A salt will dissolve in water completely, existing as the individual ions that make up the compound. These ions can be thought of as the conjugate bases and acids that result from the dissociation of the reactant acids and bases. If the conjugate acid of a weak base is present in the solution, then it will become deprotonated, releasing protons into the water and lowering the pH. To answer this question, we are looking for a salt in which one of the ions created is a conjugate acid of a weak base. Let's look at a few examples in order to find the right answer:

1. $\displaystyle NaF$ will dissociate into sodium and fluoride ions. Sodium is the metal found on sodium hydroxide. Since sodium hydroxide is considered a strong base, the sodium ions will NOT attach to any hydroxide ions in solution. Fluoride ions, on the other hand, are the conjugate base of hydrofluoric acid, a weak acid. This means that some of the fluoride ions will attach to protons in solution, effectively raising the pH. This results in a basic solution.

2. $\displaystyle LiBr$ is an example of a salt that will result in a neutral solution. Because lithium ions come from the strong base lithium hydroxide, and bromide ions come from the strong acid hydrobromic acid, neither of these ions will be involved in an acid/base reaction. This results in a solution with a pH of 7.

$\displaystyle NH_{4}Cl$ has two ions: ammonium ions and chloride ions. Chloride ions are the conjugate base of hydrochloric acid, a very strong acid. Ammonium ions, however, are the conjugate acid of ammonia, a weak base. This means that some of the ammonium ions will become deprotonated, and release protons into the solution. this results in an acidic solution.

A Brønsted-Lowry acid is an ionic compound that donates a proton, when the compound is placed in water.

HF and NaOH are the only ionic compounds of the given answer options; all the others are covalent compounds. When dissolved in water, only HF will donate protons.

$\displaystyle HF\rightarrow {\color{Red} H^+}+F^-$

Thus, HF is the only given Brønsted-Lowry acid.

An acid is classified as strong if it completely dissociates into ions in water. Some examples of common strong acids: $\displaystyle HI, H_2SO_4, HNO_3, HBr, HCl, HClO_4$

$\displaystyle HF$ is not a strong acid because it doesn't ionize completely in solution.

By definition, and Arrhenius acid will dissociate in water to release $\displaystyle H^+$. $\displaystyle HF$ will dissociate into $\displaystyle H^+$ and $\displaystyle F^-$ in solution. The increased concentration of $\displaystyle H^+$ causes a drop in pH of the solution. In general, if a hydrogen atom is bound to a very electronegative atom, like in the case of $\displaystyle HF$, fluorine tends to take the electron away from hydrogen, resulting in ions.

$\displaystyle NH_3$ or ammonia is a complicated molecule. It's a kind of molecule that we call amphoteric, meaning that it can be an acid or a base. However, because ammonia plays the role of a very weak base, it's mostly thought of as having basic qualities. It has the ability to bind with acids to create an ammonium salt or act as a proton acceptor and become ammonium (acid). 

Also, commonly, acids have the $\displaystyle HX$ format, where $\displaystyle X$ represents a halogen. In another construct, acids ideally have protons that they can donate - this can be observed in $\displaystyle H_3O$ and $\displaystyle NH_4^+$.

Commonly, bases have the $\displaystyle XOH$ format, where $\displaystyle X$ is any metallic element from the first two columns of the periodic table. This is seen with $\displaystyle Ca(OH)_2$, $\displaystyle KOH$, and $\displaystyle NaOH$. The remaining two options are essentially different "versions" of the same molecule - one just happens to be the protonated form ($\displaystyle NH_4^+$).  

While $\displaystyle NH_3$ is technically amphoteric, it's more so thought of in terms of its basic qualities. When it does act as an acid, it's a very weak acid. Ammonia can easily become ammonium, a proton donator, due to the lone pair of electrons that continue to orbit around the nitrogen center. 

The equation to use here is:

$\displaystyle (M_A)(V_A)=(M_B)(V_B)$

Here, $\displaystyle M_A$ is the molarity of the acid, $\displaystyle V_A$ is the volume of the acid, $\displaystyle M_B$ is the molarity of the base, and $\displaystyle V_B$ is the volume of the base. Don't forget to convert the volume to liters!

$\displaystyle (1.2M)V_A=(3M)(0.050L)$

$\displaystyle V_A=\frac{(3M)(0.050L)}{1.2M}=0.125L=125mL$

Below is a generic acid-base neutralization reaction:

$\displaystyle HA+BOH\rightarrow H_2O + AB$

The products are always water, and a salt. This salt is produced from the resulting ions $\displaystyle A^-$ and $\displaystyle B^+$. The $\displaystyle H$ from the acid replaces the $\displaystyle B$ from the base, and the $\displaystyle A$ from the acid replaces the $\displaystyle OH$ from the base. Since there are two replacements, acid-base neutralizations are classified as double-replacement reactions.

Arrhenius acid/base theory was created by Swedish chemist Svante Arrhenius, and is the oldest acid/base classification. According to his classification, acids are compounds that increase the concentration of $\displaystyle H^+$ ions in a solution, while bases are compounds or elements that either decrease the concentration of $\displaystyle H^+$ ions in solution or increase the concentration of $\displaystyle OH^-$ ions in a solution. The other two answers describe the Brønsted–Lowry theory of acids and bases.

For every acidic or basic solution, the product of the hydroxide ion concentration and the hydronium ion concentration will be equal to $\displaystyle 1*10^{-14}$, the dissociation constant for water. In other words:

$\displaystyle 2H_2O\rightleftharpoons H_3O^++OH^-$

$\displaystyle K_{sp}=[H_{3}O^{+}][OH^{-}] = 1*10^{-14}$

Since we are given the hydroxide ion concentration, we can determine the hydronium ion concentration using this equation.

$\displaystyle [H_{3}O^{+}][OH^{-}] = 1*10^{-14}$

$\displaystyle [H_{3}O^{+}] = \frac{1*10^{-14}}{[OH^{-}]} = \frac{1*10^{-14}}{6.6*10^{-6}} = 1.5*10^{-9}M$