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ACH3701 ASSIGNMENT 2
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ACH3701 ASSIGNMENT 2MASTER KYALO
GUARANTEED SUCCESS
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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