A Brønsted-Lowry base is any compound that accepts protons in solution. Lewis acids and bases refer to the accepting or donating of an electron pair, respectively.

\(\displaystyle H^{+} + A^{-}\leftrightharpoons HA\)

In terms of the passage, the lone proton can be considered a proton donor and would, therefore, be a Brønsted-Lowry acid. This is not an answer choice.

The third acid-base definition is the Lewis definition, which states that acids are electron acceptors and bases are electron donors. The negative charge on the \(\displaystyle A^-\) signifies that it is a Lewis base with available electrons to donate. The proton is accepting these electrons from \(\displaystyle A^-\), and is thus acting as a Lewis acid.

Lewis acids are electron pair acceptors. Molecules and ions that have a full octet cannot act a Lewis acid, therefore \(\displaystyle CO_2\) and \(\displaystyle NH_3\) are not lewis acids. \(\displaystyle AgCl\) is very stable and insoluble and cannot accept an electron pair. \(\displaystyle NH_3\) is a well known base and has extremely weak acidity. \(\displaystyle Cu^{2+}\) is a transition metal ion. Transition metal are known to be Lewis acids because of their positive charge which gives them the ability to accept electron pairs.

One of the easiest ways of determining if a molecule is an electrophile is by the presence of a positive charge. Electrophiles are in need of electrons, therefore they are electron deficient and can be attacked by nucleophiles (compounds that are electron rich). A nucleophile is a compound that provides a pair of electrons to form a new covalent bond. Nucleophiles are electron rich and one of the easiest types of nucleophiles to recognize are ones carrying a negative charge. \(\displaystyle OH^{-}\) is the only option given that contains a negative charge and therefore is not an electrophile.

In the neutralization reaction between NaOH and HCl, NaCl salt is formed. When the solution is evaporated, this salt is left behind.

3.0g of NaCl is equivalent to 0.05mol NaCl. Since the titration is between a strong acid and a strong base, all of the NaOH in the original solution is converted to NaCl in a one-to-one ratio, meaning that mol NaCl = mol NaOH.

We now know that there was 0.05mol NaOH in the 100mL solution, so the concentration must have been \(\displaystyle \dpi{100} \small 0.5M\left ( \frac{0.05}{0.1}=0.5 \right )\).

This question requires use of the simple titration equation M₁V₁ = M₂V₂. The key is to identify that sulfuric acid has two equivalents of acidic hydrogens while NaOH has only one hydroxide equivalent. All wrong answer choices result from making this mistake or other calculation errors.

\(\displaystyle M_1V_1=M_2V_2\)

\(\displaystyle (0.125M)(0.5L)=2(0.375M)(V_2)\)

\(\displaystyle V_2=0.0833L=83.3mL\)

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:

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