[chilli]

hello again
need some help with a question, I know the answer but not quite sure how they got there:
the question is:
how much ml of H₂SO₄ C=8 we would need in order to make 1L of solution where
C of H₃O⁺ will be 2M:
I know the answer is 125ml
I tried like this:
H₂SO₄ + 2H₂O 2H₃O⁺ + SO₄^-
if the acid is 8M then H₃O⁺ will have to be 16M ratio is 1:2 then if we need final solution 2M then acid needs to be 1M
and from here I'm stuck can anyone help?

[Ben Bob2]

Are you aware of the molarity by dilution equation?

M₁V₁ = M₂V₂

Where M₁ and V₁ are the initial concentration (in mols/L) and initial volume (in L). M₂ and V₂ are the desired concentration and volume.

You're almost on the right track otherwise; [H₃O⁺] is double [H₂SO₄]

• Therefore [H₃O⁺] is 2M, so [H₂SO₄] aka M₂, will be 1M
• You are told that M₁ is 8M
• You are also told that V₂ is 1L

Now just plug in the given values, solve the equation for V₁, and convert the answer to mL.

You good now?

[chilli]

unbelievable! how cool is that!
ty very much, so useful!