Question #f076a

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Answer 1 · 2014-07-02T22:42:33

You always start with the balanced chemical equation, an ICE table, and the $K_{s p}$ $K_(sp)$ expression.

The balanced chemical equation is

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

$K_{s p}$ $K_(sp)$ = [Ba²⁺][ SO₄²⁻] = $x \times x = x^{2}$ $x × x = x^2$ = 1.07 ×10⁻¹⁰

$x = \sqrt (1.07 \times 10 ⁻ ¹ ⁰)$ $x = √(1.07 ×10⁻¹⁰)$ = 1.03 × 10⁻⁵

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

Hope this helps.