The ionic substance T3U2 ionizes to form
T2+ and U3− ions. The solubility of T3U2 is 4.07×10−20 mol/L. What is the value of
the solubility-product constant?
To determine the value of the solubility-product constant, we need to first write the balanced chemical equation for the ionization of T3U2.
The chemical equation is: T3U2 ⇌ T2+ + U3-
From the equation, we see that 1 mole of T3U2 ionizes to form 1 mole of T2+ and 1 mole U3- ions.
Now, we need to use the given information about the solubility of T3U2 to calculate the concentration of the ions in the solution.
The solubility of T3U2 is given as 4.07×10^(-20) mol/L. Since T3U2 dissolves completely, the concentration of T2+ and U3- ions will be the same as the solubility of T3U2.
Therefore, the concentration of T2+ and U3- ions is also 4.07×10^(-20) mol/L.
The solubility-product constant (Ksp) is found by multiplying the concentrations of the ions raised to the power of their stoichiometric coefficients:
Ksp = [T2+] * [U3-]
Since the concentration of T2+ is 4.07×10^(-20) mol/L and the concentration of U3- is also 4.07×10^(-20) mol/L, we can substitute these values into the equation:
Ksp = (4.07×10^(-20) mol/L) * (4.07×10^(-20) mol/L)
Simplifying the equation, we get:
Ksp = 1.65649×10^(-39) mol^2/L^2
Therefore, the value of the solubility-product constant is 1.65649×10^(-39) mol^2/L^2.