Question

Consider the following balanced equation for the dissociation of barium hydroxide in an aqueous solution.$\small Ba(OH)_{2}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2OH^{-}_{(aq)}$$K_{sp}$ = 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

Answer 1

Answer:

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!