# Question #ec0eb

Feb 8, 2015

This is a very simple dilution problem, all you have to do is apply the formula

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

You start with $\text{5.0 mL}$ of a $\text{3.0 M NaOH}$ solution; these values will be ${V}_{1}$ and ${C}_{1}$. ${V}_{2}$ will be the final volume of the solution, which in your case is $\text{50.0 mL}$. Therefore,

${C}_{1} {V}_{1} = {C}_{2} {V}_{2} \implies {C}_{2} = {C}_{1} \cdot {V}_{1} / {V}_{2}$

${C}_{2} = \text{3.0 M" * ("5.0 mL")/("50.0 mL") = "0.3 M}$

This is the same is diluting the solution by a factor of 10. The volume is ten times bigger, so the concentration must be ten times smaller.

If, for instance, you were asked to determine the molarity for a final volume of $\text{1.0 mL}$, you would have had

${C}_{2} = {C}_{1} \cdot {V}_{1} / {V}_{2} = 3.0 \cdot \frac{5.0}{1.0} = \text{15 M}$

The final volume is five times smaller, so the concetration must be five times greater.