Chapter 17: Problem 93
Calculate the \(\mathrm{pH}\) of a \(0.15 \mathrm{M}\) aqueous solution of aluminum chloride, \(\mathrm{AlCl}_{3}\). The acid ionization of hydrated aluminum ion is \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) and \(K_{a}\) is \(1.4 \times 10^{-5}\).
Short Answer
Step by step solution
Understanding the Problem
Setting Up the Equation for Equilibrium
Solving for Hydronium Ion Concentration
Calculating the pH
Concluding the Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acid Ionization
- Acid ionization in aqueous solutions involves the breaking of bonds to release protons.
- The equilibrium established from this process is key to determining the solution's acidity.
Equilibrium Constant
- The magnitude of \({ K_a }\) tells us whether the acid is strong or weak; smaller numbers indicate weaker acids.
- For our problem, \({ K_a }\)'s numerical value suggests the forward reaction of aluminum ion hydrolysis is not very extensive.
Hydrolysis Reaction
- The reaction product includes hydronium ions, directly increasing the solution's acidity.
- Hydrolysis-related pH changes are significant in solutions involving metal salts, like aluminum chloride.
Aluminum Chloride Solution
- Initial dissociation leads to the formation of \({ ext{Al}( ext{H}_2 ext{O})_6^{3+} }\) in water.
- The subsequent release of \({ ext{H}_3 ext{O}^+ }\) ions is pivotal for the observed low pH.