Preparing standards for a calibration curve.

(a) How much ferrous ethylene diammonium sulfate$\left(\mathrm{Fe}\left(H_3{\mathrm{NCH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3\right){\left({\mathrm{SO}}_4\right)}_2\cdot 4H_{2}O,\mathrm{FM}382.15\right)$should be dissolved in a 500mL volumetric flask with$\mathbf{1}{\mathbf{MH}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$to obtain a stock solution with$\mathbf{~}\mathbf{500}\mathit{\mu }\mathbf{gFe}\mathbf{/}{\mathbf{mL}}^{\mathbf{2}}$?

(b) When making stock solution (a), you actually weighed out 1.627 g of reagent. What is the Fe concentration in role="math" localid="1668357579931" $\mathbf{500}\mathit{\mu }\mathbf{gFe}\mathbf{/}{\mathbf{mL}}^{\mathbf{2}}$?

(c) How would you prepare 500 mL of standard containing 1,2,3,4, and$\mathbf{5}\mathit{\mu }\mathbf{gFe}\mathbf{/}{\mathbf{mL}}^{\mathbf{2}}\mathbf{\text{in}}\mathbf{0}\mathbf{.1}{\mathbf{MH}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$infrom stock solution (b) using any Class A pipets from Table 2-4 with only 500 -mL volumetric flasks?

(d) To reduce the generation of chemical waste, describe how you could prepare 50 mL of standard containing, andinfrom stock solution (b) by serial dilution using any Class A pipets from Table 2-3 with only 50mL volumetric flasks?

First we calculate mass of Fe in 500mL:

$\begin{array}{r}\mathrm{m}=500\cdot {10}^{-6}\mathrm{g}/\mathrm{mL}\cdot 500\mathrm{mL}\\ \mathrm{m}=0.25\mathrm{g}\end{array}$

The n is:

$\begin{array}{r}\mathrm{n}\left(\mathrm{Fe}\right)=\frac{0.25\mathrm{g}}{55.85\mathrm{g}/\mathrm{mol}}\\ \mathrm{n}\left(\mathrm{Fe}\right)=4.48\cdot {10}^{-3}\mathrm{mol}\end{array}$

$\begin{array}{r}\mathrm{m}=4.26\cdot {10}^{-3}\mathrm{g}\cdot 55.85\mathrm{g}/\mathrm{mol}\\ \mathrm{m}=0.2379\mathrm{g}\end{array}$

b) First, we calculate n of reagent

$\begin{array}{r}\mathrm{n}=\frac{1.627\mathrm{g}}{382.15\mathrm{g}/\mathrm{mol}}\\ \mathrm{n}=4.26\cdot {10}^{-3}\mathrm{mol}\end{array}$

The mass of Fe is:

$\begin{array}{r}m=4.26\cdot {10}^{-3}\mathrm{g}\cdot 55.85\mathrm{g}/\mathrm{mol}\\ \mathrm{m}=0.2379\mathrm{g}\end{array}$

In one mL is:

$\frac{0.2379\mathrm{g}}{500\mathrm{mL}}=4.758\cdot {10}^{-4}\mathrm{g}/\mathrm{mL}$

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

$\gamma =475.8\mathrm{\mu g}/\mathrm{mL}$

c) To prepare 500mL of$1\mathrm{\mu g}/\mathrm{mL}$,we need to dilute 1mL of stock solution to 500mL with $0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$ to obtain:

$\frac{1\mathrm{mL}}{500\mathrm{mL}}\times 475.8\mu \mathrm{g}/\mathrm{mL}=0.9516\mu \mathrm{g}/\mathrm{mL}$

To prepare 500mL of$2\mathrm{\mu g}/\mathrm{mL}$ , we need to dilute 2mL of stock solution to 500mL with$0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$ to obtain:

$\frac{2\mathrm{mL}}{500\mathrm{mL}}\times 475.8\mu \mathrm{g}/\mathrm{mL}=1.9032\mu \mathrm{g}/\mathrm{mL}$

To prepare 500 of $3\mathrm{\mu g}/\mathrm{mL}$, we need to dilute 3mL of stock solution to 500mL with$0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$ to obtain:

$\frac{3\mathrm{mL}}{500\mathrm{mL}}\times 475.8\mu \mathrm{g}/\mathrm{mL}=2.855\mu \mathrm{g}/\mathrm{mL}$

To prepare 500 of $4\mathrm{\mu g}/\mathrm{mL}$, we need to dilute 4mL of stock solution to 500mL with$0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$ to obtain:

To prepare 500 of$5\mathrm{\mu g}/\mathrm{mL}$ ,we need to dilute 5mL of stock solution to 500mL with$0.1{\mathrm{MH}}_2{\mathrm{SO}}_4$ to obtain:

$\frac{5\mathrm{mL}}{500\mathrm{mL}}\times 475.8\mu \mathrm{g}/\mathrm{mL}=4.758\mu \mathrm{g}/\mathrm{mL}$