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

May 18, 2014

The concentration of the solution would be 0.64 mol/L.

Method 1

One way to calculate the concentration of a diluted solution is to use the formula

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

${c}_{1}$ = 4.2 mol/L; ${V}_{1}$ = 45.0 mL = 0.0450 L
${c}_{2}$ = ?; ${V}_{2}$ = (250 + 45.0) mL = 295 mL = 0.295 L

Solve the formula for ${c}_{2}$.

 c_2 = c_1 × V_1/V_2 = 4.2 mol/L × $\left(45.0 \text{ mL")/(295" mL}\right)$ = 0.64 mol/L

Method 2

Remember that the number of moles is constant

${n}_{1} = {n}_{2}$

${n}_{1} = {c}_{1} {V}_{1}$ = 4.2 mol/L × 0.0450 L = 0.19 mol

${n}_{2} = {c}_{2} {V}_{2} = {n}_{1}$ = 0.19 mol

${c}_{2} = {n}_{2} / {V}_{2} = \left(0.19 \text{ mol")/(0.295" L}\right)$ = 0.64 mol/L

This makes sense. You are increasing the volume by a factor of about 7. So the concentration should decrease by a factor of about 7.

$\frac{4.2}{7}$ = 0.6