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

1 Answer
Apr 23, 2017

The final concentration will be "64 M KOH"64 M KOH, or "64 mol/L KOH"64 mol/L KOH.

Explanation:

Molarity, "M"M, is a measure of concentration of a solution, and represents "mol solute"/"1L solution"mol solute1L solution. Milliliters need to be converted to liters. The following is the equation needed to answer this question:

C_1V_1=C_2V_2C1V1=C2V2

where concentration is CC and volume is VV.

Given/Known Information
C_1="4.2 M KOH"="4.2 mol KOH"/"L"C1=4.2 M KOH=4.2 mol KOHL
V_1="45.0 mL"xx"1 L"/"1000 mL"="0.0450 L"V1=45.0 mL×1 L1000 mL=0.0450 L
V_2="250 mL+45.0 mL"="295 mL"xx"1 L"/"1000 mL"="0.295 L"V2=250 mL+45.0 mL=295 mL×1 L1000 mL=0.295 L

Unknown: C_2C2

Solution
Rearrange the equation to isolate C_2C2. Insert the known values and solve.

C_2=(C_1V_1)/V_2C2=C1V1V2

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