At 25 °C only 0.0990 mol of the generic salt AB is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C?
...AB(s) ==> A^+ + B^-
...x.........x.......x
x = 0.0990
Ksp = (A^+)(B^-) = ?
I also posted the AB3 answer but I don't see it anywhere now.
I posted a few questions and they all disappeared so i reposted.
I answered them and my answers disappeared, also. I've re-answered them.
thank you
To find the Ksp (solubility product constant) of a salt, we need to know the concentration of the ions in the saturated solution. In this case, we are given the solubility of the salt AB in moles per liter (mol/L).
The formula of the salt AB suggests that it dissociates into two ions, A+ and B-. Therefore, the solubility in moles per liter can be considered as the concentration of both A+ and B- ions.
Given that the solubility of AB is 0.0990 mol/L, the concentration of both A+ and B- ions in the saturated solution would also be 0.0990 mol/L.
Since the salt AB dissociates into two ions, the Ksp expression would be:
Ksp = [A+] * [B-]
In this case, both [A+] and [B-] are equal to 0.0990 mol/L. Substituting the values, we get:
Ksp = (0.0990) * (0.0990)
Simplifying the expression, the Ksp of the salt AB at 25 °C is:
Ksp = 0.009801
Therefore, the Ksp of the salt AB at 25 °C is 0.009801.