The solubility product for barium sulfate is 1.1 × 10–10. calculate the molar solubility of barium sulfate.

When it comes to equilibrium reactions in chemistry, there are a lot of equilibrium constants that can be used. In the case of solubility, the appropriate one to use is the equilibrium constant of solubility product denotes as Ksp. This is the concentration of products raised to their coefficients. For example,

 cC ⇔ aA + bB 

Ksp = {[A^a][B^b]}

Now, for the this problem, the reaction is

BaSO₄ ⇔ Ba²⁺ + SO₄²⁻

The reaction is already balanced. Since we don't know the value of Ba²⁺ and SO₄²⁻, let's denote this at x.

1.1 × 10⁻¹⁰ = [x][x] =[x²]
[x] = [Ba²⁺] = [SO₄²⁻] = [BaSO₄] = 1.049 × 10⁻⁵ M

Is a logical series of steps

Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o