[hope35]

I just wanna make sure on my calculations, because I'm always second guessing myself in these kind of calculations. So my problem is that I have 12 M HCl and I want to dilute that into 50 mL 5M HCl.

So I use the equation C1V1 = C2V2

C1V1 = C2V2
(12M HCl) x V1 = (5M HCl) x (50mL total volume)
V1 =  (5M HCl) x (50mL total volume) / (12M HCl)
V1 = 20.83 mL

So this means I would mix 20.83 mL of 12M HCl and 29.17mL of water to get a diluted solutions of 50mL 5M HCl. Right?? Can someone double check my work???

[Borek]

So this means I would mix 20.83 mL of 12M HCl and 29.17mL of water to get a diluted solutions of 50mL 5M HCl. Right??

Initially your approach was correct, but the final result is a little bit off. It is not 29.17 mL water that you should add, but you should fill up to 50 mL.

Mixing 20.83 mL of 12M HCl with 29.17 mL of water you will get 49.68 mL of 5.03M acid. Difference is small and in most cases of no practical meaning, but as volumes are not additive you can't calculate amount of water simply subtracting volume of acid from the final volume.