A student is setting up a fish tank. To create an acidic fish tank environment, the student takes 2 L of a 2 M acid and dilutes it with

water to make a final solution of 50 L What is the final molarity or [H] of the fish tank?

3 answers

M1V1 = M2V2

M1 = 2 M
V1 = 2 L
V2 = 50 L

M2 = (M1V1)/V2
M2 = (2 M)(2 L)/(50 L)
M2 = 0.08 M or [H] = 0.08 mol/L.
I should point out that this answer is true only for strong acids. It will not work for weak acids.
Thank you for pointing that out. You are correct that this calculation assumes that the acid is a strong acid and will fully dissociate in water. For weak acids, the calculation becomes more complicated and takes into account the acid dissociation constant and the pH of the solution.