A solution is prepared by mixing 100.0 mL of 0.0500 M AgNO3 with 75.0 mL of 0.0500 M CaCl2. What is the equilibrium concentration of Ag+ in solution? Ksp for AgCl is 1.6 x 10^-10
To find the equilibrium concentration of Ag+ in solution, we need to determine the concentration of Ag+ ions that remain in solution after the reaction between AgNO3 and CaCl2.
First, let's write the balanced chemical equation for the reaction between AgNO3 and CaCl2:
AgNO3 + CaCl2 -> AgCl + Ca(NO3)2
Since both AgNO3 and CaCl2 have the same concentration (0.0500 M) and are present in a 1:1 molar ratio, they will react completely, forming AgCl.
Next, we need to determine the amount of Ag+ that will react with Cl- to form AgCl.
The number of moles of AgNO3 can be calculated using the formula:
moles = concentration (M) x volume (L)
Moles of AgNO3 = 0.0500 M x 0.1000 L = 0.00500 mol
Since AgNO3 and CaCl2 have a 1:1 molar ratio, this means that 0.00500 mol of Ag+ ions will react with 0.00500 mol of Cl- ions to form AgCl.
Since AgCl is insoluble, it will undergo a dissociation to a very small extent. This can be represented by the equilibrium expression for the solubility product constant (Ksp):
Ksp = [Ag+][Cl-]
We are given the value of Ksp for AgCl, which is 1.6 x 10^-10.
Assuming that the concentration of Cl- is in excess (which is reasonable considering the stoichiometry of the reaction), we can use the equilibrium expression to find the concentration of Ag+:
[Ag+] = Ksp / [Cl-]
Since Ksp = 1.6 x 10^-10 and the concentration of Cl- is the same as the concentration of CaCl2, which is 0.0500 M, we can calculate:
[Ag+] = (1.6 x 10^-10) / (0.0500 M)
[Ag+] = 3.2 x 10^-9 M
Therefore, the equilibrium concentration of Ag+ in solution is 3.2 x 10^-9 M.