9. You are asked to measure out 0.1960g of ferrous ammonium sulfate hexahydrate

Question

9. You are asked to measure out 0.1960g of ferrous ammonium sulfate hexahydrate to produce a 0.0050 M solution for the primary solution for a calibration curve, which will be diluted down as in question 7. What if you are unable to measure out 0.1960g but instead were only able to measure 0.1749g? What can you do so that you can use this solution, as is. There is no more material to start over and you do not have time to order any. You must dilute this to 100 ml. 10. When using a class A pipette that is calibrated TD, do you let the solution run out of the pipette until it stops on its own or do you blow out the pipette with your pipette bulb?

Explanation / Answer

Ans. #9. Given-

            Concentration of primary standard = 0.0050 M

            Primary standard is diluted “as in question 7”    

            Available mass of salt (NH4)2Fe(SO4)2·6H2O = 0.1749 g

# The dilution in “question 7” is not mentioned. Let the concertation of finally diluted salt after diluting the primary standard as in question 7 is 0.0005 M.

Note: You can use the exact value mentioned in question and follow the step 1 and 2 as provided below.

# The procedure to proceed with 0.1749 g salt -

# Step 1: Prepare 100.0 mL primary standard using available 0.1749 g salt

Moles of (NH4)2Fe(SO4)2·6H2O = Mass / Molar mass

= 0.1749 g / (392.125 g/ mol)

= 4.460312 x 10-4 mol

Given, volume of primary standard to be made = 100.0 mL = 0.100 L

Now,

Molarity of available primary standard = Moles of salt / Volume of soln. in L

                                                = 4.460312 x 10-4 mol / 0.100 L

                                                = 0.004460312 M                          

Step 2: Let the molarity of salt after dilution in question 7 = 0.0005 M and you have to prepare 100.0 mL of it.

Using-             C1V1 (primary standard) = C2V2 (final soln. after dilution in question 7)

            Or, 0.004460312 M x V1 = 0.0005 M x 100.0 mL

            Or, V1 = (0.0005 M x 100.0 mL) / 0.004460312 M

            Hence, V1 = 11.21 mL

Therefore, required volume of available primary standard = 11.21 mL

# Transfer 11.21 mL of available primary standard to a clean 100-mL standard volumetric flask. Make the final volume up to the mark with distilled water. It is the desired solution as after dilution in question 7.

#10. Given- the pipette is class A, calibrated TD (To Deliver).

The TD pipettes deliver the marked volume spontaneously. There is no need to “BLOW” it with pipette bulb or mouth.

So, correct option is “let the solution run out of the pipette until it stops on its own”

# The pipettes marked TC (To Contain) are blown to dispense the exact desired volumes.

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