Answer:
C2H5NH3Cl: 0.25 M
C2H5NH3+: x (very small)
Cl-: x (very small)
Step-by-step explanation:
When ethylammonium chloride dissolves in water, it dissociates into its constituent ions, ethylammonium cation (C2H5NH3+) and chloride anion (Cl-). The balanced chemical equation for this dissociation is:
C2H5NH3Cl(s) → C2H5NH3+(aq) + Cl-(aq)
The initial concentration of ethylammonium chloride is 0.25 M. At equilibrium, the concentration of ethylammonium cation and chloride anion will be the same as they are produced in a 1:1 ratio. Let's represent the equilibrium concentration of C2H5NH3+ and Cl- as x. Then the equilibrium concentrations of all species will be:
C2H5NH3Cl: 0.25 M - x
C2H5NH3+: x
Cl-: x
To solve for x, we can use the equilibrium constant expression for the dissociation of ethylammonium chloride in water, which is:
Kc = [C2H5NH3+][Cl-]/[C2H5NH3Cl]
The value of Kc for this dissociation has not been provided, but we can assume it is sufficiently large, such that the dissociation can be considered to go to completion.
Substituting the equilibrium concentrations into the expression for Kc, we get:
Kc = [C2H5NH3+][Cl-]/[C2H5NH3Cl]
Kc = x*x / (0.25 - x)
Since the dissociation is complete, the value of x will be very small compared to the initial concentration of ethylammonium chloride. This means that the term 0.25 - x will be approximately equal to 0.25. Substituting this approximation, we get:
Kc = x*x / 0.25
Solving for x, we get:
x = sqrt(Kc * 0.25)
Since the value of Kc has not been provided, we cannot obtain a numerical value for x. However, we can determine the concentrations of all species at equilibrium once we know the value of Kc.
If Kc is very large, as we assumed earlier, then the dissociation will be complete and the concentrations of C2H5NH3+ and Cl- will be equal to x, which is a very small value. The concentration of ethylammonium chloride will be equal to the initial concentration minus x, which is essentially equal to the initial concentration. Therefore, the concentrations of all species at equilibrium will be approximately:
C2H5NH3Cl: 0.25 M
C2H5NH3+: x (very small)
Cl-: x (very small)
Again, this assumes that Kc is very large, such that the dissociation goes to completion. If Kc is not very large, the dissociation will be less complete and the concentrations of all species will be different.