# If "250 mL" of water is added to "45.0 mL" of a "4.2 M KOH" solution, what will be the final concentration?

Apr 23, 2017

The final concentration will be $\text{64 M KOH}$, or $\text{64 mol/L KOH}$.

#### Explanation:

Molarity, $\text{M}$, is a measure of concentration of a solution, and represents $\text{mol solute"/"1L solution}$. Milliliters need to be converted to liters. The following is the equation needed to answer this question:

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

where concentration is $C$ and volume is $V$.

Given/Known Information
${C}_{1} = \text{4.2 M KOH"="4.2 mol KOH"/"L}$
${V}_{1} = \text{45.0 mL"xx"1 L"/"1000 mL"="0.0450 L}$
${V}_{2} = \text{250 mL+45.0 mL"="295 mL"xx"1 L"/"1000 mL"="0.295 L}$

Unknown: ${C}_{2}$

Solution
Rearrange the equation to isolate ${C}_{2}$. Insert the known values and solve.

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

C_2=(4.2"mol KOH"/"L"xx0.0450color(red)cancel(color(black)("L")))/(0.295color(red)cancel(color(black)("L")))="64 mol KOH"/"L"="64 M KOH" rounded to two significant figures