Question

From Le Chatlier's Principle Equilibrium Lab

Fill a second cuvette about ¾ full with 0.10 M Co(NO3)2 .6H2O.

Prepare 100 mL of a 0.010 M Co(NO3)2 . 6H2O solution. Add exactly 10.0 mL of 0.10 M Co(NO3)2 .6H2O to a 100 mL volumetric flask and dilute with deionized water to the mark.

5. Use the 10.0 ml graduated cylinder to prepare a solution of [C oCl4]2-. To the 10 mL graduated cylinder add 1.0 mL of 0.010 M Co(NO3)2 * 6H2O (prepared in step 4).   Carefully pipette 4.0 mL of 12 M HCl into the graduated cylinder so that the total volume is 5 mL.  

With this information, I have to calculate the concentration of Co(NO3)2. 6 H2O and do not know how. Thank you.

Answer 1

In step 5, you are taking 1 mL of solution prepared in step 4 and diluting it to a final volume of 5 mL

This corresponds to a 5 times dilution.

The solution you prepared in step 4 was 0.01 M Co(NO₃)₂, which gets diluted by 5 times in step 5, so its conc in step 5 cylinder becomes 0.01/5 = 0.002 M

Also, in this cylinder in step 5, Co(NO₃)₂ reacts with chloride ion to form [CoCl₄]^2-.

So conc of [CoCl₄]^2- is also 0.002 M

Hope this helps ! Revert back if you have any queries.

answered by: anonymous