Question

Hello, in the book: Quantitative Chemical Analysis (9th Edition)

Ch. 18 question number 24. All parts I need the steps/answers to check. Thanks.

Preparing standards for a calibration curve.

Part A) How much ferrous ammonium sulfate (Fe(NH4)2(SO4)2 * 6H2O, FM 392.15) should be dissolved in a 500-mL volumetric flask with 1M H2SO4 to obtain a stock solution with 1000 microgram Fe/mL?

Part B) When making, stock solution (a), you weighed out 3.627g of reagent. What is the Fe concentration in microgram Fe/mL?

Part C) How would you prepare 250mL of standard containing ~,1,2,3,4,5,7,8, and 10 (+-20%) microgram Fe/mL in 0.1 M H2SO4 from stock solution (b) using only 5mL and 10mL Class A pipets, only 250mL volumetric flasks, and only two consecutive dilutions of the stock solution? For example, to prepare a solution with ~4 microgram Fe/mL, you could first dilute 15mL ( = 10+5 mL) of stock solution up to 250 mL to get ~(15/250)(1000 microgram Fe/mL) = ~60microgram Fe/mL. Then dilute 15 mL of the new solution up to 250 mL again to get ~(15/250)(60 microgram Fe/mL) = ~3.6 microgram Fe/mL.

Answer 1

Given data: Molecular mass of FAS = 392.15g, Volume = 500ml.

392.15g of FAS contains 56g Fe

x g salt  = 1g Fe

$\therefore x=\left (\frac{1}{56} \right )392 = 7g$

thus 7g of FAS contains 1g of Fe

to prepare 500ml of 1000 microgram/ml (or 1mg/ml) solution, = 1X500 = 500mg

thus 500mg of Fe is needed.

As 7g of FAS is equivalent to 1g, to get 500mg 3.5g of FAS is needed.

Thus by dissolving 3.5g of FAS in 500ml of distilled water, 1000microgram/ml solution can be prepared.

b. from the above calculation it is clear that 7g of FAS contains 1g of iron

Thus 3.627g of FAS contains

$\therefore x=\left (\frac{1}{7} \right )3.627 = 0.518g = 518mg = 518X10^{3} microgram$

thus 518X10³ microgram of Fe is dissolved in 500ml thus the concentration would be

$=\left (\frac{518X10^{3} }{500} \right ) = 1036microgram/ml$