[OP] I have read that temperature and pressure dictate how much water vapor can exist in the air.
Considering water+air as a system of two components in vapor-liquid equilibrium, the number of intensive variables to specify its thermodynamic state is given by the Gibbs phase-rule for non reacting systems
\begin{align}
F &= 2 + \text{[Num. of phases]} - \text{[Num. of components]} \\
&= 2 + 2 - 2 = 2 \rightarrow F=2 \tag{1}
\end{align}
So, you are correct, and we need two variables (which you may choose $p$ and $T$, but we can choose other ones). Note that this derivation is "lumping" air in one component, but this is not true! Air is made of many components, but since this approximation yields good results, we get away with this (tiny) lie.
[OP] Air is a mixture, so shouldn’t the addition of water vapor just add to the mixture regardless the temperature and pressure? And since mixture don’t need a specific amount of each of it’s components
You can add all the water you like to the mixture, but only a part of it can exist in the gas phase. We will determine rapidly this amount accepting two hypotheses:
- The liquid and the gas phase behave as an ideal solution and ideal gas, respectively.
- The air is almost insoluble in the liquid phase, so that the molar fraction of water in the liquid phase is essentially $1$. You could relaxate this hypothesis and model air according to Henry's law, but is not important right now.
Denoting $1$ as water and $2$ as air, VLE equilibrium for 1 resumes to Raoult's law
\begin{align}
y_1 p &= x_1 P_1^\pu{sat}(T) \approx P_1^\pu{sat}(T) \\
y_1 &= \frac{P_1^\pu{sat}(T)}{p} \tag{2} = \text{function}(p,T)
\end{align}
Eq. (2) is in agreement with Eq. (1). If you want to know the molar fraction in the gas phase, you need to give me two things: $p$ and $T$. However, in everyday life, you check the news and it seems at first glance that temperature determines this. This is so, because the news may take the ambient pressure as a fixed value throughout the centuries, so we just "need" to know $T$.
Here, $P_1^\pu{sat}(T)$ is the saturation pressure of water as a pure substance. We can plot Eq. (2) if we have data of this value, which you may find in the references. I present you the results for temperatures representative of planet earth:
Pressure decreases as you go up in the atmosphere, so what we say is approximately true at sea-level. At $\pu{25} \;^\circ$C, if you could watch the molecules through atmosphere, $\ce{97} \%$ is made of air and only $\ce{3} \%$ of water. Say your atmosphere has $n$ moles, then, at most $0.03n$ moles is water. What happens to the rest? They end up accumulating in the liquid phase.
[OP] Why is there a limit even though the atmosphere is so vast?
If you think in terms of amount of substance, well, for the atmosphere $0.03n$ is a lot to count, and you are correct. However, in terms of relative proportions of water and air, the proposition "the atmosphere, in its vast extension, is scarce of water relative to air" is also true regarding what we have just said.
[OP] I have also read that when air is ‘’saturated’’ it means that it reaches the maximum amount of water vapor it can hold at that temperature and pressure
This is what I showed in the graph.
[OP] How can water dissolve in air if air is a neutral mixture and water is polar, considering that polar only dissolves polar?
This is a famous rule that has it shortcomings. But here in this case I think it applies rather well, and your statement in reverse is true: air is non-polar, water is polar, so air doesn't let water come in easily. As you can see in the Figure, this is true.
References
The saturation pressure of water was obtained as a function of temperature in the following book:
- The Properties of Gases and Liquids, J.M. Prausnitz, J. P. O'Connell, McGraw-Hill, New York, 2001.
that has an Antoine's equation functionality:
$$
\ln\left(\frac{p^\pu{sat}}{\pu{kPa}}\right) = \pu{16.3872} -
\frac{3885.79}{T/{^\circ}C + 230.170} \hspace{1 cm}
(0 \; ^\circ C \leq T \leq 200 \; ^\circ C)
$$
