Solid Al(NO3)3 is added to distilled water to produce a solution in which the concentration of nitra — Solutions Chemistry Question

STEPS:

1. Identify the Dissociation of the Salt: Aluminum nitrate, $Al(NO_3)_3$, is a strong electrolyte that dissociates completely when added to water. The balanced chemical equation for this process is:
$$Al(NO_3)_3(s) \rightarrow Al^{3+}(aq) + 3\ NO_3^-(aq)$$
2. Determine the Molar Ratio: From the balanced equation, observe the stoichiometry: for every 1 mole of $Al(NO_3)_3$ that dissolves, it produces 1 mole of $Al^{3+}$ ions and 3 moles of $NO_3^-$ ions.
3. Relate the Ion Concentrations: Because the ions come from the same source, their concentrations are directly related by their molar ratio. Specifically, the concentration of nitrate ions is three times the concentration of aluminum ions:
$$[NO_3^-] = 3 \times [Al^{3+}]$$
4. Perform the Calculation: The problem states that the concentration of nitrate, $[NO_3^-]$, is 0.10 M. Substitute this value into the relationship:
$$0.10\ M = 3 \times [Al^{3+}]$$
$$[Al^{3+}] = \frac{0.10\ M}{3} \approx \mathbf{0.033\ M}$$
5. Conclusion: The concentration of aluminum ions in the solution is approximately 0.033 M, which corresponds to option B.

WHY_OTHERS_WRONG:

• A) 0.010 M: This value results from an incorrect mathematical operation (such as dividing by 10) rather than using the 1:3 stoichiometric ratio.
• C) 0.066 M: This would be the result if the ratio of $Al^{3+}$ to $NO_3^-$ were 2:3, which is not supported by the formula $Al(NO_3)_3$.
• D) 0.10 M: This assumes a 1:1 ratio between aluminum and nitrate ions, which ignores the subscript "3" in the chemical formula.
• E) 0.30 M: This is the result of multiplying the nitrate concentration by 3 instead of dividing by 3; it incorrectly suggests there are more aluminum ions than nitrate ions.