The molar solubility of BaCrO₄ in pure water is 1.4 × 10⁻⁵ M, while the molar solubility in 1.6 × 10⁻³ M Na₂CrO₄ is 1.3 × 10⁻⁷ M.
Let's consider the solution of BaCrO₄.
BaCrO₄(s) ⇄ Ba²⁺(aq) + CrO₄²⁻(aq)
To find the molar solubility (S) in pure water, we need to make an ICE chart.
BaCrO₄(s) ⇄ Ba²⁺(aq) + CrO₄²⁻(aq)
I 0 0
C +S +S
E S S
The solubility product constant is:
[tex]Ksp = 2.1 \times 10^{-10} = [Ba^{2+} ][CrO_4^{2-} ]= S^{2} \\S = \sqrt{2.1 \times 10^{-10}} = 1.4 \times 10^{-5} M[/tex]
To calculate the molar solubility of BaCrO₄ in 1.6 × 10⁻³ M Na₂CrO₄, we need to consider the Na₂CrO₄ that is a strong electrolyte. Thus, when making the ICE chart, the initial concentration of CrO₄²⁻ will be 1.6 × 10⁻³ M.
BaCrO₄(s) ⇄ Ba²⁺(aq) + CrO₄²⁻(aq)
I 0 1.6 × 10⁻³
C +S +S
E S 1.6 × 10⁻³ + S
The solubility product constant is:
[tex]Ksp = 2.1 \times 10^{-10} = [Ba^{2+} ][CrO_4^{2-} ]= (S)(S+1.6\times 10^{-3} )[/tex]
Since 1.6 × 10⁻³ >>> S, 1.6 × 10⁻³ + S ≅ 1.6 × 10⁻³
[tex]Ksp = 2.1 \times 10^{-10} = (S)(1.6\times 10^{-3} )\\\\S = 1.3 \times 10^{-7} M[/tex]
The molar solubility of BaCrO₄ in pure water is 1.4 × 10⁻⁵ M, while the molar solubility in 1.6 × 10⁻³ M Na₂CrO₄ is 1.3 × 10⁻⁷ M.
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