What volume of 2.00 M HCl must you use to prepare 100. mL of 0.500 M HCl?
(Hint: Just use the dilution equation! MiVi = MfVf)
So what's wrong with "just use the dilution equation?"
To solve for the volume of 2.00 M HCl needed, we can use the dilution equation: MiVi = MfVf.
Given:
Mi (Initial concentration) = 2.00 M
Vi (Initial volume) = ? (Unknown)
Mf (Final concentration) = 0.500 M
Vf (Final volume) = 100. mL = 0.100 L
Plugging in the known values into the equation, we can solve for Vi:
(2.00 M)(Vi) = (0.500 M)(0.100 L)
Rearranging the equation to solve for Vi:
Vi = (0.500 M)(0.100 L) / 2.00 M
Vi = (0.050 mol/L)(0.100 L) / 2.00 M
Vi = 0.005 mol
Therefore, you would need to use 0.005 L (or 5.00 mL) of 2.00 M HCl to prepare 100 mL of 0.500 M HCl.
To find the volume of 2.00 M HCl that you must use to prepare 100 mL of 0.500 M HCl, you can use the dilution equation MiVi = MfVf, where:
- Mi is the initial concentration (2.00 M)
- Vi is the initial volume that you want to find (unknown)
- Mf is the final concentration (0.500 M)
- Vf is the final volume (100 mL or 0.100 L)
Rearranging the equation to solve for Vi, we have:
Vi = (Mf * Vf) / Mi
Plugging in the given values, we have:
Vi = (0.500 M * 0.100 L) / 2.00 M
Vi = (0.050 mol/L) / 2.00 M
Vi = 0.025 L or 25 mL
Therefore, you need to use 25 mL of 2.00 M HCl to prepare 100 mL of 0.500 M HCl.