Question

eneric anion, B−(aq). What is the equilibrium constant, Ksp, of the generic salt AB2(s)? Ksp=At 25 °C, an aqueous solution has an equilibrium concentration of 0.00253 M for a generic cation, A2+(aq), and 0.00506 M for a generic anion, B−(aq). What is the equilibrium constant, Ksp, of the generic salt AB2(s)? Ksp=

Answer 1

Answer:

The equilibrium constant Ksp of the generic salt AB2 =  6.4777 *10^-8 M

Explanation:

Step 1: The balanced equation

AB2 ⇒ A2+ + 2B-

Step 2: Given data

Concentration of A2+ = 0.00253 M

Concentration of B- = 0.00506 M

Step 3: Calculate the equilibrium constant

Equilibrium constant Ksp of [AB2] = [A2+][B-]²

Ksp = 0.00253 * 0.00506² = 6.4777 *10^-8 M

The equilibrium constant Ksp of the generic salt AB2 =  6.4777 *10^-8 M

Answer 2

Answer:

$6.477\times 10^{-7}$ is the equilibrium constant, $K_{sp}$, of the generic salt $AB_2$.

Explanation:

Solubility product constant : It is defined as the product of the concentration of the ions present in a solution raised to the power by its stoichiometric coefficient in a solution of a salt. This takes place at equilibrium only. The solubility product constant is represented as, $K__{sp}$.

$A_xB_y\rightleftharpoons xA^{y+}+yB^{x-}$

$K_{sp}=[A^{y+}]^x\times [B^{x-}]^y$

Equilibrium concentration for a generic cation = $[A^{2+}]=0.00253 M$

Equilibrium concentration for a generic anion = $[B^{-}]=0.00506 M$

$AB^2\rightleftharpoons A^{2+}+2B^-$

The expression of solubility product is given as:

$K_{sp}=[A^{2+}][[B^-]]^2$

$K_{sp}=0.00253 M\times (0.00506 M)^2=6.477\times 10^{-7}$

$6.477\times 10^{-7}$ is the equilibrium constant, $K_{sp}$, of the generic salt $AB_2$.