If you start with 10M 31.45% muriatic acid in one container, and in the other you have distilled water, after a week you end up with a solution of 5M muriatic acid (at what percentage now??) and 5M reagent grade HCL.

Right. 5 M hydrochloric acid is about 16.9%

So how does the molarity come into play here? Which of the two end products is stronger? Wouldn't this effectively mean I'm making 2 solutions of 5M, 15.72% muriatic acid??

Once the system is in equilibrium concentration of acid is identical in both beakers, so they have both exactly the same strength. Whether their concentration is exactly half of the initial concentration, depends on the amount of distilled water put in the second beaker.

Actually this is kind of calculations where being precise about "amount of water" is paramount.

Let's assume we start with 1.000 L of 10 M HCl. It contains 10 moles (364.6 g) of HCl. Solution density is 1.1569 g/mL, so the mass of the solution is 1157 g, of this 1157-364.6=792.4 g is water.

After the procedure we want to end with 2.000 L of 5 M acid (in separate beakers, but we can treat them combined here). Density of 5 M acid solution is 1.0819 g/mL, so the total mass of the solution needs to be 2164 g. We already have 1157 g of the solution and all HCl we need is there, so we need to add 2164-1157=1007 g of water.

Things get tricky when we try to calculate volumes of both acids - the pure one (reagent grade) and the leftover (muriatic, technical, stock, whatever you want to call it). New beaker contained 1007 g of water, now it contains 5 M HCl solution with a density of 1.0819 g/mL. How much HCl does it contain? Let's call HCl mass m

_{HCl} - then the mass of the solution is

[tex]1007+m_{HCl}[/tex]

volume of the solution is mass over density

[tex]\frac {1007+m_{HCl}} {1000 \times 1.0819}[/tex]

(1000 is a conversion factor between mL and L) and number of moles of HCl is

[tex]\frac {m_{HCl}}{36.46}[/tex]

(36.46 g/mol being molar mass of HCl).

By definition molar concentration is

[tex]C = \frac n V[/tex]

We know concentration to be 5 M, so we can write

[tex]5 = \frac {\frac {m_{HCl}}{36.46}}{\frac{1007+m_{HCl}}{1000\times 1.0819}}[/tex]

solving for m

_{HCl} yields 204 g, so mass of the reagent grade solution is 1007+204=1211 g (or 1211/1.0819=1119 mL), the other beaker contains now 1157-204=953 g (or 953/1.0819=880 mL) of the technical, 5M solution.