To answer this question you need to use Le Chatelier’s principle. Adding sodium acetate to the solution will cause it to dissociate as follows:

\(\displaystyle NaCH_3COO \rightleftharpoons CH_3COO^- \:+\: Na^+\)

The dissociation reaction will produce more acetate ions. According to Le Chatelier’s principle, the increase in acetate ions will shift the equilibrium of the reaction (given in the question) to the left. This means that \(\displaystyle CH_3COO^-\)and \(\displaystyle H_3O^+\) will be utilized to form \(\displaystyle CH_3COOH\). This will cause a decrease in the amount of hydronium ions in solution. Recall that pH is increased when the concentration of hydrogen ions (or hydronium ions, \(\displaystyle H_3O^+\)) is decreased; therefore, adding sodium acetate will increase the pH of the solution.

Increasing acetic acid concentration will shift the equilibrium to the right and produce more hydronium ions, thereby decreasing the pH. Recall that you can never change the pKa of an acid. The pKa of acetic acid is around 4.75, and it cannot be altered. Le Chatelier’s principle only applies when there is a change in amount of aqueous or gaseous substances; liquid and solid substances will not shift the equilibrium. Changing the volume of liquid water will not change the concentration of hydronium ions.

Sulfuric acid (\(\displaystyle H_{2}SO_{4}\)) is considered a polyprotic acid because it has two ionizable protons in its molecular formula. The protons dissociate in an aqueous solution according to the acid-base equilibria below:

\(\displaystyle H_{2}SO_{4} \leftrightarrow H^{+} + HSO_{4}^{-}\)

\(\displaystyle HSO_{4}^{-} \leftrightarrow H^{+} + SO_{4}^{2-}\)

All the other acids listed in the answer choices are monoprotic.

Carbonic acid (\(\displaystyle H_{2}CO_{3}\)) is considered a polyprotic acid because it has two ionizable protons (\(\displaystyle H\) atoms) in its molecular formula. The protons dissociate in an aqueous solution according to the acid-base equilibria below:

\(\displaystyle H_{2}CO_{3} \leftrightarrow H^{+} + HCO_{3}^{-}\)

\(\displaystyle HCO_{3}^{-} \leftrightarrow H^{+} + CO_{3}^{2-}\)

The other acids are all monoprotic acids.

The equilibrium governing the dissolution of \(\displaystyle HClO_2\) in water is:

\(\displaystyle HClO_{2}(aq)\ +\ H_{2}O(l)\rightleftharpoons\ H_{3}O^{+}(aq)\ +\ HClO_{2}^{-}(aq)\)

\(\displaystyle HClO_2\) is the conjugate acid of \(\displaystyle ClO_2^{-}\). In other words, \(\displaystyle ClO_2^{-}\) is the conjugate base of \(\displaystyle HClO_2\) .

Using the relationship, \(\displaystyle K_{w}=K_{a}K_{b}\), we can calculate the K_b.

Rearrange the equation and solve:

\(\displaystyle K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0*10^{-14}}{1.1*10^{-2}}=9.1*10^{-13}\) 

The equilibrium governing the dissolution of \(\displaystyle HBrO\) in water is:

\(\displaystyle HBrO(aq)\ +\ H_{2}O(l)\rightleftharpoons\ H_{3}O^{+}(aq)\ +\ BrO^{-}(aq)\)

\(\displaystyle HBrO\) is the conjugate acid of \(\displaystyle BrO^-\). In other words, \(\displaystyle BrO^-\) is the conjugate base of \(\displaystyle HBrO\).

Using the relationship, \(\displaystyle K_{w}=K_{a}K_{b}\), we can calculate the K_b.

Rearrange the equation and solve:

\(\displaystyle K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0*10^{-14}}{2.5*10^{-9}}=4*10^{-6}\) 

A base is a substance that can accept a proton. The conjugate base of an acid is formed when the acid donates a proton. In this case, \(\displaystyle Cl^{-}\) is the conjugate base to the acid \(\displaystyle HCl\). This is because \(\displaystyle HCl\) donates a hydrogen ion to the organic molecule to form \(\displaystyle Cl^{-}\), the conjugate base.

\(\displaystyle HF\) (hydrofluoric acid) is the weakest acid. Fluoride ion is the most electronegative ion. Among the other halogens, its atomic radius is smaller, and therefore bonds more strongly with hydrogen and therefore does not completely dissociate in solution as compared to \(\displaystyle HBr\), \(\displaystyle HI\), and \(\displaystyle HCl\). Perchloric acid is a strong acid and dissociates completely in solution.

The reaction given is called a neutralization reaction because the acid and base components react to counterbalance each other making them chemically neutral. This type of reaction occurs between an acid and base to form a water and a salt. A neutralization reaction between a strong acid and a strong base react to form a neutral solution of pH 7.

We need to convert milliliters to liters:

\(\displaystyle 25.0\ mL \times\frac{1L}{1000mL}=0.0250\ L\)

We need to determine the moles of \(\displaystyle H_{2}SO_{4}\) using dimensional analysis the concentration as a conversion factor:

\(\displaystyle Moles\ of\ H_{2}SO_{4}=0.0250\ L \times \frac{0.100\ moles}{L}=0.00250\ moles\ H_{2}SO_{4}\)

Based on the chemical equation given:

\(\displaystyle 1\ mole\ of H_{2}SO_{4}=2\ moles\ of\ NaOH\)

We can used the relationship of moles of \(\displaystyle NaOH\) and moles of \(\displaystyle H_{2}SO_{4}\) as a conversion factor to determine the moles of \(\displaystyle NaOH\):

\(\displaystyle Moles\ of\ NaOH=0.00250\ moles\ H_{2}SO_{4} \times\frac{2\ moles\ NaOH}{1\ mole\ H_{2}SO_{4}}=0.00500\ moles\ NaOH\)

Concentration in molarity can be calculated using the following formula:

\(\displaystyle Molarity\ of NaOH=\frac{moles\ of\ NaOH}{Liters\ of\ solution}\)

Let's convert liters of \(\displaystyle NaOH\) to the mL:

\(\displaystyle Liters\ of\ NaOH\ solution=15.0\ mL \times\frac{1L}{1000mL}=0.0150\ L\)

Therefore, the concentration of \(\displaystyle NaOH\) is:

\(\displaystyle Molarity\ of NaOH=\frac{0.00500\ moles\ NaOH}{0.0150\ L}=0.333\ M\)

We need to convert milliliters of \(\displaystyle HNO_{3}\) to liters:

\(\displaystyle 20.0\ mL \times\frac{1L}{1000mL}=0.0200\ L\)

We need to determine the moles of \(\displaystyle HNO_{3}\) using dimensional analysis and the concentration as a conversion factor:

\(\displaystyle Moles\ of\ HNO_{3}=0.0200\ L \times \frac{0.100\ moles}{L}=0.00200\ moles\ HNO_{3}\)

 Based on the chemical equation given:

\(\displaystyle 1\ mole\ of HNO_{3}=1\ mole\ of\ NaNO_{3}\)

We can use the relationship of moles of \(\displaystyle HNO_{3}\) and moles of \(\displaystyle NaNO_{3}\) as a conversion factor to determine the moles of \(\displaystyle NaNO_{3}\):

\(\displaystyle Moles\ of\ NaNO_{3}=0.00200\ moles\ HNO_{3} \times\frac{1\ moles\ NaNO_{3}}{1\ mole\ HNO_{3}}=0.00200\ moles\ NaNO_{3}\)