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For AgCl, Ksp = 1.8 x 10^-10 at 25°C. What is the solubility of AgCl at this temperature?

2007-03-16 10:25:27 · 2 answers · asked by MEB 2 in Science & Mathematics ➔ Chemistry

2 answers

If s is the solubility (in mole/L) then

.. .. .. .. .. AgCl <=> Ag+ +Cl-
Dissolve .. s
Produce .. .. .. .. .. .. s .. .. . s
At equil .. ... .. ... .. .. s .. ... . s

Ksp= [Ag+][Cl-]= s*s =s^2 =>
s= squareroot (Ksp)= SQRT(1.8*10^-10) =1.34*10^-5 M

2007-03-16 10:36:44 · answer #1 · answered by bellerophon 6 · 0⤊ 0⤋

Ksp = 1.8 * 10^-10 = [Ag][Cl] = [x][x] = x^2

Solubility of a salt made of only two ions (one cation, one anion) is square root of the Ksp expression so
S = sq root of Ksp

Go for it.

2007-03-16 17:36:31 · answer #2 · answered by ? 4 · 0⤊ 0⤋