# What would be the concentration of a solution made by diluting 45.0 mL of 4.2 M KOH to 250 mL?

Aug 10, 2014

The concentration would be 0.76 mol/L.

The most common way to solve this problem is to use the formula

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

${c}_{1}$ = 4.2 mol/L; ${V}_{1}$ = 45.0 mL
${c}_{2}$ = ?; ${V}_{2}$ = 250 mL

c_2 = c_1 × V_1/V_2 = 4.2 mol/L × $\left(45.0 \text{mL")/(250"mL}\right)$ = 0.76 mol/L

This makes sense. You are increasing the volume by a factor of about 6, so the concentration should be about ¹/₆ of the original (¹/₆ × 4.2 = 0.7).