Describe how to prepare \(1.00 \mathrm{~L}\) of \(0.646 \mathrm{M} \mathrm{HCl}\) solution, starting with a \(2.00 \mathrm{M} \mathrm{HCl}\) solution.

Short Answer

Expert verified
To prepare a \(1.00 \mathrm{~L}\) of \(0.646 \mathrm{M} \mathrm{HCl}\) solution, one would need to take \(0.323 \mathrm{L}\) (or \(323 \mathrm{mL}\)) of the \(2.00 \mathrm{M} \mathrm{HCl}\) solution and then dilute it with water until the total volume of the solution becomes \(1.00 \mathrm{L}\).

Step by step solution

01

Identify the known values

In this case, we know the initial molarity \(M1 = 2.00 \mathrm{M}\), final molarity \(M2 = 0.646 \mathrm{M}\) and final volume \(V2 = 1.00 \mathrm{L}\). The initial volume \(V1\) for the more concentrated solution is what we need to find.
02

Apply the dilution formula

Apply the formula \(M1V1 = M2V2\) by substituting the known values into it. This would look like \(2.00 \mathrm{M} * V1 = 0.646 \mathrm{M} * 1.00 \mathrm{L}\)
03

Solve for \(V1\)

To get the volume \(V1\), you would then divide both sides of the equation by \(2.00 \mathrm{M}\). This yields: \(V1 = (0.646 \mathrm{M} * 1.00 \mathrm{L}) / 2.00 \mathrm{M}\)
04

Calculate the Volume

After performing the division, the resulting volume \(V1\) is therefore the amount of the \(2.00 \mathrm{M} \mathrm{HCl}\) solution required to make \(1.00 \mathrm{L}\) of a \(0.646 \mathrm{M} \mathrm{HCl}\) solution.

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