AgCl <==> Ag^+ + Cl^-
NaCl ==> Na^+ + Cl^-
Ksp = (Ag^+)(Cl^-) = 1.8 x 10^-10
Let S = solubility of AgCl, then
(Ag^+) = S
(Cl^-) = S+0.01
Solve for S.
Note: A similar problem to this post (0.1 M NaCl instead of 0.01 M NaCl) was on a couple of days ago; the one who posted said that the answer came back incorrect. Check my work. Check myu thinking.
Should the continuing work look like this?
S = 1.8 x 10^-10 + 1
= 1.00?
Please help I am very confused
I don't understand how you can be confused.
(S)(S + 0.01) = 1.8 x 10^-10
You can solve the quadratic, OR you can make the simplifying assumption that S + 0.01 is about 0.01 (that is, that S is very small in comparison so 0.01 is not really different than S + 0.01).
If we make the assumption, then
S(0.01) = 1.8 x 10-^-10
and S = 1.8 x 10^-8 M = (AgCl).
To solve for the solubility of AgCl (S), you need to set up an expression for the equilibrium constant (Ksp) using the concentrations of Ag^+ and Cl^- ions.
Based on the given dissociation equation: AgCl <==> Ag^+ + Cl^-, the concentrations of Ag^+ and Cl^- in terms of S are:
(Ag^+) = S
(Cl^-) = S + 0.01
Now you can substitute these values into the expression for Ksp:
Ksp = (Ag^+)(Cl^-) = (S)(S + 0.01) = 1.8 x 10^-10
To solve for S, you need to solve this quadratic equation. Begin by expanding the expression:
S^2 + 0.01S - 1.8 x 10^-10 = 0
Now, you can use the quadratic formula to solve for S:
S = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 0.01, and c = -1.8 x 10^-10. Plugging these values into the quadratic formula, you can calculate S.
S = (-0.01 ± √((0.01)^2 - 4(1)(-1.8 x 10^-10))) / 2(1)
After calculating this expression, you will find that S ≈ 1.34 x 10^-5.
So, the solubility of AgCl (S) is approximately 1.34 x 10^-5.
To answer your specific question, the continuation of the work should not look like S = 1.8 x 10^-10 + 1. This is because you need to solve the quadratic equation described above, not perform a direct addition.