Asked by Marcus 2nd question
Calculate the solubility of silver chloride in a 0.1 mol/L solution of sodium chloride at 25°C.?
At SATP, Ksp AgCl(s) = 1.8 x 10^-10
What did I do wrong?
AgCl <----> Ag+ + Cl-
[Ag+] = x
[Cl-] = x + 0.1
Ksp = [Ag+] [Cl-]
1.8 x 10^-10 = (x) ( x + 0.1)
x = molar solubility = 1.8 x 10^-9 M
At SATP, Ksp AgCl(s) = 1.8 x 10^-10
What did I do wrong?
AgCl <----> Ag+ + Cl-
[Ag+] = x
[Cl-] = x + 0.1
Ksp = [Ag+] [Cl-]
1.8 x 10^-10 = (x) ( x + 0.1)
x = molar solubility = 1.8 x 10^-9 M
Answers
Answered by
DrBob222
I don't see an error. SATP refers to 25<sup>o</sup> and that's what you have for Ksp. The solubility is certainly smaller than x + 0.1 so x + 0.1 certainly is almost 0.1. The only thing I can think of is that perhaps the problem wanted grams/L instead of molarity. And finally, I don't know the level of the class but sometimes one must make a correction for ionic strength and that can make a SMALL difference.
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