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Calculate the molar solubility of baso4 in water, ksp=1.0*10^-10

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Final answer:

The molar solubility of BaSO4 in water, with a Ksp of 1.0 x 10^-10, is calculated by setting up the equilibrium expression Ksp = x^2 and solving for x, obtaining a molar solubility of 1.0 x 10^-5 M.

Step-by-step explanation:

To calculate the molar solubility of BaSO4 in water with a given Ksp of 1.0 x 10-10, we'll write the equilibrium expression for its dissolution:

Ksp = [Ba2+][SO42-]

Since the stoichiometric coefficients of BaSO4 dissolution into Ba2+ and SO42- are a 1:1 ratio, we can express the concentrations of the ions in terms of a single variable, x, where x represents the molarity of Ba2+ and SO42-. Therefore, the Ksp expression becomes:

Ksp = x2

Solving for x:

x2 = 1.0 x 10-10
x = \(\sqrt{1.0 x 10-10}\)
x = 1.0 x 10-5 M

Hence, the molar solubility of BaSO4 in water at 25°C is 1.0 x 10-5 M.

User Kolin
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when BaSO₄ dissolves in water it dissociates as follows;
BaSO₄ ---> Ba²⁺ + SO₄²⁻
molar solubility of compound is the amount of moles that can be dissolved in 1 L solution
if molar solubility of BaSO₄ is x then molar solubility of Ba²⁺ is x and SO₄²⁻ is x.
the formula for solubility product constant - ksp is as follows
ksp = [Ba²⁺][SO₄²⁻]
ksp = (x)(x)
ksp = x²
and ksp = 1 x 10⁻¹⁰
therefore
x² = 1 x 10⁻¹⁰
x = 1 x 10⁻⁵ M
the molar solubility of BaSO₄ is 1 x 10⁻⁵ M
User WeZZard
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