# How do you do dilution calculations?

Jun 17, 2014

You use the formula ${V}_{1} {c}_{1} = {V}_{2} {c}_{2}$.

In any dilution, the number of moles of solute stays the same. You are simply increasing the amount of solvent in the solution.

You know that

Moles = $\text{litres" × "moles"/"litres}$ = volume × molarity = V× c.

So, if moles before dilution = moles after dilution,

${V}_{1} {c}_{1} = {V}_{2} {c}_{2}$.

Example 1. Calculating new concentration

A chemist starts with 50.0 mL of a 0.400 mol/L NaCl solution and dilutes it to 1000. mL. What is the concentration of NaCl in the new solution?

Solution 1

${V}_{1} {c}_{1} = {V}_{2} {c}_{2}$

${V}_{1}$ = 50.0 mL; ${c}_{1}$ = 0.400 mol/L
${V}_{2}$ = 1000. mL; ${c}_{2}$ = ?

c_2 = c_1 × V_1/V_2 = 0.400 mol/L × $\left(50.0 \text{mL")/(1000"mL}\right)$ = 0.0200 mol/L

Note: You don't have to convert the volumes to litres, but you must use the same units on each side of the equation.

Example 2. Calculating initial volume

A chemist wants to make 500. mL of 0.0500 mol/L HCl by diluting a 6.00 mol/L HCl solution. How much of that solution should she use?

Solution 2

${V}_{1}$ = 500. mL; ${c}_{1}$ = 0.0500 mol/L
${V}_{2}$ = ?; ${c}_{2}$ = 6.00 mol/L

V_2 = V_1 × c_1/c_2 = 500. mL × $\left(0.0500 \text{mol/L")/(6.00"mol/L}\right)$ = 4.17 mL

Hope this helps.