Diluting Solutions!!

First and foremost, we, the beryllium chemists, would like to welcome you back from the (never-long-enough) winterbreak! It still feels kind of hazy being the first week back...

Today, we will be talking about diluting solutions to prepare in workable solutions.

RECALL:
solution: homogeneous mixture where one substance is dissolved in another
solute: the chemical/substance being dissolved in the solvent
solvent: the substance that does the dissolving


Have you ever had a juice that said 100% juice or 100% concentration? (Concentration: amount of solute that exists in a given volume of solution)

Well that means it is consisted of a high concentration of the natural juice, without any other added liquids (besides what is stated). It will also have a stronger taste. In other words, it is made up of mostly the natural juice - which is the solute in this case - that is dissolved in this solution.
If we add water (solvent) into the concentration, it will dilute, and therefore will create a lower concentration of the juice because there is less juice that is being dissolved with the added water.
For example, adding large amounts of water would decrease (or weaken) the amount of concentrated juice. This is the process of dilution.

It is important to note that the number of moles of the solute always remains the same. The only difference is that the volume increases and consists of more water (solvent) in the less concentrated solution.
Thus,

moles of solute before = moles of solute after

In terms of molarity and volume..

­M­_1­L₁ = M₂L₂

_{The subscript 1 represents the "before" and the suscript 2 represents "after"}

Eg. A 0.230M HCl solution is diluted to a final volume of 2.50L and a concentration of 0.100 M. How would you make up the diluted solution?

Plug in the information into the equation:

(0.230M)(L₁₎ = (0.100M)(2.50L)

L₁ = (0.100M)(2.50L)/0.230M

    = 1.09L

You would take 1.09L of the concentrated HCl and add 1.41L of the solvent (2.50 - 1.09) to make up a total of 2.50L of 0.100M HCl.

Eg. 120.0 mL of 0.150M HCl is added to 200.0 mL of water. What's the new molarity?

Convert millilitres to litres.

(0.150M)(0.1200L)/ 0.3200L = new molarity

M2 = 0.0562M

Always remember to round to the desired sig figs!

Here's a video if you need further help or practice!
Goodluck! :)