Imagine an experiment analogous to this one involving the hypothetical salt AB_2
ID: 474636 • Letter: I
Question
Imagine an experiment analogous to this one involving the hypothetical salt AB_2. When pure AB_2(s) was equilibrated with pure water, the concentration of A^2+ was found to be 6.5 times 10^-5 M. Calculate K_Sp for the salt AB_2. In another experiment to determine K_sp of the salt AB_2, 500.0 mL of 0.100 M A^2+ was added to 500.0 mL of 0.205 M B^-. After the precipitate settled, a portion of the supernatant liquid was filtered and analyzed and found to contain 2.3 rimes 10^7 -M A^2+. Calculate K_sp; compare this result to that of Problem 1 and discuss briefly any difference.Explanation / Answer
A^2+ + 2B- ---> AB2
In this reaction B- is the limiting reagent as it is completely consumed and only thing left is A.
Ksp = [A][B]^2
moles of A added = 0.5 L * 0.1 M = 0.05 moles
moles of A left after the reaction = 2.3*10^-7 M * (500mL + 500mL) = 2.3*10^-7 moles
moles of A reacted = 0.05 moles - 2.3*10^-7 moles = 0.0499 moles
Molarity of A that has reacted = 0.0499 moles * 1000mL/500mL = 0.0998 M
Moles of B that has reacted = 0.5 * 0.205 M =0.1025 moles
Molarity of B in the solution mixture = 0.1025 M
Ksp = (0.0988) * (0.1025)^2 =1.03*10^-3
Correct method for determination of Ksp is 1 and not 2. In the second experiment, there could be some side reacts leading to the formation of undesired products. Also, the reactant may not even react completely. You data calculated from 1 is much more correct