Consider the following balanced equation for the dissociation of barium hydroxide in an aqueous solution.[tex]k_{sp}[/te x] = 5 x 10⁻³a 1m solution of barium nitrate is added to the solution. what is the solubility of barium hydroxide after this addition? a. 0.035mb. 0.108mc. 0.050md. 0.118me. 0.003m

a. 0,035M

Explanation:

For the reaction:

Ba(OH)₂(s) ⇄ Ba²⁺(aq) + 2OH⁻(aq)

Ksp is defined as:

Ksp = [Ba²⁺] [OH⁻]²

5x10⁻³ = [Ba²⁺] [OH⁻]²

if is added a solution of 1M of Ba²⁺:

5x10⁻³ = [1M] [OH⁻]²

The addition of barium hydroxide Ba(OH)₂ gives:

[Ba²⁺] = 1M + x

[OH⁻]² = 2x

Replacing:

5x10⁻³ = [1 + x] [2x]²

5x10⁻³ = 4x² + 4x³

Solutions are:

x = -1,00 M

x = -0,036 M

x = 0,035 M → Right answer, there are not negative concentrations.

Thus, solubility is

a. 0,035M

I hope it helps!

it's all four options.