Why is NaCl less soluble in water?

But for sodium chloride, the hydration energy of sodium chloride is greater than its lattice energy. Hence, it is highly soluble in water. And the size of sodium ions is smaller than potassium ions. And when the sodium chloride is mixed with water, due to the covalent bonds, the salts will dissolve in water.

Will NaCl be soluble in water?

Table salt, or sodium chloride (NaCl), the most common ionic compound, is soluble in water (360 g/L). Recall that NaCl is a salt crystal composed not of discrete NaCl molecules, but rather of an extended array of Na+ and Cl- ions bound together in three dimensions through electrostatic interactions.

Does the presence of a common ion increase or decrease the solubility of the insoluble salt?

Common Ion Effect on Solubility Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product.

Is NaCl or KCl more soluble?

NaCl is more soluble in water than KCl. The solubility of NaCl is 359 g/L at 200C and the solubility of KCl is 344 g/L at 200C.

Why does NaCl dissolve better than KCl?

Sodium ion has smaller size than potassium ion.So sodium ion can more easily form a hydration sphere around it than potassium.So NaCl is more soluble than KCl.

Is NaCl a sparingly soluble salt?

Recall that NaCl is highly soluble in water. Strategy: Write the balanced equilibrium equation for the precipitation reaction and the expression for Ksp. Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product (Q).

Does the presence of a common ion increase or decrease the solubility of the insoluble salt sup port your answer with evidence from Model 1?

Does the presence of a common ion increase or decrease the solubility of the insoluble salt? Sup- port your answer with evidence from Model 1. The presence of a common ion decreases the solubility of the insoluble salt. The change in concentration of calcium ion to reach equilibrium is 0.11 M.