Q24P Preparing standards for a calibr... [FREE SOLUTION]

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Chapter 18: Q24P (page 458)

Preparing standards for a calibration curve.
(a) How much ferrous ammonium sulfate $(Fe {({NH}_{4})}_{2} {({SO}_{4})}_{2} 鈰� 6 H_{2} OFM 392 .15)$ should be dissolved in a 500mL volumetric flask withto obtain a stock solution with $1000 渭 gFe / mL$ ?
(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 $~ 1, 2, 3, 4, 5, 6, 7, 8 and 10 in 0.1 {MH}_{2} {SO}_{4}$ 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 $~ 4 渭 gFe / mL,$ , you could first dilute 15mL(=10+5mL) of stock solution up to 250 mL $mL to get ~ (\frac{15}{25}) (1000 渭 gFFe / mL) =鈭� 60 渭 gFe / mL$ Then dilute 15mL of the new solution up to 250 mL again to get $~ (\frac{15}{250}) (60 渭 gFe / mL) =鈭� 3 .6 渭 gFe / mL$

Short Answer

Expert verified

a) m = 3.51 1g

b) y = 1033 $渭驳 / mL$

c) 9.919 $渭驳 / mL$

Step by step solution

Find mass of ferrous ethylenediammonium sulfate

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

$\begin{aligned} m = 1000 鈰� 10^{鈭� 6} g / mL 鈰� 500 mL \\ m = 0.5 g \end{aligned}$

The n is:

$\begin{aligned} n (Fe) = \frac{0.5 g}{55.85 g / mol} \\ n (Fe) = 8.95 鈰� 10^{鈭� 3} mol \end{aligned}$

The mass of ferrous ethylenediammonium sulfate is:

$\begin{matrix} m = 8.95 鈰� 10^{鈭� 3} mol 鈰� 392.15 g / mol \\ m = 3.511 g \end{matrix}$

Find y

b) First we calculate n of reagent

$\begin{aligned} n = \frac{3.627 g}{392.15 g / mol} \\ n = 9.25 鈰� 10^{鈭� 3} mol \end{aligned}$

The mass of Fe is:

$\begin{aligned} m = 9.25 鈰� 10^{鈭� 3} g 鈰� 55.85 g / mol \\ m = 0.5166 g \end{aligned}$

In one mL is:

$\frac{0.5166 g}{500 mL} = 1.033 鈰� 10^{鈭� 3} g / mL$

In $渭驳 / mL$ is:

$y = 1033.2 渭驳 / mL$

Find solution for different dilute

c) To prepare 250mL of $1 渭驳 / mL$ ,there is requirement to dilute 10mL of stock solution to 250mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain:

$\frac{10 mL}{250 mL} 鈰� 1033.2 渭 g / mL = 41.328 渭 g / mL$

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

$\frac{5 mL}{250 mL} 鈰� 41.328 渭 g / mL = 0.826 渭 g / mL$

To prepare 250mL of $2 渭驳 / mL$ ,there is requirement to dilute 10mL of stock solution to 250mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain:

$\frac{10 mL}{250 mL} 鈰� 1033.2 渭 g / mL = 41.328 渭 g / mL$

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

$\frac{10 mL}{250 mL} 鈰� 41.328 渭 g / mL = 1.653 渭 g / mL$

To prepare 250mL of $3 渭驳 / mL$ ,there is requirement to dilute to dilute 15mL of stock solution to 250mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain

$\frac{10 mL}{250 mL} 鈰� 1033.2 渭 g / mL = 41.328 渭 g / mL$

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

$\frac{15 mL}{250 mL} 鈰� 41.328 渭 g / mL = 2.479 渭 g / mL$

To prepare 250mL of $4 渭驳 / mL$ ,there is requirement to dilute 15mL of stock solution to 250mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain

$\frac{15 mL}{250 mL} 鈰� 61.992 渭 g / mL = 3.719 渭 g / mL$

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

$\frac{15 mL}{250 mL} 鈰� 61.992 渭 g / mL = 3.719 渭 g / mL$

To prepare 250mL of $5 渭驳 / mL$ ,there is requirement to dilute 20mL of stock solution to 250mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain

$\frac{15 mL}{250 mL} 鈰� 1033.2 渭 g / mL = 61.992 渭 g / mL$

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

$\frac{20 mL}{250 mL} 鈰� 61.992 渭 g / mL = 4.959 渭 g / mL$

To prepare 250mL of role="math" localid="1668406953716" $7 渭驳 / mL$ ,there is requirement to dilute 20mL of stock solution to 250mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain

$\frac{20 mL}{250 mL} 鈰� 1033.2 渭 g / mL = 82.656 渭 g / mL$

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

$\frac{20 mL}{250 mL} 鈰� 82.656 渭 g / mL = 6.612 渭 g / mL$

To prepare 250mL of $8 渭驳 / mL$ ,we need to dilute 25mL of stock solution to 250mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain

$\frac{20 mL}{250 mL} 鈰� 1033.2 渭 g / mL = 82.656 渭 g / mL$

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

$\frac{25 mL}{250 mL} 鈰� 82.656 渭 g / mL = 8.266 渭 g / mL$

To prepare 250mL of $10 渭驳 / mL$ ,we need to dilute 30mL of stock solution to 250mL with $0.1 {MH}_{2} {SO}_{4}$ to obtain

$\frac{20 mL}{250 mL} 鈰� 1033.2 渭 g / mL = 82.656 渭 g / mL$

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

$\frac{30 mL}{250 mL} 鈰� 82.656 渭 g / mL = 9.919 渭 g / mL$

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Short Answer

Step by step solution

Find mass of ferrous ethylenediammonium sulfate

Find y

Find solution for different dilute

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