Quantitative dilution of solutions from higher to lower concentrations is a crit
ID: 535116 • Letter: Q
Question
Quantitative dilution of solutions from higher to lower concentrations is a critical skill in this lab course. You have already diluted HCl from higher to lower concentrations using two different approaches requiring slightly different calculations. For example one procedure could use a 100.00 mL (V2) volumetric flask and a 10.00 mL (V1) volumetric pipet or pipettor to prepare a 0.500 M (M2) solution of HCl. A 10.00 mL volumetric pipet or pipettor is used to transfer 5.00 M HCl (M1) to a 100.00 mL (V2) volumetric flask and then the flask is filled exactly to 100.00 mL with distilled water and mixed well. The same dilution can be done using 5.00 mL (V1) and a 50.00 mL (V2) volumetric flask if less solution is needed. One way to make different concentrations of HCl (M2) in a 100.00 mL (V2) volumetric flask from a stock 5.00 M (M1) solution would be to use different volumetric pipets. What would be the molar (M2) of the 100.00 mL (V2) of dilute HCl if you used a 70.00 (V1) mL volumetric pipet? Less quantitative dilutions can also be done without a volumetric flask, The precision of concentration is dependent on how precise the volumes are transferred. For example a dilute solution of HCl can be made using a precise amount of 5.00 M HCl and distilled water. In this case the total volume of dilute HCl (V2) is the total volume of 5.00 M HCl plus the volume of water. Determine the volume of water and the volume (V1) of 5.00 M (M1) HCl in mL needed to prepare 5.00 mL (V2) of 4.20 M (M2) HCl solution: Volume of 5.00 M HCl == > Volume of distilled water delivered using a graduated cylinder, Mohr pipet, or pipettor == >Explanation / Answer
We know that
V1M1 = V2M2
where V and M are the volume and molarity of the solution before and after dilution.
Given M1 = 5M
V2 =100 mL and V1 = 70 mL
Thus M2 = 70*5/100 = 3.5 M
Second part
Here M1=5M
V2= 5mL
and M2=4.2 M
V1= 4.2*5/5 = 4.2 mL stock HCl and 0.8 mL water.