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Determine the solubility for cuc2o4(s) in pure water. Ksp for is 2. 9 × 10^-8

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Parrain

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

Find out more on solubility at https://brainly.com/question/23659342.

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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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