Dilution of Solutions

– In this subject, we will discuss the Dilution of Solutions and Dilution Law.

Dilution of Solutions

– Concentrated solutions are often stored in the laboratory stockroom for use as needed.

– Frequently we dilute these “stock” solutions before working with them. 

– Dilution of solution is the procedure for preparing a less concentrated solution from a more concentrated one.

– Suppose that we want to prepare 1 L of a 0.400 M KMnO₄ solution from a solution of 1.00 M KMnO₄.

– For this purpose, we need 0.400 mole of KMnO₄.

– Because there is 1.00 mole of KMnO₄ in 1 L of a 1.00 M KMnO₄ solution, there is 0.400 mole of KMnO₄ in 0.400 L of the same solution:

– Therefore, we must withdraw 400 mL from the 1.00 M KMnO₄ solution and dilute it to 1000 mL by adding water (in a 1-L volumetric flask).

– This method gives us 1 L of the desired solution of 0.400 M KMnO₄

Dilution law 

– In carrying out a dilution process, it is useful to remember that adding more solvent to a given amount of the stock solution changes (decreases) the concentration of the solution without changing the number of moles of solute present in the solution.

moles of solute before dilution = moles of solute after dilution

– Molarity is defined as moles of solute in one liter of solution, so the number of moles of solute is given by:

– Because all the solute comes from the original stock solution, we can conclude that n remains the same; that is,

– M_i and M_f are the initial and final concentrations of the solution in molarity, respectively

– V_i and V_f are the initial and final volumes of the solution, respectively.

– Of course, the units of V_i and V_f must be the same (mL or L) for the calculation to work.

– To check the reasonableness of your results, be sure that M_i > M_fand V_f > V_i.

Solved problems on Dilution of Solutions

Describe how you would prepare 5.00 × 10² mL of a 1.75 M H₂SO₄ solution, starting with an 8.61 M stock solution of H₂SO₄.

Strategy:

– Because the concentration of the final solution is less than that of the original one, this is a dilution process.

– Keep in mind that in dilution, the concentration of the solution decreases but the number of moles of the solute remains the same.

Solution:

– We prepare for the calculation by tabulating our data:

– Thus, we must dilute 102 mL of the 8.61 M H₂SO₄ solution with sufficient water to give a final volume of 5.00 × 10² mL in a 500-mL volumetric flask to obtain the desired concentration.

Check:

– The initial volume is less than the final volume, so the answer is reasonable.

• Reference: Chemistry / Raymond Chang, Williams College /(10th edition).