Diluting Solutions

"Hi, fellow students... unfortunately, the blogger today had an accident with her experiment involving diluting solutions! In the mean time, I will be teaching you guys all about this lesson because obviously I know more than her to start with. She made many dumb mistakes in preparing for workable solutions indeed!"

Diluting solutions is actually a simple math involving the formula we have learned....Molarity M = mole/ litre

The KEY CONCEPT of diluting solutions is:

[MOLES of Solutes is constant before, after and forever...]

Moles of solute b4 = moles of solute after

How to get moles?

M x L !

thus, M1L1 = M2L2

(careful, 2 meaning the second solute, not times 2)

I just don't understand how she could not do this simple calculation to get the solution right...keke...I am sure all of you can do the following examples as fast as I  can do.

Eg. I have 2 Litre of 10 M of Hcl, and I need 0.5 Litre of 1 M Hcl.

Consider if I need all of 10 M of Hcl? No..

I need to solve for how much Volume to start with ...L1

M1L1 = M2L2

10 M x L1 = 1M x 0.5L

L1 = 0.05 L

So if I need 0.05 L of the 10 M Hcl solution to make 0.5 L of 1M Hcl... I would take 0.05 L of concentrated Hcl and add how much water?

0.5 - 0.05 = 0.45 L

These are useful practices to get familiar with the calculation:

Calculate the new concentration (molarity) if enough water is added to 100mL of 0.25M sodium chloride to make up 1.5L.
 
note: here V = the volume in Litre!

• M₂=(M₁V₁) ÷ V₂
• M₁ = 0.25M
• V₁ = 100mL = 100 ÷ 1000 = 0.100L (volume must be in litres)
• V₂ = 1.5L
• [NaCl(aq)]_new = M₂ = (0.25 x 0.100) ÷ 1.5 = 0.017M
      (or 0.0.017 mol/L or 0.0.17mol L^-1)

Calculate the volume to which 500mL of 0.02M coppper sulfate solution must be diluted to make a new concentration of 0.001M.

• V₂=(M₁V₁) ÷ M₂
• M₁ = 0.02M
• V₁ = 500mL = 500 ÷ 1000 = 500 x 10^-3L = 0.500L (since there are 1000mL in 1L)
• M₂ = 0.001M
• V(CuSO₄)_new = V₂ = (0.02 x 0.500) ÷ 0.001 = 10.00L

Thank you for keeping up with me so far... now relax ....