Determine Ksp if the equilibrium concentration of X is 0.02 M and Y is 0.05 M given that X2Y(s)<-->2X^+(aq)+Y^(2-)(aq).
The 2 in X2Y is a subscript.
.......X2Y(s)<-->2X^+(aq)+Y^(2-)(aq).
Ksp = (X^2+)^2(Y^2-)
X = 0.02
Y = 0.05
Ksp = (0.02)^2(0.05) = ?
the answer is supposed to be 7.5x10^-5 but I don't know how
No, the answer is 2.0E-5. I've looked at the problem again and it appears not to have any "tricks of the trade" attached to it so I believe if you have a database version or an answer sheet that those are incorrect.
To determine the Ksp (solubility product constant) for the given equilibrium, we need to first write the balanced chemical equation and then set up an expression for the Ksp.
From the equation X2Y(s) ⇌ 2X^+(aq) + Y^(2-)(aq), we can see that one X2Y molecule dissociates into two X^+ ions and one Y^2- ion in the aqueous solution.
The balanced chemical equation also implies that the solubility of X2Y(s) is equal to the concentration of X^+ ions and Y^2- ions in the solution. We can express this as:
Ksp = [X^+]^2 * [Y^2-]
Given the equilibrium concentrations of X and Y as 0.02 M and 0.05 M, respectively, we can substitute these values into the Ksp expression:
Ksp = (0.02 M)^2 * (0.05 M)
Simplifying the expression:
Ksp = 0.0004 M^2 * 0.05 M
= 0.00002 M^3
Therefore, the Ksp for the given equilibrium is 0.00002 M^3.