

Ba+2>Na+>Al+3

Al+3>Na+>Ba+2

Al+3>Ba+2>Na+

Ba2+​>Al3+​>Na+

According to this law the coagulating effect of an ion on dispersed phase of opposite charge increases with the valency of the ion. The precipitating power of Al3+,Ba++,Na+ ions is in order Al3+>Ba2+>Na+.

$B a^{+ 2} > N a^{+} > A l^{+ 3}$

$A l^{+ 3} > N a^{+} > B a^{+ 2}$

$A l^{+ 3} > B a^{+ 2} > N a^{+}$

$B a_{2}^{+} > A l_{3}^{+} > N a^{+}$

According to this law the coagulating effect of an ion on dispersed phase of opposite charge increases with the valency of the ion. The precipitating power of $A l^{3 +}, B a^{++}, N a^{+}$ ions is in order $A l^{3 +} > B a^{2 +} > N a^{+}$ .