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

Question

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

Answer 1

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.

Answer 2

I should point out that this answer is true only for strong acids. It will not work for weak acids.

Answer 3

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.