Le Chatelier’s principle states that a chemical system will respond to changes in the environment and maintain equilibrium by changing the direction of reaction. A reaction will shift to the right if the ratio of concentration of products to reactants goes down and will shift to the left if the ratio goes up. Reaction quotient, \(\displaystyle Q\), characterizes the state of a reaction in a nonequilbrium state. It is calculated the same way as equilibrium constant.

\(\displaystyle Q = \frac{[products]}{[reactants]}\)

The question is asking about a shift to the right. Reaction shifts to the right (and approaches equilbrium) when the nonequilbrium reaction has more reactants than products. This occurs when the ratio of products to reactants (\(\displaystyle Q\)) is lower than the ratio of products to reactants at equilibrium (\(\displaystyle K\)). 

Hess's Law allows us to combine the two given equation to create the equation that we need. Hess's Law also tells us that when combining equilibrium constants, the value is the product of K₁ and K₂. 

Remark: a common misconception in this type of problem is to find the sum of the values. This use of Hess's Law is only valid for thermodynamic values. 

According to Le Chatelier's principle, a change in pressure will cause a shift in order to counteract the pressure change. When the pressure of a vessel is increased, the side of the reaction with fewer gas molecules will be preferred in order to minimize contact between gas molecules. For this reaction, there are four gas molecules on the left side and two gas molecules on the right side. As a result, the products side will be preferred and more ammonia will be created.

When a system is at equilibrium, it is possible to predict how a system will respond to sudden changes using Le Chatelier's principle. In this scenario, ammonia has been removed from the system. This will cause the reaction to produce more ammonia, and proceed to the right in order to reestablish equilibrium.

High pressure will favor the side of the reaction with fewer moles of gas, which in this case is the product side. 

The reaction is exothermic, meaning that it generates heat. By removing heat from the reaction, we are constantly stressing the system towards the product side.

Finally, the reaction should occur under a lamp. This is due to the hv symbol above the double arrows, which signifies that light promotes the reaction. Note: had the hv symbol been below the double arrows, the reaction should have occurred in the dark.  

Adding \(\displaystyle NaOH\) contributes \(\displaystyle OH^-\) ions into the solution, driving the reaction to the left, and decreasing the solubility of the calcium hydroxide. This is known as the common-ion effect. The \(\displaystyle K_{sp}\) remains unchanged. This is due to the fact that only a change in temperature can bring about a change in an equilibrium constant. 

Since a strong acid is being titrated with a strong base, we can simply subtract how many moles of base have been added from how many moles of acid were originally present.

\(\displaystyle moles = (molarity)(Liters)\)

\(\displaystyle (0.2M)(0.05L) = 0.01 mol\ HCl\)

Using the same equation, we found we have added 0.0096 moles of sodium hydroxide.

\(\displaystyle (0.2M)(0.048L)=0.0096mol\ NaOH\)

Subtracting this amount from the hydrochloric acid leaves us with 0.0004 moles of acid.

\(\displaystyle 0.01-0.0096=0.0004\)

Keep in mind that we must divide this molar amount by the new volume, after the base has been added.

\(\displaystyle molarity = \frac{moles}{liters} = \frac{0.0004moles}{0.098L} = 0.0041M\)

Since this is a strong acid, we can simply take the negative log of this concentration, and are left with a pH of 2.39.

\(\displaystyle pH = -log[H^{+}] = -log[0.0041] = 2.39\)

When the molar amount of acid equals the molar amount of base, the solution has reached what is known as the equivalence point. When a strong acid is titrated with a strong base, such as in this example, the two agents will neutralize each other completely. Because the conjugate base for the acid is incredibly weak, it will not manipulate the pH. As a result, the pH at this equivalence point is 7.00.

Since hydrofluoric acid is a weak acid, we need to consider how much of the conjugate base is created when the base neutralizes the acid. We can start by determining how many moles of acid are present initially.

\(\displaystyle moles = (molarity)(liters)= (0.2M)(0.05L) = 0.01 mol\)

Using the same equation, we find that only 0.004 moles of base have been added.

\(\displaystyle (0.2M)(0.02L)=0.004mol\)

After the acid and base cancel each other out, we are left with 0.006 moles of acid and 0.004 moles of the conjugate base. This is based on the neutralization equation:

\(\displaystyle HF + NaOH \rightarrow H_{2}O + F^{-}\)

(For simplicity, the bystander sodium ion has been omitted, although it is present).

Now that we know the amount of acid and conjugate base, we can solve for the pH using the Henderson-Hasselbalch equation.

\(\displaystyle pH = pK_a + log\frac{[A^{-}]}{[HA]} = -log[7.2*10^{-4}] + log\frac{0.004}{0.006} = 2.96\)

This question is not as easy as just assuming that the pH is 7 because there are equal amounts of acid and base. The reason is because although the acid and base have neutralized each other, there is still the issue of the fluoride ion acting as a conjugate base. The equation for this reaction is:

\(\displaystyle HF + NaOH \rightarrow H_{2}O + F^{-}\)

(The sodium ion has been omitted, as it is simply a spectator ion).

By converting each amount of acid and base to moles, we can find that 0.01 moles of each has been added, neutralizing each other in the process. According to the equation, that will leave us with 0.01 moles of fluoride ions. In order to find the concentration, we simply divide this by the new volume after all of the base has been added:

\(\displaystyle molarity = \frac{moles}{liters} = \frac{0.01 mol\ F^{-}}{0.1 L} = 0.1M\)

Next, we can determine the base dissociation constant for the fluoride ion by using the equation

\(\displaystyle K_{w} = (K_{a})(K_{b})\)

\(\displaystyle K_{b} = \frac{1*10^{-14}}{7.2*10^{-4}} = 1.38*10^{-11}\)

Finally, we use an ICE table in order to determine the hydroxide concentration in the solution.

\(\displaystyle F^{-} + H_{2}O \rightarrow HF + OH^{-}\)

The fluoride ion concentration will decrease by \(\displaystyle x\), and both the hydroxide and hydrofluoric acid concentrations will increase by \(\displaystyle x\). This leaves us with the equilibrium expression:

\(\displaystyle 1.38*10^{-11} = \frac{x^{2}}{0.1}\)

\(\displaystyle x = 1.17*10^{-6}M=[OH^-]\)

Using the pH equation, we determine the new pH of the system to be 8.07.

\(\displaystyle pH=14-pOH=14-(-log[OH^-])\)

\(\displaystyle pH=14-(-log(1.17*10^{-6}))=8.07\)