Week 36 – Dilution

Often when you are working with chemicals in the lab, the chemicals are already in solution.  For example, imagine you need to use some sodium hydroxide, NaOH, in a chemical reaction.  You have a 1.0 L bottle of 1.0 M NaOH on the shelf, but your reaction calls for a 0.25 M solution.  What to do?  Prepare a dilution by adding solvent (in this case, water) to the solution to lower the concentration of the solute (in this case, NaOH).

For starters, we need to know the volume of 0.25 M NaOH that we need.  Let’s say we need to end up with 1.0 L of the 0.25 M NaOH.  Now we can figure this out.  We know that M = mol/L.  For our 0.25 M solution, M = 0.25 and L = 1.0.  Rearranging the equation to solve for moles and we get mol = M x L = 0.25 x 1.0 = 0.25 mol.  Therefore, we need to end up with 0.25 mol of NaOH in 1.0 L of solution.

Our stock solution of NaOH has a molarity of 1.0 M, or 1.0 mol / L.  We need 0.25 moles of NaOH.  To figure out the volume of stock solution we need to obtain 0.25 moles of NaOH, we can set up a proportion: 1.0 mol / 1.0 L = 0.25 mol / x.  Solving for x, we need 0.25 L of the stock solution.

Finally, now that we know the volume of 1.0 M NaOH stock solution needed to add to prepare our 0.25 M NaOH solution (0.25 L), we need to calculate how much water to add to make the 0.25 M NaOH solution.  We need a total volume of 1.0 L, and 0.25 L is going to come from the 1.0 M NaOH stock solution.  Therefore, we need 1.0 L – 0.25 L = 0.75 L of water.  To prepare the 1.0 L 0.25 M NaOH solution, we need to add 0.75 L of water to our flask, then add 0.25 L of the 1.0 M NaOH stock solution.

Now that you have seen the math and read the reasoning behind it in painstaking detail, let’s try a practice problem.

Question: How would you prepare 2.0 L of a 0.5 M aqueous solution of CuCl2 from a stock solution of 3.0 M CuCl2?

Answer: 0.33 L of the 3.0 M stock solution + 1.67 L of water.  Why?  A 3.0 M CuCl2 solution has 3.0 mol of CuCl2 per liter of solution.  We want to prepare 2.0 L of a 0.5 M solution, so solving M = mol/L for mol, mol = M x L, so we need 2.0 x 0.5 = 1.0 mol of CuCl2 in a total volume of 2.0 L.  Our stock solution is 3.0 mol/L so 1.0 mol = 0.33 L.  Therefore, we need to add 0.33 L of the stock solution to 1.67 L of water.

One more question: Vinegar is commonly sold as a 5% acetic acid solution (the other 95% is water).  A 100% acetic acid solution is called glacial acetic acid: glacial because the freezing point is just a few degrees below normal room temperature, so the acetic acid appears like a partially frozen glacier.

The molarity of glacial acetic acid is 17.4 M.  How would you prepare 0.5 L of 1.0 M acetic acid?

Answer: You want to prepare 0.5 L of a 1.0 M acetic acid solution.  First, calculate how many moles you need: mol = M x L so 0.5 x 1.0 = 0.5 mol of acetic acid.  Next, the stock solution of glacial acetic acid has a molarity of 17.4 M, or 17.4 mol/L.  To determine the volume of glacial acetic acid needed to obtain 0.5 mol: 0.5 mol x (1 L / 17.4 mol) = 0.029 L.  Always add acid to water, so first add 0.471 L of water (0.5 L – 0.029 L) to the flask and then add 0.029 L of glacial acetic acid.

Time to show what you know!  Complete the Week 36 – Dilution Google Form assignment and then return to Week 36 – Solution Concentration and continue working.

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