[onenameless]

I'm doing an experiment where i need to dilute hydrochloric acid and it has a pH of 1. Im using water to dilute it and im also using 2mL of HCl. I need to know how much water to use to obtain these pH levels: 2, 3 and 4.

What i've calculated:

10^(-1) = 0.1mol/L   --> concentration of HCl (pH of 1)
10^(-2) = 0.01mol/L  --> concentration of HCl (pH of 2)
10^(-3) = 0.001mol/L  --> concentration of HCl (pH of 3)
10^(-4) = 0.0001mol/L  --> concentration of HCl (pH of 4)

0.1mol/1000mL * 2mL = 0.0002mol

(0.0002mol) / (0.01mol/L) = 0.02L

0.02L = 20mL

20mL - 2mL = 18mL Therefore i need to add 18mL of water to HCl with a pH of 1 to obtain HCl with a pH of 2

Is what i'm doing right?

[Borek]

So far so good

[Donaldson Tan]

This rule works as long as you are not diluting a weak acid or weak base.

[sjb]

So far so good

Well, to be pedantic, it's 2ml of acid made up to 20ml of solution. Probably more or less the same thing, though

[onenameless]

Thank you! Just needed to double check and yes, it was a strong acid not a weak one.