Question

The increase in the solubility of Sodium halides, in water at 25°C is:

• A. NaCl > NaBr > NaI
• B. NaBr > NaI > NaCl
• C. NaI > NaBr > NaCl
• D. NaCl = NaBr > NaI

Hint:

Larger ions like iodide (I-) tend to form weaker ionic bonds with sodium, making NaI more soluble in water than NaBr and NaCl.

Answer 1

The Correct Option is C

Step 1: Rule.
For sodium halides, lattice energy drops faster than hydration energy as the halide ion size increases from Cl to Br to I.

Step 2: Apply the rule.
A lower lattice energy means the ions pull apart more easily in water, so solubility should rise from NaCl to NaBr to NaI, not fall or stay flat.

Step 3: Final Answer.
Solubility rises in the order NaI > NaBr > NaCl.

Answer 2

Another way to settle this is to recall the actual water solubility values of these three salts at 25 degrees Celsius instead of reasoning from energy terms.

1. NaCl > NaBr > NaI: Measured solubility of NaCl in water is about 36 g per 100 mL, well below NaBr and NaI, so NaCl cannot be the highest.
2. NaBr > NaI > NaCl: NaBr dissolves to about 91 g per 100 mL, but NaI dissolves even more, to about 184 g per 100 mL, so NaBr cannot outrank NaI.
3. NaI > NaBr > NaCl: With roughly 184 g per 100 mL for NaI, 91 g per 100 mL for NaBr, and 36 g per 100 mL for NaCl, the measured data lines up exactly in this order.
4. NaCl = NaBr > NaI: NaCl and NaBr are nowhere near equal (36 versus 91 g per 100 mL), and both fall short of NaI, so this order fails on the numbers.

The measured solubility values confirm the order NaI > NaBr > NaCl.

So the correct answer is NaI > NaBr > NaCl.