Use an ICE table and the  equation to solve.

Initial                                

Change                

Equilibrium               

For this question, we're given the acid dissociation constant for a reaction that occurs at a given temperature. We're asked to find the standard free energy change for the reaction.

First, we're going to need to use an equation that relates standard free energy changes with an equilibrium constant, which is shown as follows.

With regards to the temperature, we will need to convert the units given in the question stem into units of Kelvin.

Knowing that  is the ideal gas constant, we have all the information we need to solve for the value of .

Acid dissociation constant which is denoted as  is the equilibrium constant for the ionization of an acid. Therefore, the numerator contains the product of the concentrations of the substances on the product side of the chemical equation. The denominator contains the product of the concentrations of the substances on the reactant side of the chemical equation.  is omitted in the acid dissociation constant expression because as the solvent it is in excess and therefore the change in its concentration is negligible in comparison to the other substances in solution.

Recall Le Chatelier's principle: A chemical system at equilibrium will shift in the direction that minimizes the disturbance to the system.

Before the addition of , the system is in equilibrium, meaning the reaction quotient is equal to the equilibrium constant; .

However, after the addition of , a product, the reaction quotient is now less than the equilibrium constant; .

In order to maintain equilibrium, the reaction will then shift left to favor the reactants in this chemical system.

Recall Le Chatelier's principle: A chemical system at equilibrium will shift in the direction that minimizes the disturbance to the system.

Before the addition of , the system is in equilibrium, meaning the reaction quotient is equal to the equilibrium constant; .

However, after the addition of , a reactant, the reaction quotient is now less than the equilibrium constant; .

In order to maintain equilibrium, the reaction will then shift right of favor the formation of more products.

Since the equation is exothermic, you can think of heat as another product of the reaction:

Apply LeChatelier's principle to this equation.

If the temperature at which the reaction is decreased, that is akin to decreasing the amount of product made, thus causing the reaction to shift towards the products.

Start by writing out the equation for the dissolution of barium fluoride.

Next, write out the chart to keep track of concentrations of each ion.

In the chart,  is the amount of barium fluoride that dissolves. Since we are putting barium fluoride in a solution of  sodium fluoride, the initial concentration of the fluoride ion is . Since there is already some fluoride in the solution, we should expect that the molar solubility of barium fluoride should be less than its molar solubility when dissolved in water.

Now, write the equilibrium expression using the chemical equation:

Plug in the given  and solve for .

At this point, use a graphing calculator to solve for .

The molar solubility of barium fluoride in sodium fluoride is .

In this question, we're shown an equilibrium expression for a precipitation reaction in solution. We're asked to identify a compound that will help calcium precipitate out of solution.

To identify a compound that will precipitate calcium out of solution, we need to consider the equilibrium expression shown in the question stem. Applying Le Chatlier's principle, we'll need to consider the direction each of the answer choices will push the reaction in.

Adding  will make the solution more acidic. As a result, there will be a decreased amount of . Consequently, the reaction will shift to the right and there will be more calcium in solution.

Adding  will drive up the concentration of calcium, which will drive the reaction to the left. Even though this will cause calcium to precipitate out of the solution, it nonetheless defeats the purpose because the amount of dissolved calcium in solution is not decreasing.

Adding  to the solution will have no effect on the reaction equilibrium because neither of the ions appears in the expression.

Adding  will cause the solution to become more basic, with an increased amount of . Consequently, the reaction equilibrium will be driven toward the left. This will, in turn, cause calcium to precipitate out of solution in the form of the solid .

The equilibrium given tells us how the solid dissolves in solution:

However, the solubility product constant (Ksp) given tells us the degree by which a solid dissolves in solution. The larger the Ksp, the more soluble a substance is in water. Writing this expression follows the same rules as other equilibrium constant expressions. Therefore solids and water (when it is the solvent) are omitted from this expression. You must raise the concentration of the substances involved to the power of its coefficient.

For the chemical reaction given, the Ksp is:

The equilibrium given tells us how the solid dissolves in solution:

However, the solubility product constant (Ksp) given tells us the degree by which a solid dissolves in solution. The larger the Ksp, the more soluble a substance is in water. Writing this expression follows the same rules as other equilibrium constant expressions. Therefore solids and water (when it is the solvent) are omitted from this expression. You must raise the concentration of the substances involved to the power of its coefficient.

For the chemical reaction given, the Ksp is: