find the molar solubility of zinc carbonate if its ksp=1.5x10^-10. (at 298K)
.........ZnCO3 ==> Zn^2+ + CO3^2-
I.......solid.......0........0
C.....x dissolves...x........x
E........solid......x........x
Substitute the E line into Ksp expression and solve for x = solubility in mols/L.
To find the molar solubility of zinc carbonate (ZnCO3) given its Ksp (solubility product constant), we need to set up an equilibrium expression based on the balanced equation for the dissolution of zinc carbonate in water.
The balanced equation for the dissolution of zinc carbonate is:
ZnCO3(s) ⇌ Zn2+(aq) + CO32-(aq)
According to the equation, one mole of zinc carbonate will produce one mole of zinc ions (Zn2+) and one mole of carbonate ions (CO32-).
Let's assume that the molar solubility of zinc carbonate is "s" mol/L. Then, the concentration of zinc ions and carbonate ions will also be "s" mol/L at equilibrium.
The Ksp expression for zinc carbonate can be written as:
Ksp = [Zn2+][CO32-]
Using the molar solubility "s", we can rewrite the Ksp expression as:
Ksp = (s)(s) = s^2
Substituting the given Ksp value into the expression, we have:
1.5x10^-10 = s^2
To solve for the molar solubility "s", take the square root of both sides of the equation:
√(1.5x10^-10) = s
Calculating this value, we find that the molar solubility of zinc carbonate at 298K is approximately 1.23x10^-5 mol/L.