Calculate the solubility of BaCO3 in mol ^-1 given the solubility product constant 5.1* 10^-9 at 25°C?

The solubility product constant (Ksp) for BaCO3 is given as 5.1 x 10^-9 at 25°C.

The equation for the solubility product constant is:

BaCO3 (s) ⇌ Ba2+ (aq) + CO3^2- (aq)

The Ksp expression for this reaction is:

Ksp = [Ba2+][CO3^2-]

Given that BaCO3 dissociates into 1 Ba2+ ion and 1 CO3^2- ion, the solubility of BaCO3 can be represented as "s" mol/L.

Therefore, the Ksp expression can be written as:

5.1 x 10^-9 = s * s

5.1 x 10^-9 = s^2

Taking the square root of both sides:

s = √(5.1 x 10^-9)

s = 7.14 x 10^-5 mol/L

Therefore, the solubility of BaCO3 in mol^-1 at 25°C is 7.14 x 10^-5 mol/L.

The solubility product (Ksp) expression for BaCO3 is:

Ksp = [Ba2+][CO32-]

where [Ba2+] and [CO32-] are the molar concentrations of barium and carbonate ions in a saturated solution.

At 25°C, Ksp for BaCO3 is 5.1 - 10^-9.

Since BaCO3 dissociates into one Ba2+ ion and one CO32- ion, the solubility (s) of BaCO3 can be expressed as:

s = [Ba2+] = [CO32-]

Substituting s into the Ksp expression, we get:

Ksp = s^2

Solving for s, we get:

s = √(Ksp)

Substituting the value of Ksp, we get:

s = √(5.1 - 10^-9) = 2.26 - 10^-5 mol/L

Therefore, the solubility of BaCO3 at 25°C is 2.26 - 10^-5 mol/L.