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Calculate the pH of a 0.010 M aluminum chloride solution. The Ka value for Al(H₂O)₆³⁺ is 1.4x10⁻⁵.

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Final answer:

To calculate the pH of a 0.010 M aluminum chloride solution, we use the hydrolysis reaction of the hydrated aluminum ion. The concentration of H₃O⁺ ions is 6 times the concentration of the hydrated aluminum ions. Using the equation pH = -log[H₃O⁺], we find the pH to be approximately 1.22.

Step-by-step explanation:

The aluminum chloride (AlCl3) dissociates in water to form hydrated aluminum ions [Al(H2O)6]³⁺ and chloride ions (Cl⁻). The hydrated aluminum ions can undergo hydrolysis to produce hydronium ions (H₃O⁺) and aluminum hydroxide.

The equation for the hydrolysis reaction is:

[Al(H2O)6]³⁺ + 3H₂O ⇌ Al(OH)3 + 6H₃O⁺

The equilibrium constant (Ka) for this reaction is given as 1.4 x 10⁻⁵.

To calculate the pH of the solution, we need to determine the concentration of H₃O⁺ ions. Since 1 mole of [Al(H2O)6]³⁺ produces 6 moles of H₃O⁺ ions, the concentration of H₃O⁺ ions is 6 times the concentration of [Al(H2O)6]³⁺.

Given that the concentration of [Al(H2O)6]³⁺ is 0.010 M, the concentration of H₃O⁺ ions is 6 x 0.010 M = 0.060 M.

Lastly, we can calculate the pH using the formula pH = -log[H₃O⁺].

pH = -log(0.060) ≈ 1.22

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