Homework Help

ConcentrationCo (II) ion is available as a stock 0.1880 M solution. A student pipettes...

user profile pic

faithlovehope | Student | Honors

Posted April 13, 2012 at 7:36 AM via web

dislike 1 like

Concentration

Co (II) ion is available as a stock 0.1880 M solution. A student pipettes 5, 10, 15 mL of the stock in a 25mL volumetric and dilutes it to the mark. What is the concentration of each solution?

1 Answer | Add Yours

user profile pic

mlsiasebs | College Teacher | (Level 1) Associate Educator

Posted April 13, 2012 at 2:56 PM (Answer #1)

dislike 1 like

When we create a dilute solution, we are adding more solvent but the moles of solute doesn't change.  As a result, if we find the moles of the solute present in the original solution, we can determine the concentration of the dilute solution by dividing by the new (increased) volume.

The equation we can use to relate these is

M1*V1 = M2*V2

Where M1 and M2 are the initial and final molarities, respectively and V1 and V2 are the volumes.  While molarity has units of mol/L, we don't have to change the volumes to liters as long as they have the same units (i.e. both mL)

Since we have a stock solution with M1 = 0.1880 wiuth a volume of 5 mL, we can set up the dilution equation knowing that V2 will be the final volume.  A volumetric flask is designed to measure a specific volume so we will add enough solvent to fill to that volume.

M1*V1 = M2*V2

(0.1880)(5 mL) = M2(25 mL)

M2 = 0.0376

This is a reasonable answer because we expect the concentration to decrease as we increase the volume because the moles of solute remains the same.  We can do the same calculation with each of the remaining solutions (10 and 15 mL of the stock solution).

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes