A chemist wants to make a solution of 3.4 M HCl. There are two solutions of HCL that he can find on the shelf. One has a concentration of 6.0 M, while the other has a concentration of 2.0 M. Which solution can the chemist use to make the desired acid?
Technically the chemist can use either; however, the 2.0M would need to be evaporated to arrive at a more concd solution and that isn't practicable.
The 6.0 M solution can be diluted to make a 3.4M soln.
If you want to calculate, use
c1v1 = c2v2, the dilution formula.
c = concn
v = volume
To determine which solution the chemist can use to make the desired 3.4 M HCl solution, we need to do a calculation based on the principles of dilution.
The formula for dilution is:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution (6.0 M or 2.0 M, in this case)
V1 = initial volume of the solution
C2 = final concentration of the solution (3.4 M, in this case)
V2 = final volume of the solution
Let's calculate the volume of the concentrated HCl solution (C1V1) needed to make the desired 3.4 M solution (C2V2) using both concentrations available.
Case 1: Concentrated HCl solution with a concentration of 6.0 M:
C1 = 6.0 M
V1 = ?
C2 = 3.4 M
V2 = 1.0 L (assuming the final volume required is 1.0 L)
Using the formula, we can rearrange it to solve for V1:
V1 = (C2V2) / C1
V1 = (3.4 M * 1.0 L) / 6.0 M
V1 = 0.57 L
In this case, for a 1.0 L final volume, the chemist would need to measure 0.57 L of the 6.0 M HCl solution and dilute it to the desired concentration.
Case 2: Concentrated HCl solution with a concentration of 2.0 M:
C1 = 2.0 M
V1 = ?
C2 = 3.4 M
V2 = 1.0 L (assuming the final volume required is 1.0 L)
Using the formula, we can rearrange it to solve for V1:
V1 = (C2V2) / C1
V1 = (3.4 M * 1.0 L) / 2.0 M
V1 = 1.7 L
In this case, for a 1.0 L final volume, the chemist would need to measure 1.7 L of the 2.0 M HCl solution and dilute it to the desired concentration.
Therefore, based on these calculations, the chemist can use the 6.0 M HCl solution to make the desired 3.4 M HCl solution by diluting it to the required concentration.