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ACH3701 ASSIGNMENT 2
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ACH3701 ASSIGNMENT 2
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ACH3701 ASSIGNMENT 2[School]
[Course title]
, 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
So, the estimated solubility of calcium carbonate (CaCO3) at 25°C in the presence of
0.09M CaCl2(aq) is approximately 1.87 x 10⁻⁸ M.
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