Chemistry · General chemistry I · Worked example
Dilute a stock solution
How would you prepare 500.0 mL of 0.300 M HCl from a 6.00 M stock solution?
Write what stays the same
The moles of HCl taken from the stock are the moles in the final solution, so M₁V₁ = M₂V₂, with 1 for the stock and 2 for the diluted solution.
Solve for the stock volume
Divide by M₁: V₁ = M₂V₂/M₁ = (0.300 M)(500.0 mL)/(6.00 M) = 25.0 mL.
Describe the procedure
Measure 25.0 mL of the 6.00 M stock with a pipet into a 500 mL volumetric flask that already holds some water, then add water to the mark and mix. With a concentrated acid, add the acid to water, not water to the acid.
Check with moles
The 25.0 mL portion holds 0.0250 L × 6.00 mol/L = 0.150 mol of HCl, and 0.150 mol in 0.5000 L is 0.300 M.
Result
Dilute 25.0 mL of the 6.00 M stock with water to a total volume of 500.0 mL.
Your turn
10.0 mL of 2.50 M NaOH is diluted to 250.0 mL. What is the new concentration?
Show the answer and explanation
0.100 M.
M₂ = M₁V₁/V₂ = (2.50 M)(10.0 mL)/(250.0 mL) = 0.100 M. The volume grew 25-fold, so the concentration fell 25-fold.
Keep exploring
Open the dilution in Solutions & concentration and change the final volume to 250.0 mL: the same 25.0 mL of stock now gives 0.600 M.
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