Solubility product and common ion effect

Correct answer: A. It decreases solubility

Consider a salt MX dissolving as MX(s) ⇌ M+ + X−. If a soluble electrolyte already supplies M+ or X−, the concentration of that ion increases. By Le Chatelier's principle, the dissolution equilibrium shifts to the left, so the salt dissolves less until the solubility-product condition is restored. The effect is not an increase or complete disappearance of solubility.

Correct answer: A. Solubility of the salt will decrease

Adding an ion already produced by the dissolving salt creates the common-ion effect. For MX(s) ⇌ M+ + X−, adding M+ or X− raises the ionic product and shifts the equilibrium toward solid MX. Consequently, the amount of salt that can remain dissolved decreases; some may precipitate. Evaporation of the salt or disappearance of solvent is unrelated to this equilibrium effect.

Correct answer: A. The salt is sparingly soluble

For a given salt type and at a fixed temperature, Ksp is the equilibrium product of the concentrations of its ions in a saturated solution. A small Ksp generally means that only a small ionic concentration product is supported, so the salt is sparingly soluble. Exact comparisons require attention to stoichiometry: Ksp values of salts with different ion ratios cannot always be compared directly without calculation.

Correct answer: A. Precipitate will form

The ionic product Q is calculated from the current ion concentrations, whereas Ksp describes the equilibrium limit for a saturated solution. If Q exceeds Ksp, the solution contains more ions than can remain dissolved at equilibrium. The excess ions combine to form solid until Q falls to Ksp. Therefore precipitation occurs; the solution is not unsaturated and neither solvent destruction nor gas formation follows.

Correct answer: A. Unsaturated

When Q, the ionic product, is less than Ksp, the current ion concentrations are below the equilibrium precipitation limit. The solution can still dissolve some of the salt without forming a precipitate, so it is unsaturated with respect to that salt. Saturation corresponds to Q = Ksp, while precipitation is favoured when Q > Ksp. The other options therefore do not fit the stated inequality.

Correct answer: A. Saturated

The equality Q = Ksp means that the ion concentrations satisfy the solubility-product equilibrium condition. The solution is saturated with respect to the salt: it contains the maximum equilibrium amount of dissolved ions at the stated temperature. No net precipitation is required at that instant, although undissolved solid may coexist with the solution. Q less than Ksp is unsaturated and Q greater than Ksp favours precipitation.

Correct answer: A. Keep ionic product less than solubility product

For a salt, precipitation is thermodynamically favoured when the current ionic product Q exceeds Ksp. Keeping Q below Ksp keeps the solution unsaturated, so the ions can remain dissolved without forming that solid phase. At Q equal to Ksp the solution is just saturated; raising Q above this value promotes precipitation. The remaining options either promote separation or are chemically meaningless.

Correct answer: A. Comparison of ionic product and solubility product

For a sparingly soluble salt, calculate Q from the current concentrations of its ions and compare it with Ksp. If Q is below Ksp, the solution is unsaturated; if Q equals Ksp, it is saturated; and if Q exceeds Ksp, the excess ions tend to form a precipitate until equilibrium is restored. This comparison directly tests the precipitation condition, unlike colour, smell or container height.