Equilibrium expression: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) KSP = 1.8 x 10-10 The solubility of AgCl in: i) pure water:
ii) a 0.50 M NaCl solution solubility
You didn't look at your previous post did you. I agreed with your answers for (AgCl) in pure water AND in 0.5 M NaCl.
See your previous post. I still don't see a question.
CAlculate molar solubility of AgCl
oooooooh tha nk you
To determine the solubility of AgCl in different solutions, we need to consider the equilibrium expression and the solubility product constant (Ksp).
The solubility product constant (Ksp) is a measure of the solubility of a compound in water. It is defined as the product of the concentrations of the ions raised to the power of their stoichiometric coefficients in the balanced equation.
For the given equilibrium expression:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
The solubility product constant (Ksp) is given as 1.8 x 10^(-10).
i) Solubility of AgCl in pure water:
When AgCl is placed in pure water, there are no additional ions present to react with AgCl. Therefore, the concentration of Ag+ and Cl- in the equilibrium expression will be zero. To calculate the solubility, we need to calculate the concentration of Ag+ and Cl- using the Ksp expression.
Ksp = [Ag+][Cl-]
1.8 x 10^(-10) = [Ag+][Cl-]
Since [Ag+] = [Cl-], let's assume x represents the solubility of AgCl in pure water. Hence, the concentration of both Ag+ and Cl- would be x.
Ksp = x * x
x^2 = 1.8 x 10^(-10)
x ≈ 1.3 x 10^(-5)
Therefore, the solubility of AgCl in pure water is approximately 1.3 x 10^(-5) M.
ii) Solubility of AgCl in a 0.50 M NaCl solution:
In this case, we have a 0.50 M NaCl solution. When NaCl is dissolved in water, it dissociates into Na+ and Cl- ions.
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
NaCl(aq) → Na+(aq) + Cl-(aq)
The additional Cl- ions from the NaCl solution will shift the equilibrium position to the left, decreasing the solubility of AgCl. The concentration of Cl- in the 0.50 M NaCl solution is 0.50 M.
Let's assume the solubility of AgCl in the 0.50 M NaCl solution is y.
Ksp = [Ag+][Cl-]
1.8 x 10^(-10) = [Ag+][Cl-]
1.8 x 10^(-10) = y * 0.50
Solving for y, we find:
y = 1.8 x 10^(-10) / 0.50
y ≈ 3.6 x 10^(-10)
Therefore, the solubility of AgCl in a 0.50 M NaCl solution is approximately 3.6 x 10^(-10) M.