Assuming the final concentration of chloride anion after the addition of HCl (precipitation step) was 0.1M, what is the remaining concentration of Ag+ in the solution? (pKsp for AgCl = 9.74)
pKsp = 9.74 = -log Ksp
Ksp = 1.82E-10
.............AgCl ==> Ag^+ + Cl^-
I............solid....0.......0
C............solid....x.......x
E............solid....x...... 0.1
The 0.1 for Cl^- comes from the problem
Ksp AgCl =1.82E-10 = (Ag^+)(Cl^-)
1.82E-10 = (x)(0.1)
Solve for x = (Ag^+)
To find the remaining concentration of Ag+ in the solution, we need to consider the solubility product constant (Ksp) for AgCl.
The formula for the solubility product constant for AgCl is:
Ksp = [Ag+][Cl-]
Since the final concentration of chloride anion ([Cl-]) after the addition of HCl is given as 0.1 M, we can substitute this value into the Ksp equation:
0.1M = [Ag+][Cl-]
But from the equation, we know that the concentrations of Ag+ and Cl- ions are equal, so we can substitute [Cl-] = 0.1 M with [Ag+]:
0.1M = [Ag+]^2
Taking the square root of both sides gives us:
√(0.1M) = [Ag+]
Calculating this, we get:
[Ag+] ≈ 0.316 M
Therefore, the remaining concentration of Ag+ in the solution is approximately 0.316 M.
To determine the remaining concentration of Ag+ in the solution, we need to utilize the solubility product constant (Ksp) and the concept of ionic equilibrium.
The equation for the solubility product constant (Ksp) of AgCl is as follows:
AgCl ⇌ Ag+ + Cl-
The Ksp expression for AgCl is given by:
Ksp = [Ag+][Cl-]
We are given the pKsp for AgCl, which is the negative logarithm of the solubility product constant:
pKsp = -log10(Ksp)
Therefore, we can determine the value of Ksp by taking the antilog (10 raised to the power of -pKsp). Hence, in this case, Ksp = 10^(-pKsp).
First, let's calculate the Ksp for AgCl using the given pKsp value:
Ksp = 10^(-9.74)
Now, we know that when AgCl dissolves, it dissociates to form Ag+ and Cl-. However, in the presence of excess chloride ions (Cl-) from the addition of HCl, the Ag+ ions react with the chloride ions to form AgCl precipitate according to the equation:
Ag+ + Cl- ⇌ AgCl(s)
Since the concentration of chloride ions after the addition of HCl is given as 0.1 M, we can assume that all the Ag+ ions have reacted to form AgCl precipitate. Therefore, the concentration of Ag+ remaining in the solution is zero.
Hence, the remaining concentration of Ag+ in the solution is zero.