What concentration of SO32– is in equilibrium with Ag2SO3(s) and 2.10 × 10-3 M Ag ? The Ksp of Ag2SO3=1.50×10–14
...........Ag2SO3 ==> 2Ag^+ + SO3^2-
I.......................0.......0
C......................2x.......x
E.....................2x........x
Ksp = (Ag^+)^2(SO3^2-)
The problem gives you (Ag+) = 0.0021; substitute that into Ksp expression along with Ksp and solve for x = (SO3^2-)
To find the concentration of SO32– in equilibrium with Ag2SO3(s) and 2.10 × 10-3 M Ag, we need to use the solubility product constant (Ksp) equation.
The chemical equation for the dissociation of Ag2SO3 is:
Ag2SO3(s) ⇌ 2Ag+(aq) + SO32–(aq)
The Ksp expression for this reaction is:
Ksp = [Ag+]^2 [SO32–]
Given the value of Ksp as 1.50 × 10–14, and that the concentration of Ag+ ions ([Ag+]) is 2.10 × 10-3 M, we can substitute these values into the Ksp expression to find the concentration of SO32–.
1.50 × 10–14 = (2.10 × 10-3)^2 [SO32–]
Rearranging the equation to solve for [SO32–]:
[SO32–] = (1.50 × 10–14) / (2.10 × 10-3)^2
Calculate the value:
[SO32–] = 4.08 × 10–9 M
Therefore, the concentration of SO32– in equilibrium with Ag2SO3(s) and 2.10 × 10-3 M Ag is 4.08 × 10–9 M.