Preparing standards for a calibration curve.

(a) How much ferrous ammonium sulfate$\left(\mathrm{Fe}{\left({\mathrm{NH}}_4\right)}_2{\left({\mathrm{SO}}_4\right)}_2\cdot 6H_2\mathrm{OFM}392.15\right)$should be dissolved in a 500mL volumetric flask withto obtain a stock solution with$\mathbf{1000}\mathit{\mu }\mathbf{gFe}\mathbf{/}\mathbf{mL}$?

(b) When making stock solution (a), you weighed out 3.627 g of reagent. What is the Fe concentration in?

(c) How would you prepare 250mL of standard containing containing $\mathbf{~}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{4}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{7}\mathbf{,}\mathbf{8}\mathbf{and}\mathbf{10}\mathbf{}\mathbf{in}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{1}{\mathbf{MH}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$infrom stock solution (b) using only 5- and 10-mL Class A pipets, only 250mL volumetric flasks, and only two consecutive dilutions of the stock solution? For example, to prepare a solution with $\mathbf{~}\mathbf{4}\mathit{\mu }\mathbf{gFe}\mathbf{/}\mathbf{mL}\mathbf{\text{,}}$, you could first dilute 15mL(=10+5mL) of stock solution up to 250 mL $\mathbf{mL}\mathbf{\text{to get}}\mathbf{~}\left(\frac{15}{25}\right)\mathbf{(}\mathbf{1000}\mathit{\mu }\mathbf{gFFe}\mathbf{/}\mathbf{mL}\mathbf{)}\mathbf{=\sim }\mathbf{60}\mathit{\mu }\mathbf{gFe}\mathbf{/}\mathbf{mL}$ Then dilute 15mL of the new solution up to 250 mL again to get$\mathbf{~}\left(\frac{15}{250}\right)\mathbf{(}\mathbf{60}\mathit{\mu }\mathbf{gFe}\mathbf{/}\mathbf{mL}\mathbf{)}\mathbf{=\sim }\mathbf{3}\mathbf{.6}\mathit{\mu }\mathbf{gFe}\mathbf{/}\mathbf{mL}$

a) First we calculate mass of Fe in 500mL:

$\begin{array}{r}\mathrm{m}=1000\cdot {10}^{-6}\mathrm{g}/\mathrm{mL}\cdot 500\mathrm{mL}\\ \mathrm{m}=0.5\mathrm{g}\end{array}$

The n is:

$\begin{array}{r}\mathrm{n}\left(\mathrm{Fe}\right)=\frac{0.5\mathrm{g}}{55.85\mathrm{g}/\mathrm{mol}}\\ \mathrm{n}\left(\mathrm{Fe}\right)=8.95\cdot {10}^{-3}\mathrm{mol}\end{array}$

The mass of ferrous ethylenediammonium sulfate is:

$\begin{array}{c}\mathrm{m}=8.95\cdot {10}^{-3}\mathrm{mol}\cdot 392.15\mathrm{g}/\mathrm{mol}\\ \mathrm{m}=3.511\mathrm{g}\end{array}$

b) First we calculate n of reagent

$\begin{array}{r}\mathrm{n}=\frac{3.627\mathrm{g}}{392.15\mathrm{g}/\mathrm{mol}}\\ \mathrm{n}=9.25\cdot {10}^{-3}\mathrm{mol}\end{array}$

The mass of Fe is:

$\begin{array}{r}\mathrm{m}=9.25\cdot {10}^{-3}\mathrm{g}\cdot 55.85\mathrm{g}/\mathrm{mol}\\ \mathrm{m}=0.5166\mathrm{g}\end{array}$

In one mL is:

$\frac{0.5166\mathrm{g}}{500\mathrm{mL}}=1.033\cdot {10}^{-3}\mathrm{g}/\mathrm{mL}$

In $\mathrm{\mu g}/\mathrm{mL}$is:

$\mathrm{y}=1033.2\mathrm{\mu g}/\mathrm{mL}$

c) To prepare 250mL of $1\mathrm{\mu g}/\mathrm{mL}$,there is requirement to dilute 10mL of stock solution to 250mL with$0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$ to obtain:

$\frac{10\mathrm{mL}}{250\mathrm{mL}}\cdot 1033.2\mu \mathrm{g}/\mathrm{mL}=41.328\mu \mathrm{g}/\mathrm{mL}$

Then we dilute 5 mL of resulting solution to 250 mL to obtain:

$\frac{5\mathrm{mL}}{250\mathrm{mL}}\cdot 41.328\mu \mathrm{g}/\mathrm{mL}=0.826\mu \mathrm{g}/\mathrm{mL}$

To prepare 250mL of $2\mathrm{\mu g}/\mathrm{mL}$,there is requirement to dilute 10mL of stock solution to 250mL with$0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$ to obtain:

$\frac{10\mathrm{mL}}{250\mathrm{mL}}\cdot 1033.2\mu \mathrm{g}/\mathrm{mL}=41.328\mu \mathrm{g}/\mathrm{mL}$

Then there is requirement to dilute 10mL of resulting solution to 250 mL to obtain:

$\frac{10\mathrm{mL}}{250\mathrm{mL}}\cdot 41.328\mu \mathrm{g}/\mathrm{mL}=1.653\mu \mathrm{g}/\mathrm{mL}$

To prepare 250mL of $3\mathrm{\mu g}/\mathrm{mL}$,there is requirement to dilute to dilute 15mL of stock solution to 250mL with $0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$to obtain

$\frac{10\mathrm{mL}}{250\mathrm{mL}}\cdot 1033.2\mu \mathrm{g}/\mathrm{mL}=41.328\mu \mathrm{g}/\mathrm{mL}$

Then there is requirement to dilute 15mL of resulting solution to 250 mL to obtain:

$\frac{15\mathrm{mL}}{250\mathrm{mL}}\cdot 41.328\mu \mathrm{g}/\mathrm{mL}=2.479\mu \mathrm{g}/\mathrm{mL}$

To prepare 250mL of$4\mathrm{\mu g}/\mathrm{mL}$ ,there is requirement to dilute 15mL of stock solution to 250mL with$0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$ to obtain

$\frac{15\mathrm{mL}}{250\mathrm{mL}}\cdot 61.992\mu \mathrm{g}/\mathrm{mL}=3.719\mu \mathrm{g}/\mathrm{mL}$

Then there is requirement to dilute 15mL of resulting solution to 250 mL to obtain:

$\frac{15\mathrm{mL}}{250\mathrm{mL}}\cdot 61.992\mu \mathrm{g}/\mathrm{mL}=3.719\mu \mathrm{g}/\mathrm{mL}$

To prepare 250mL of$5\mathrm{\mu g}/\mathrm{mL}$ ,there is requirement to dilute 20mL of stock solution to 250mL with $0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$to obtain

$\frac{15\mathrm{mL}}{250\mathrm{mL}}\cdot 1033.2\mu \mathrm{g}/\mathrm{mL}=61.992\mu \mathrm{g}/\mathrm{mL}$

Then there is requirement to dilute 20mL of resulting solution to 250 mL to obtain

$\frac{20\mathrm{mL}}{250\mathrm{mL}}\cdot 61.992\mu \mathrm{g}/\mathrm{mL}=4.959\mu \mathrm{g}/\mathrm{mL}$

To prepare 250mL of role="math" localid="1668406953716" $7\mathrm{\mu g}/\mathrm{mL}$,there is requirement to dilute 20mL of stock solution to 250mL with $0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$to obtain

$\frac{20\mathrm{mL}}{250\mathrm{mL}}\cdot 1033.2\mu \mathrm{g}/\mathrm{mL}=82.656\mu \mathrm{g}/\mathrm{mL}$

Then there is requirement to dilute 20mL of resulting solution to 250 mL to obtain

$\frac{20\mathrm{mL}}{250\mathrm{mL}}\cdot 82.656\mu \mathrm{g}/\mathrm{mL}=6.612\mu \mathrm{g}/\mathrm{mL}$

To prepare 250mL of $8\mathrm{\mu g}/\mathrm{mL}$,we need to dilute 25mL of stock solution to 250mL with $0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$to obtain

$\frac{20\mathrm{mL}}{250\mathrm{mL}}\cdot 1033.2\mu \mathrm{g}/\mathrm{mL}=82.656\mu \mathrm{g}/\mathrm{mL}$

Then we dilute 25mL of resulting solution to 250 mL to obtain

$\frac{25\mathrm{mL}}{250\mathrm{mL}}\cdot 82.656\mu \mathrm{g}/\mathrm{mL}=8.266\mu \mathrm{g}/\mathrm{mL}$

To prepare 250mL of $10\mathrm{\mu g}/\mathrm{mL}$,we need to dilute 30mL of stock solution to 250mL with $0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$to obtain

$\frac{20\mathrm{mL}}{250\mathrm{mL}}\cdot 1033.2\mu \mathrm{g}/\mathrm{mL}=82.656\mu \mathrm{g}/\mathrm{mL}$

Then we dilute 30mL of resulting solution to 250 mL to obtain

$\frac{30\mathrm{mL}}{250\mathrm{mL}}\cdot 82.656\mu \mathrm{g}/\mathrm{mL}=9.919\mu \mathrm{g}/\mathrm{mL}$