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Sagot :
Based on the elements and charges in Copper (II) Oxalate, CuC₂O₄(s), the solubility in pure water is 1.7 x 10⁻⁴ M.
What is the solubility of Copper (II) Oxalate in pure water?
The solubility equilibrium (Ksp) is 2.9 x 10⁻⁸ so the solubility can be found as:
Ksp = [Cu²⁺] [C₂O₄²⁻]
Solving gives:
2.9 x 10⁻⁸ = S x S
S² = 2.9 x 10⁻⁸
S = 1.7 x 10⁻⁴ M
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The value of solubility for cuc2o4(s) in pure water when the value of solubility product, Ksp for it is 2.9 ×10⁻⁸, is 1.703 × 10⁻⁴.
What is solubility?
The solubility of a chemical compound or an element is the property or ability of it to be dissolved, specially in water.
The solubility for CuC₂O₄(s) in pure water has to be found out. The solubility product is,
[tex]K_{sp}=2. 9 \times 10^{-8}[/tex]
Let suppose x is the molar solubility of CuC₂O₄(s). From the ICE approach solution,
[tex]K_{sp}=x^2\\2. 9 \times 10^{-8}=x^2\\x=\sqrt{2. 9 \times 10^{-8}}\\x=1.703 \times 10^{-4}\rm\; M[/tex]
Thus, the value of solubility for cuc2o4(s) in pure water when the value of solubility product, Ksp for it is 2.9 ×10⁻⁸, is 1.703 × 10⁻⁴.
Learn more about the solubility here;
https://brainly.com/question/4529762
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