Question

43 sep interpret data suppose you added some solid potassium nitrate (kno₃) to a saturated solution of kno₃ at 20°c and then warmed the mixture to 40°c. what would happen to the added kno₃? what would happen if you repeated the procedure, except with sodium chloride (nacl)?

Explanation:

For \( \text{KNO}_3 \): A saturated solution at \( 20^\circ\text{C} \) can't dissolve more \( \text{KNO}_3 \) at that temp, so added solid remains. But solubility of \( \text{KNO}_3 \) increases significantly with temperature. Warming to \( 40^\circ\text{C} \) makes the solution unsaturated (can hold more), so the added \( \text{KNO}_3 \) will dissolve.

For \( \text{NaCl} \): A saturated \( \text{NaCl} \) solution at \( 20^\circ\text{C} \) won't dissolve added \( \text{NaCl} \) initially. Solubility of \( \text{NaCl} \) increases slightly with temperature. Warming to \( 40^\circ\text{C} \) increases its solubility a small amount, so a tiny portion of the added \( \text{NaCl} \) might dissolve, but most likely, since the solubility change is minimal, most of the added \( \text{NaCl} \) will still remain undissolved (or only a very small amount dissolves).

Answer:

• For added \( \text{KNO}_3 \): At \( 20^\circ\text{C} \), it remains undissolved (saturated solution); when warmed to \( 40^\circ\text{C} \), it dissolves (due to increased solubility with temperature).
• For added \( \text{NaCl} \): At \( 20^\circ\text{C} \), it remains undissolved; when warmed to \( 40^\circ\text{C} \), only a small amount (if any) dissolves (since \( \text{NaCl} \)’s solubility has a small increase with temperature).