I'm confused about this ratio. Help

So, I have some muriatic acid that I need to dilute to a ratio of one part acid to six parts water. This would be really easy if I had 100% muriatic acid, but that would be really dangerous, so it's already diluted to 31.45%, which is still pretty dangerous. Anyway, I don't need to be extremely exact, so we can just say that right now it's at 31% and I need to get it down to 14% (which is 1/7). And by the way, I know you pour the acid into the water. But can anyone give me some ideas on measurements of acid and water? The finished mixture would preferably stay under 500 ml. Thanks.

Dilution problems are readily solved using

C1V1 = C2V2. Let's say you want to make 100 mL of the 14% acid.

C1 = concentration of original acid = 31%
V1 = volume of original acid = ?
C2 = concentration of diluted acid = 14%
V2 = volume of diluted acid = 100 mL

(31%)(V1) = (14%)(100 mL)
V1 = (14%)(100 mL) / (31%) = 45 mL

So you would take 45 mL of the 31% acid and add it to 55 mL of water (total volume = 100 mL) with plenty of mixing. Or 90 mL of 31% acid + 110 mL of water if you want twice as much.

Let`s assume that 100 mL muriatic acid contains 31.45 g HCl.
You require 500 mL of 14 g / 100 mL which is 5 * 14 g = 70 g.
70 / 31.45 * 100 mL = 222.6 mL. If you add some water to the
container say c. 200 mL then carefully transfer 226.6 mL acid
mixing all the while then top up to the 500 mL mark with water
you`ll have 500 mL 14% (w/v) HCl.