A strong electrolyte is a compound that will conduct electricity while in solution. This is because it creates many charged ions in the solution. Because we want a weak electrolyte, the answer will be the compound that does not form many ions when in solution. NH₃ is a weak base which does not dissociate completely. As a result, it creates few ions in solution, making it a weak electrolyte.

The other answer choices represent either ionic compounds or strong acids, and will dissociate completely in solution.

An electrolyte is a substance that creates ions when in solution. These compounds conduct electricity very well by donating ions to an aqueous solution. The worst electrolyte will be the compound that creates the smallest amount of ions in solution. Silver chloride is relatively insoluble in solution, meaning that it will make the smallest amount of ions out of the given options.

Van't Hoff factor can also be useful in determining electrolyte effectiveness, but is irrelevant in this particular question.

Salts are a type of strong electrolyte, as are strong acids and strong bases. Strong electrolytes break up completely into ions (or "ionize") when dissolved in water. A solution of sodium cyanide would contain almost entirely sodium and cyanide ions, and no salt molecules.

When 0.5M of NaF is added, it increases the amount of fluoride ions in the solution. As a result, we can expect the solubility of the salt to decrease (a result of the common ion effect).

To determine by how much the solubility will decrease, we must use the solubility product constant expression with the newfound fluoride concentration. The expression for the equation is  . Remeber that pure solids and liquids (such as solid barium flouride are not included in this expression).

Since there is a starting concentration of 0.5M for the fluoride, the predicted increases in the ion concentrations for Ba²⁺ and F^- are now X and (0.5 + X) respectively. We can plug these variables into out equation.

Since the value of X will be much smaller than 0.5, it is acceptable to simplify the fluoride portion of the equation to just (0.5)².

When the solution is saturated, we can determine the solubility of the salt in the solution. We can determine the solubility by using the solubility product constant and comparing it to the amount of ions predicted to be created in the dissolving of the salt. Since the dissolving of BaF₂ will yield 1 Ba²⁺ ion and 2 F^- ions, we can express their increases as X and 2X respectively. Plugging this in to the equation for the solubility product constant yields the calculations below.

Scandium hydroxide dissociates according to the following reaction.

The  for this dissociation is written as follows. Note that the solid is left out of the equation.

If  is the amount of  that forms when  dissociates, then  is the amount of  formed, based on stoichiometry.

Solving this equation gives us the value of .

This value is the solubility of scandium hydroxide.

Though we are asking about solubility in a strong acid and in a strong base, this is not an acid-base chemistry question.

Instead, we are dealing with solubility, which is directly affected by the solubility product constant. The solubility product constant for each answer choice is given below.

Hydrochloric acid will contribute to hydrogen and chloride ion concentrations, while sodium hydroxide will contribute to sodium and hydroxide ion concentrations. The solubility product constant for hydroiodic acid incorporates the hydrogen ions, while the product constant for lead (II) hydroxide incorporates the hydroxide ions.

Increasing the ion concentration results in the common ion effect. Compounds are less soluble in solutions already containing ions that contribute to their solubility product constant. We can thus conclude that lead (II) hydroxide will be less soluble in sodium hydroxide because the sodium hydroxide contributes a common ion.

The K_sp is a standard equilibrium constant for the dissolution reaction of a solute. As such, it follows the standard formula for equilibrium constants, and excludes pure liquids and solids (like water and calcium sulfate!).

When writing the solubility product for a reaction, it is important to remember that solid and liquid compounds are not included in the expression. As a result, barium fluoride would not be included, because it is a solid. Just like the law of mass action, coefficients are the exponents in the expression.

Since only the products are used in the expression (the reactant is a pure solid), it is written as .

A saturated solution is a solution that is at equilibrium. The ionization product is equal to the K_sp in a saturated solution, and the rate of the forward reaction is equal to the rate of the reverse reaction. In an unsaturated solution, the solute continues to dissolve until equilibrium is reached. In a supersaturated solution, salt will "crash out," or precipitate, from solution.