Approximate time for pure water pH to equilibrate in open air? I'm a complete non-chemist, so I don't know how to calculate this,  or even what to google for (my searches yielded esoteric research papers). 
I'm doing some hobby experiments, and I'd  like to know what approximate timeframe to expect for relatively pure (distilled or double distilled) water left in open air at around 24 °C/75F to reach whatever the pH equilibrium point is (I read around 5.5 somewhere). 

I'm not looking for a giant equation please,  just a quick approximate number.
Thanks
Assume you have poured your distilled water in a 1 liter container. 

We write the flux of $n_{\ce{CO2}}$ between air towards your container, as 
$$
J_{\ce{CO2}}=\frac{dn_{\ce{CO2}}}{dt}\times\frac{1}{A}=\frac{d[\ce{H2CO3}]^{*}}{dt}\times\frac{V}{A}
$$
where $n_{\ce{CO2}}$ is the number of ${\ce{CO2}}$ moles in water, $A$ the water surface and $V$ the water volume.
Assuming $\ce{CO2}$ atmospheric pressure equal to $10^{-3.5}\,atm$, we obtain the concentration $[\ce{CO2}]_{air}$ approximately equal to $10^{-5}\,M$ (the latter value obtained via $PV=nRT$).

We call $[\ce{H2CO3}]^{*}$ the sum of $[{\ce{CO2}}]_{water}$ and $[\ce{H2CO3}]_{water}$ concentrations in water: at the beginning, we set it equal to $0$. Note that $[\ce{H2CO3}]^{*}$ could be reasonably written as:

$$
[\ce{H2CO3}]^{*}\approx[{\ce{CO2}}]_{water}
$$

indicating that the majority of carbon dioxide in water is not converted into carbonic acid, being the kinetics for such a conversion very slow, as pointed out by @Nicolau Saker Neto, .

By expressing $J_{\ce{CO2}}$ as:

$$
J_{\ce{CO2}}=\frac{D_{{\ce{CO2}}}}{Z}\times ([\ce{CO2}]_{air}-[\ce{H2CO3}]^{*})
$$

with $D_{\ce{CO2}}$ the diffusion coefficient of ${\ce{CO2}}$ equal to $7.2\times10^{-4}\,dm^{2}h^{-1}$and $Z$ the thickness of superficial layer through which the exchange occurs. For the latter, we take $40\times10^{-5}\,dm$.

Putting previous equations together (with $A/V=1\,dm^{-1}$), the following is obtained:

$$
\frac{d[\ce{H2CO3}]^{*}}{dt}=1.8\times10^{-5}-1.8\times[\ce{H2CO3}]^{*}
$$
which yields:

$$
[\ce{H2CO3}]^{*}=1\times10^{-5}(1-e^{-1.8t})
$$

The graph below shows that about $2$ hours later, a plateau is established, for $[\ce{H2CO3}]^{*}= 1\times10^{-5} M$. 

By considering the acid-base equilibrium ($pK_{a}=6.3$):
$$
\text{H}_{2}\text{C}\text{O}_{3}^{*} \ce{&lt;=&gt; HCO3- + H+}
$$
we obtain a $\text{pH}=5.7$, at the plateau.


UPDATE 1: the value for $Z$ has of course a big impact on time. A larger value will increase the time needed to reach the plateau. I have used $40\times10^{-5}\,dm$, considered a "typical value" by the authors of this book (sorry in French, page 181, 3rd edition). Any more accurately derived value for $Z$ is welcome !

UPDATE 2: I used the notation $\text{H}_{2}\text{C}\text{O}_{3}^{*}$ to represent the two species $\ce{CO2}$ and $\ce{H2CO3}$, in water.