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QUESTION

# How can I calculate the molar solubility of BaSO4 given a Ksp of 1.07×10−10?

You always start with the balanced chemical equation, an ICE table, and the K_(sp) expression.

The balanced chemical equation is

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) I/mol·L⁻¹: -; 0; 0 C/mol·L⁻¹: -; +x; +x E/mol·L⁻¹: -; x; x

K_(sp) = [Ba²⁺][ SO₄²⁻] = x × x = x^2 = 1.07 ×10⁻¹⁰

x = √(1.07 ×10⁻¹⁰) = 1.03 × 10⁻⁵

Solubility = x mol/L = 1.03 × 10⁻⁵ mol/L

Hope this helps.

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