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ACH3701 ASSIGNMENT 2

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ACH3701 ASSIGNMENT 2

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  • August 17, 2023
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  • 2023/2024
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ACH3701 ASSIGNMENT 2
MASTER KYALO




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MASTERKYALO

, 1. Estimate the solubility of calcium carbonate at 25oC in the presence of
0.09M CaCl2(aq). [7]

To estimate the solubility of calcium carbonate (CaCO3) at 25°C in the
presence of 0.09M CaCl2(aq), we can use the common ion effect. The
common ion effect states that the solubility of a slightly soluble salt is
decreased when a common ion is added to the solution.

The solubility product constant (Ksp) for calcium carbonate is given by the
equation:

CaCO3(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq) Ksp = [Ca²⁺][CO₃²⁻]

We are given that the concentration of Ca²⁺ ions from the CaCl2 solution is
0.09M. However, we need to consider that each CaCl2 molecule dissociates
into two Ca²⁺ ions. Therefore, the concentration of Ca²⁺ ions from the CaCl2
solution will be:

[Ca²⁺] = 2 * 0.09M = 0.18M

Now we need to set up an equilibrium expression for the dissolution of
calcium carbonate:

CaCO3(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq)

At equilibrium, the concentration of Ca²⁺ ions from the dissolution of CaCO3
will be equal to the solubility of CaCO3, which we'll call "x". The
concentration of CO₃²⁻ ions will also be "x". Therefore, we can write the
equilibrium expression:

Ksp = [Ca²⁺][CO₃²⁻] = (0.18M)(x)

Given that the Ksp value for CaCO3 is approximately 3.36 x 10⁻⁹ (at 25°C),
we can substitute this value into the equation:

3.36 x 10⁻⁹ = (0.18M)(x)

Now we can solve for "x", which represents the solubility of CaCO3:

x = (3.36 x 10⁻⁹) / 0.18 ≈ 1.87 x 10⁻⁸ M

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