At 25 °C, an aqueous solution has an equilibrium concentration of 0.00147 M for a generic cation, A (aq), and 0.000735 M for a generic anion, B2–(aq). What is the equilibrium constant, Ksp, of the generic salt A2B(s)?

I don't believe the above answer is right. It should be

A2B ==> 2A^+ + B^2-
Ksp = (A^+)^2(B^2-) =
(0.00147)^2(0.000735) = ?

if that's so am sorry thanks for the correct answer pls help me with my ques as well

Why did the chemist break up with their bacteria girlfriend? She had too much culture for them to handle!

Well, to calculate the equilibrium constant, Ksp, of the generic salt A2B(s), we need to remember that the equilibrium constant expression is given by Ksp = [A]^2[B].

Given that [A] = 0.00147 M and [B] = 0.000735 M, we can plug these values into the equation to get:

Ksp = (0.00147)^2 * (0.000735)

Now go grab your calculator to get the final answer.

To calculate the equilibrium constant, Ksp, of the generic salt A2B(s), you need to know the concentrations of the ions present in the solution at equilibrium.

In this case, you are given the equilibrium concentrations of the cation A (aq) and the anion B2–(aq). These concentrations are 0.00147 M and 0.000735 M, respectively.

The equilibrium constant expression for the reaction A2B(s) ⇌ 2A(aq) + B2–(aq) can be written as:

Ksp = [A]^2[B]^1

Where [A] and [B] are the concentrations of A and B ions, respectively, at equilibrium.

In this case, since the stoichiometric coefficient of A2B(s) is 1, the concentration of A ions in the equilibrium expression is the same as the concentration of A.

Substituting the given values into the equation, we get:

Ksp = (0.00147)^2 * (0.000735)^1

Calculating this expression, we find:

Ksp = 1.90 x 10^(-9)

equilibrium constant- (0.00147)(0.000735)^2=7.94*10^-10