A 0.0035 M aqueous solution of a particular compound has pH = 2.46. The compound is (A) a weak base (B) a weak acid (C) a strong
1 answer:
Answer:
(a) A strong acid
Explanation:
We have given the pH of the solution is 2.46
pH=2.46
So the concentration of
solution having H+ concentration more than is acidic
Since in the given solution, H+ concentration is 0.00346 M which is more than 10^{-7}[/tex] so this is an acidic solution
Note-The concentration of decide the behavior of the solution that is, it is acidic or basic
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The rate of Formation of Carbocation mainly depends on two factors'
1) Stability of Carbocation:
The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.
2) Ease of detaching of Leaving Group:
The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,
R-I > R-Cl > R-F
B > C > A
1) Stability of Carbocation:
The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.
2) Ease of detaching of Leaving Group:
The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,
R-I > R-Cl > R-F
B > C > A