ChemixaLEARN CHEMISTRY

Preparing a Dilute Solution from Stock

Easy
Problem

Describe how to prepare 250.0 mL of 0.100 M solution from a 1.00 M stock solution. What volume of stock is needed?

Given

  • Target: V2 = 250.0 mL of M2 = 0.100 M solution

  • Stock concentration: M1 = 1.00 M

  • Dilution with water adds no solute, so moles of solute are conserved.

Find

The volume of 1.00 M stock solution required.

Principle

Diluting a solution adds solvent but no solute, so the moles of solute before and after are equal: M1V1 = M2V2. Solving for the unknown volume gives the amount of stock to measure out, which is then brought up to the final volume with solvent.

Formula

Solution

  1. Write M1V1 = M2V2 with M1 = 1.00 M (stock), M2 = 0.100 M (target), V2 = 250.0 mL (target).

  2. Solve for V1: V1 = M2V2/M1 = (0.100 M x 250.0 mL) / 1.00 M = 25.0 mL.

  3. Procedure: pipet 25.0 mL of 1.00 M stock into a 250.0 mL volumetric flask and add distilled water up to the calibration mark, then mix.

Measure 25.0 mL of the 1.00 M stock and dilute with water to a total volume of 250.0 mL.

Unit check

Molarity (mol/L) times volume (mL) gives millimoles on both sides; the units match, and since both volumes use the same unit they can stay in mL — V1 comes out in mL as required.

What this means physically

Only 25.0 mL of the concentrated stock is needed because it is ten times stronger than the target — a 1:10 dilution. In the lab this is done with a pipet and a 250.0 mL volumetric flask for accuracy.

Common mistake

Students often add 250.0 mL of water to the 25.0 mL of stock, making 275 mL total. Dilution means bringing the final total volume to 250.0 mL — add water to the stock in a volumetric flask up to the mark.