Q23P Preparing standards for a calibr... [FREE SOLUTION]

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Chapter 18: Q23P (page 458)

Preparing standards for a calibration curve.
(a) How much ferrous ethylene diammonium sulfate $(Fe (H_{3} {NCH}_{2} {CH}_{2} {NH}_{3}) {({SO}_{4})}_{2} 鈰� 4 H_{2} O, FM 382 .15)$ should be dissolved in a 500mL volumetric flask with $1 {MH}_{2} {SO}_{4}$ to obtain a stock solution with $~ 500 渭 gFe / {mL}^{2}$ ?
(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" $500 渭 gFe / {mL}^{2}$ ?
(c) How would you prepare 500 mL of standard containing 1,2,3,4, and $5 渭 gFe / {mL}^{2} in 0 .1 {MH}_{2} {SO}_{4}$ 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?

Short Answer

Expert verified

a) m = 1.712g

b) $纬$ = 475.8 $渭驳 / mL$

c) $4.75 渭驳 / mL$

Step by step solution

Find mass of ferrous ethylenediammonium sulfate

First we calculate mass of Fe in 500mL:

$\begin{aligned} m = 500 鈰� 10^{鈭� 6} g / mL 鈰� 500 mL \\ m = 0.25 g \end{aligned}$

The n is:

$\begin{aligned} n (Fe) = \frac{0.25 g}{55.85 g / mol} \\ n (Fe) = 4.48 鈰� 10^{鈭� 3} mol \end{aligned}$

$\begin{aligned} m = 4.26 鈰� 10^{鈭� 3} g 鈰� 55.85 g / mol \\ m = 0.2379 g \end{aligned}$

The mass of ferrous ethylenediammonium sulfate is:

Find y (concendration)

b) First, we calculate n of reagent

$\begin{aligned} n = \frac{1.627 g}{382.15 g / mol} \\ n = 4.26 鈰� 10^{鈭� 3} mol \end{aligned}$

The mass of Fe is:

$\begin{aligned} m = 4.26 鈰� 10^{鈭� 3} g 鈰� 55.85 g / mol \\ m = 0.2379 g \end{aligned}$

In one mL is:

$\frac{0.2379 g}{500 mL} = 4.758 鈰� 10^{鈭� 4} g / mL$

In $渭驳 / mL$ is:

$纬 = 475.8 渭驳 / mL$

Find concentration for different dilution

c) To prepare 500mL of $1 渭驳 / mL$ ,we need to dilute 1mL of stock solution to 500mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain:

$\frac{1 mL}{500 mL} 脳 475.8 渭 g / mL = 0.9516 渭 g / mL$

To prepare 500mL of $2 渭驳 / mL$ , we need to dilute 2mL of stock solution to 500mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain:

$\frac{2 mL}{500 mL} 脳 475.8 渭 g / mL = 1.9032 渭 g / mL$

To prepare 500 of $3 渭驳 / mL$ , we need to dilute 3mL of stock solution to 500mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain:

$\frac{3 mL}{500 mL} 脳 475.8 渭 g / mL = 2.855 渭 g / mL$

To prepare 500 of $4 渭驳 / mL$ , we need to dilute 4mL of stock solution to 500mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain:

To prepare 500 of $5 渭驳 / mL$ ,we need to dilute 5mL of stock solution to 500mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain:

$\frac{5 mL}{500 mL} 脳 475.8 渭 g / mL = 4.758 渭 g / mL$

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Short Answer

Step by step solution

Find mass of ferrous ethylenediammonium sulfate

Find y (concendration)

Find concentration for different dilution

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