What is the expression and numerical value for the equilibrium constant for the following:
Ag+ + Cl- = AgCl(s)
using the Ksp= 1.8*10^-10 of AgCl
please show how its calculated.
Your question really isn't a question. The Ksp expression is
Ksp = (Ag^+)(Cl^-) = Ksp
But I don't know how to show how Ksp is calculated using Ksp.
To determine the equilibrium constant for the reaction Ag+ + Cl- → AgCl(s), we need to consider the solubility product constant, Ksp, of AgCl. The Ksp represents the equilibrium constant for a solid compound dissolving in a solution.
The balanced equation tells us that one Ag+ ion and one Cl- ion combine to form AgCl(s). Thus, the expression for the equilibrium constant, K, can be written as:
K = [Ag+][Cl-]
Here, [Ag+] represents the concentration of Ag+ ions, and [Cl-] represents the concentration of Cl- ions. Since AgCl(s) is a solid, its concentration remains constant and is not included in the expression.
Given that the Ksp of AgCl is 1.8 * 10^-10, we can use this value to determine the concentrations of Ag+ and Cl- ions at equilibrium.
Since the reaction involves a 1:1 stoichiometric ratio between Ag+ and Cl-, at equilibrium, both concentrations will be the same, let's call this value x.
Therefore, the expression for the equilibrium constant becomes:
K = [Ag+][Cl-] = x * x = x^2
Now, we can substitute the given Ksp value to solve for x:
Ksp = 1.8 * 10^-10 = x^2
To solve for x, we take the square root of both sides:
sqrt(Ksp) = sqrt(1.8 * 10^-10) = 1.34 * 10^-5
Now, we have determined the concentration of Ag+ and Cl- ions at equilibrium as x = 1.34 * 10^-5 M.
Therefore, the expression for the equilibrium constant becomes:
K = (1.34 * 10^-5 M)^2 = 1.8 * 10^-10