Question

A laboratory experiment requires 4.8 L of a 2.5 M solution of sulfuric acid (H2SO4), but the only available H2SO4 is a 6.0 M stock solution. How could you prepare the solution needed for the lab experiment? Show all the work used to find your answer.

Answer 1

v₁=4.8 l
c₁=2.5 moles/l
c₀=6.0 moles/l

v₀-?
v(H₂O)-?

n(H₂SO₄)=v₁c₁=v₀c₀

v₀=v₁c₁/c₀

v₀=4.8·2.5/6=2.0 l

v(H₂O)=v₁-v₀

v(H₂O)=4,8-2.0=2.8 l

you need to take 2.0 liters of solution (6.0 M H₂SO₄) and 2.8 liters of water

Answer 2

Answer:

Take 2.0 L of the H2SO4 stock solution of 6.0 M and dilute it to 4.8 L

Explanation:

For The exercise we are going to need the next equation:

(Final concentration)(final volume) = (initial concentration)(initial volume) or CfVf=CiVi

The final solution is the one required to the experiment, the one with a concentration of 2.5 M and a volume of 4.8 L, and the initial solution is the one available of concentration 6.0 M. So, we need to know the volume of the initial solution we need to take to prepare the require solution. For the equation above we have:

$Vi=\frac{(Cf)(Vf)}{Ci}$

Replacing the values:

$Vi=\frac{(2.5M)(4.8L)}{6.0M}=2.0L$

Then, to prepare the solution needed for the experiment we must take 2.0 L of the H2SO4 stock solution of 6.0 M and dilute it to 4.8 L of solution, that is to say, add 2.8 L of water, so to get a 2.5 M H2SO4 solution.