Question

In potash alum, the ratio of \(\text{K}^+\) and \(\text{SO}_4^{2-}\) ions is

• A. \(3:2\)
• B. \(1:2\)
• C. \(2:1\)
• D. \(2:3\)

Hint:

Always remember the empirical formula of potash alum can also be written in its simplified form as \(\text{KAl}(\text{SO}_4)_2 \cdot 12\text{H}_2\text{O}\). In this format, you can read the ratio directly from the subscripts: there is $1$ potassium ion for every $2$ sulfate ions, giving the answer $1:2$ instantly.

Answer 1

The Correct Option is B

Concept: Potash alum is a classic example of a double salt. A double salt is a crystalline molecular compound that exists as a single stable solid structure but dissociates completely into its individual constituent simple ions when dissolved in water or any other aqueous solvent. To evaluate the ionic ratios correctly, one must look at the exact stoichiometric coefficients of the ions present in its balanced chemical formula.

Step 1: Write down the chemical formula of potash alum.
The chemical composition of potash alum (potassium aluminum sulfate dodecahydrate) is represented by the formula: \[ \text{K}_2\text{SO}_4 \cdot \text{Al}_2(\text{SO}_4)_3 \cdot 24\text{H}_2\text{O} \]

Step 2: Trace the ionization behavior in aqueous solution.
When potash alum is dissolved in water, the crystal lattice breaks apart fully, releasing all of its constituent cations and anions into the medium. The balanced ionization equation can be written as follows: \[ \text{K}_2\text{SO}_4 \cdot \text{Al}_2(\text{SO}_4)_3 \cdot 24\text{H}_2\text{O} \xrightarrow{\text{H}_2\text{O}} 2\text{K}^+_{\text{(aq)}} + 2\text{Al}^{3+}_{\text{(aq)}} + 4\text{SO}_4^{2-}_{\text{(aq)}} + 24\text{H}_2\text{O}_{(\text{l})} \] Let us carefully count the total number of specific ions released from one single formula unit of the double salt:

• Number of potassium ions (\(\text{K}^+\)) = 2 (coming entirely from the \(\text{K}_2\text{SO}_4\) unit)

• Number of aluminum ions (\(\text{Al}^{3+}\)) = 2 (coming entirely from the \(\text{Al}_2(\text{SO}_4)_3\) unit)

• Number of sulfate ions (\(\text{SO}_4^{2-}\)) = \(1 + 3 = 4\) (1 from \(\text{K}_2\text{SO}_4\) and 3 from \(\text{Al}_2(\text{SO}_4)_3\))

Step 3: Compute the required ionic ratio.
The problem specifically asks for the ratio of the number of \(\text{K}^+\) ions to the number of \(\text{SO}_4^{2-}\) ions. Using our counted values: \[ \text{Ratio} = \frac{\text{Number of }\text{K}^+\text{ ions}}{\text{Number of }\text{SO}_4^{2-}\text{ ions}} = \frac{2}{4} \] Simplifying this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2, gives: \[ \text{Ratio} = \frac{1}{2} = 1:2 \] Hence, the stoichiometric ratio of \(\text{K}^+\) to \(\text{SO}_4^{2-}\) ions in an aqueous solution of potash alum is \(1:2\).