K_a represents the equilibrium constant for the acid dissociation in water, HA → H⁺ + A^–, so it is a measure of the products divided by the reactants. As this value increases, the reaction favors the products (the ions) more, meaning that the acid dissociates more. The definition of a strong acid is one that fully dissociates in water, so as K_a increases the strength of the acid increases, and the strength of the conjugate base decreases. pK_a is defined as the –log(K_a), so as K_a increases pK_a decreases. All of these answer choices are correct.

We know that pK_a is equal to –log(K_a). Thus, pK_a of acetic acid is –log(1.8 * 10^–5). This is not an easy problem to solve in your head, but there is a trick.

We know that 1 * 10^–4 > 1.8 * 10^–5 > 1 * 10^–5, and we know that –log(1 * 10^–4) = 4 and –log(1 * 10^–5) = 5. Now we can conclude that our pK_a is somewhere between 4 and 5.

Two answer choices fall in this range: 4.2 and 4.7. 

1.8 * 10^–5 is closer to 1 * 10^–5 than it is to 1 * 10^–4, so we can pick the answer closer to 5 than to 4 : 4.7.

Remember that the K_a for the acid and the K_b for the conjugate base, when multiplied will equal the autoionization of water constant (K_w).

\(\displaystyle K_{a} K_{b} = K_{w}=\small 1*10^{-14}\)

\(\displaystyle \small K_{b} = \frac{K_w}{K_a}=\frac{1*10^{-14}}{6.2*10^{-10}}\)

\(\displaystyle \small K_{b}= 1.6*10^{-5}\)

The strongest acid of the group will also have the smallest K_b value. As the weakest base (smallest K_b), compound 4 will only partially dissociate in solution because it has a fairly strong conjugate acid. Were this question asking which base was the strongest, compound 1 would be the answer, due to its large K_b value.

The base reaction will essentially be the reverse of the acid reaction. In the question, the aqueous conditions mean that the base, \(\displaystyle A^{-}\), reacts with water to give the following reaction:

\(\displaystyle A^{-} +H_{2}O \leftrightharpoons HA + OH^{-}\)

Following normal equilibrium convention, we omit water from the equation because it is a pure liquid.

\(\displaystyle K_{eq}=\frac{[product]}{[reactant]}\)

\(\displaystyle K_b=\frac{[HA][OH^-]}{[A^-]}\)

The acid dissociation constant, K_a, describes how strongly an acid tends to break apart into hydrogen ions (H⁺) and its conjugate base (A^-). The higher the dissociation constant, the stronger the acid. HF has the largest K_a of these acids, making it the strongest, and H₂CO₃ has the smallest K_a, making it the weakest.

Since hydrofluoric acid is a weak acid, an ICE table needs to be set up in order to determine the hydronium ion concentration. Since both fluoride ion and hydronium ion concentrations will increase by \(\displaystyle x\), while the acid concentration will decrease by \(\displaystyle x\), the equilibrium expression comes out to be:

\(\displaystyle K_a=\frac{[H^+][F^-]}{[HF]}\)

\(\displaystyle 7.2*10^{-4}=\frac{x^{2}}{0.3-x}\)

Note that the \(\displaystyle x\) in the denominator will have a negligible effect and can be ignored.

\(\displaystyle 7.2*10^{-4}=\frac{x^{2}}{0.3}\)

\(\displaystyle x = 0.015\)

Since \(\displaystyle x\) is equal to the hydronium ion concentration, we can calculate the pH by taking the negative log of the concentration:

\(\displaystyle -log[0.015] = 1.83\)

At the equivalence point, there are equimolar amounts of acid and base. This means that all weak acid has been neutralized, and only the conjugate base remains. Since the conjugate base of a weak acid will affect the pH, we need to use an ICE table in order to find the pH. First, we start by finding the base dissociation constant of the conjugate base, using the equation:

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

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

\(\displaystyle K_{b} = 1.39*10^{-11}\)

The balanced equation for the conjugate base dissociation is:

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

As the hydroxide ion and acid concentrations increase by \(\displaystyle x\), the fluoride ion concentration will decrease by \(\displaystyle x\). This makes the equilibrium expression:

\(\displaystyle \frac{x^{2}}{0.3-x} = 1.39*10^{-11}\)

\(\displaystyle x = 2.04*10^{-6}\)

Since this is the hydroxide concentration, we can find the pH by taking the negative log of this value, then subtracting from 14:

\(\displaystyle pH = 14 - (-log(2.04*10^{-6})) = 14-5.69 = 8.31\)

\(\displaystyle K_{w} = [H_{3}O^{+}] \cdot [OH^{-}] = 1 \cdot 10^{-14}\)

Since the product of the cation and anion is \(\displaystyle 1\cdot10^{-14}\), the only true statement is that the concentration of the cation \(\displaystyle (H_{3}O^{+})\) is the square root of this number:

\(\displaystyle \sqrt{1 \cdot 10^{-14}} =1 \cdot 10^{-7} M\)

Electrolytic cells have a negative electromotive force and require an outside energy source to power a non-spontaneous reaction. Galvanic cells, in contrast, have positive potentials and facilitate spontaneous reactions without the need of a power source.

Regardless of cell type, however, oxidation always takes place at the anode and reduction always takes place at the cathode. The flow of electrons is always from the anode to cathode.