You have 505 mL of a 0.125 M #HCl# solution and you want to dilute it to exactly 0.100 M. How much water should you add?

Question

Assume volumes are additive.

Naomi Aronson · Accepted Answer

Approx. #100*mL# of water should be added.

#"Moles of HCl"=505*mLxx10^-3L*mL^-1xx0.125*mol*L^-1=0.0631*mol.# We require a #"concentration"# of #0.100*mol*L^-1#. But #"Concentration"="Moles of solute"/"Volume of solution"# Thus #"Volume of solution"# #=# #"Moles of solute"/"Concentration"# #=(0.0631*mol)/(0.100*mol*L^-1)=0.631*L,# or #631*mL#. And thus we dilute the original #505*mL# volume to #631*mL#. #"IMPORTANT EXPERIMENTAL NOTE:"# Ordinarily, we would never add water to an acid. Why not? Well, because if you spit in conc. acid, it spits back at you. Because here we deal with fairly dilute acids (i.e. #~0.1*mol*L^-1#), we can relax this rule a bit and add water to the acid.

Natalie Anton · Answer

undefined To dilute the 0.125 M HCl solution to 0.100 M, you need to use the formula for dilution:

$C_1V_1 = C_2V_2$

Where:
- $C_1$ is the initial concentration (0.125 M)
- $V_1$ is the initial volume (505 mL)
- $C_2$ is the final concentration (0.100 M)
- $V_2$ is the final volume (unknown, to be calculated)

Rearranging the formula to solve for $V_2$, you get:

$V_2 = \frac{C_1V_1}{C_2}$

Substituting the given values:

$V_2 = \frac{(0.125 \, \text{M})(505 \, \text{mL})}{0.100 \, \text{M}}$

$V_2 = \frac{63.125}{0.100} \, \text{mL}$

$V_2 = 631.25 \, \text{mL}$

To find the volume of water to add, subtract the initial volume from the final volume:

$Volume \, of \, water = V_2 - V_1$

$Volume \, of \, water = 631.25 \, \text{mL} - 505 \, \text{mL}$

$Volume \, of \, water = 126.25 \, \text{mL}$

So, you should add 126.25 mL of water.