help with problem 6B

[annabelchen2a]

6B.3) A careless laboratory technician wants to prepare 200.0 mL of a 0.025 m HCl(aq) solution but uses a volumetric flask of vol- ume 250.0 mL by mistake. (a) What would the pH of the desired solution have been? (b) What will be the actual pH of the solution as prepared?

I'm a little confused on how exactly I should go about solving this problem. I've done this two ways and I'm unsure of what the right way is:

my 1st way:
a) 0.025 mol HCl/0.2L * 1mol H/1mol HCl = 0.125 mol H
pH = -log(0.125) = 0.903
b) 0.025mol HCl/0.25L * 1 mol H/1mol HCl = 0.1 mol H
pH = 1

my 2nd way:
a) pH = -log (0.025M) = 1.602
b) M1V1 = M2V2 --> (0.025M)(200mL) = (250mL)(M)
M = 0.02
pH = -log(0.02) = 1.7

I'm leaning towards the 2nd way being the right way of working out this problem, but I'm not really sure why. And if I'm right, why would my first way be wrong?

[Chem_Mod]

You are right that the second method is correct! The reason that the first method is not correct is that HCl is a strong acid so it completed dissociates. Therefore [HCl]=[H3O+].

[annabelchen2a]

Okay, thank you!!

[Valerie Doan 3I]

The second method is correct! The reason why the first method you used was wrong was that you solved for moles of H+. However, the pH equation is equal to -log[H+], with [H+] being the molarity or moles/L of hydronium ions. Because you just used moles, your answer was wrong.