ipie Results 2- CO, + H* +→ HCO;¯+ H* + Н-СОЗ Known Sample #1: 29.95% CO3 Obtained mass of known #1: 0.5086g According to the figure 1 below: 1st equivalence point: VHCF 10.00ML, pH= 7.62 2nd equivalence point: VHCI = 20.00 mL, pH = 3.48 Concentration of prepared HCl: 0.0844M 0.0844 mol тol HCl 3D [HCI] * Vнсі 0.002000 L = 0.00169 mol HCI Mole to mole ratio of CO, :H* is 1:2 тol CO — 0.5 тol HCІ — 0.5 * 0.00169 тol HCI — 0.000844 mol C03- Formula Weight of Na,CO3: 105.988g/mole g Na2CO3 = mol CO * FW = 0.000844 mol CO3 * 105.988; 0.0894 g Na2CO3 mol 2- Percentage of CO3 g Co3- g sample 0.0894 g CO3 0.5086 g sample %CO?- 100 = * 100 = 17.6% 2- Percentage recovery of CO; 17.6% CO?- 29.95% CO3- Experimental % % Recovery * 100 = z= * 100 = 58.7% %D Кпown % Initial concentration of Carbonate mol CO- L solution 0.000844 mol co [CO} ]; = = 0.00844 M co?- 0.100 L solution Concentration of Carbonate at the Equivalence Point [CO} ]; * Vi Veg 0.00844 M CO * 100 mL [CO? leg 0.00703 M CO?- 120.00 mL Finding the Concentration of Hydronium at the Equivalence Point [H*1 = 10 pH = 10-3.48 = 3.31x10¬ªM [H*] Finding the pKai 2 HC1 + Na,CO3 + H2CO3 +2NaCı [H*1? (F – [H*]) (3.31x10-4 M )² Ka1 1.64 x10-5 0.00703M – 3.31x10-4M pKa1 = – log(Ka1) = – log(1.64x10-5) = 4.79 pH = 1/2 (pKa1 + pKa2) pKa2 = 2 * pH – pKa2 = 2 * 7.62 – 4.79 = 10.45 Unknown 243 27.1% Percent Recovery Correction % CO32¯ Experiment 27.1% % CO32 Corrected 100 = 46.5% * 100 = % Recovery 58.3%
ipie Results 2- CO, + H* +→ HCO;¯+ H* + Н-СОЗ Known Sample #1: 29.95% CO3 Obtained mass of known #1: 0.5086g According to the figure 1 below: 1st equivalence point: VHCF 10.00ML, pH= 7.62 2nd equivalence point: VHCI = 20.00 mL, pH = 3.48 Concentration of prepared HCl: 0.0844M 0.0844 mol тol HCl 3D [HCI] * Vнсі 0.002000 L = 0.00169 mol HCI Mole to mole ratio of CO, :H* is 1:2 тol CO — 0.5 тol HCІ — 0.5 * 0.00169 тol HCI — 0.000844 mol C03- Formula Weight of Na,CO3: 105.988g/mole g Na2CO3 = mol CO * FW = 0.000844 mol CO3 * 105.988; 0.0894 g Na2CO3 mol 2- Percentage of CO3 g Co3- g sample 0.0894 g CO3 0.5086 g sample %CO?- 100 = * 100 = 17.6% 2- Percentage recovery of CO; 17.6% CO?- 29.95% CO3- Experimental % % Recovery * 100 = z= * 100 = 58.7% %D Кпown % Initial concentration of Carbonate mol CO- L solution 0.000844 mol co [CO} ]; = = 0.00844 M co?- 0.100 L solution Concentration of Carbonate at the Equivalence Point [CO} ]; * Vi Veg 0.00844 M CO * 100 mL [CO? leg 0.00703 M CO?- 120.00 mL Finding the Concentration of Hydronium at the Equivalence Point [H*1 = 10 pH = 10-3.48 = 3.31x10¬ªM [H*] Finding the pKai 2 HC1 + Na,CO3 + H2CO3 +2NaCı [H*1? (F – [H*]) (3.31x10-4 M )² Ka1 1.64 x10-5 0.00703M – 3.31x10-4M pKa1 = – log(Ka1) = – log(1.64x10-5) = 4.79 pH = 1/2 (pKa1 + pKa2) pKa2 = 2 * pH – pKa2 = 2 * 7.62 – 4.79 = 10.45 Unknown 243 27.1% Percent Recovery Correction % CO32¯ Experiment 27.1% % CO32 Corrected 100 = 46.5% * 100 = % Recovery 58.3%
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
Can you please explain to me how they the got 27.1% under the percent recovery correction?
![ipie
Results
2-
CO, + H* +→ HCO;¯+ H* +
Н-СОЗ
Known Sample #1: 29.95% CO3
Obtained mass of known #1: 0.5086g
According to the figure 1 below:
1st equivalence point: VHCF 10.00ML, pH= 7.62
2nd equivalence point: VHCI = 20.00 mL, pH = 3.48
Concentration of prepared HCl: 0.0844M
0.0844 mol
тol HCl 3D [HCI] * Vнсі
0.002000 L = 0.00169 mol HCI
Mole to mole ratio of CO,
:H* is 1:2
тol CO
— 0.5 тol HCІ — 0.5 * 0.00169 тol HCI — 0.000844 mol C03-
Formula Weight of Na,CO3: 105.988g/mole
g Na2CO3 = mol CO * FW = 0.000844 mol CO3 * 105.988;
0.0894 g Na2CO3
mol
2-
Percentage of CO3
g Co3-
g sample
0.0894 g CO3
0.5086 g sample
%CO?-
100 =
* 100 = 17.6%
2-
Percentage recovery of CO;
17.6% CO?-
29.95% CO3-
Experimental %
% Recovery
* 100 =
z= * 100 = 58.7%
%D
Кпown %](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20a2f328-bced-4431-a632-3ebcfa0e6361%2F96da7ed1-b3b1-4c6e-9403-ed82ace1d07d%2Fh0guj0k.png&w=3840&q=75)
Transcribed Image Text:ipie
Results
2-
CO, + H* +→ HCO;¯+ H* +
Н-СОЗ
Known Sample #1: 29.95% CO3
Obtained mass of known #1: 0.5086g
According to the figure 1 below:
1st equivalence point: VHCF 10.00ML, pH= 7.62
2nd equivalence point: VHCI = 20.00 mL, pH = 3.48
Concentration of prepared HCl: 0.0844M
0.0844 mol
тol HCl 3D [HCI] * Vнсі
0.002000 L = 0.00169 mol HCI
Mole to mole ratio of CO,
:H* is 1:2
тol CO
— 0.5 тol HCІ — 0.5 * 0.00169 тol HCI — 0.000844 mol C03-
Formula Weight of Na,CO3: 105.988g/mole
g Na2CO3 = mol CO * FW = 0.000844 mol CO3 * 105.988;
0.0894 g Na2CO3
mol
2-
Percentage of CO3
g Co3-
g sample
0.0894 g CO3
0.5086 g sample
%CO?-
100 =
* 100 = 17.6%
2-
Percentage recovery of CO;
17.6% CO?-
29.95% CO3-
Experimental %
% Recovery
* 100 =
z= * 100 = 58.7%
%D
Кпown %
![Initial concentration of Carbonate
mol CO-
L solution
0.000844 mol co
[CO} ]; =
= 0.00844 M co?-
0.100 L solution
Concentration of Carbonate at the Equivalence Point
[CO} ]; * Vi
Veg
0.00844 M CO * 100 mL
[CO? leg
0.00703 M CO?-
120.00 mL
Finding the Concentration of Hydronium at the Equivalence Point
[H*1 = 10 pH = 10-3.48 = 3.31x10»M [H*]
Finding the pKai
2 HC1 + Na,CO3 +
H2CO3 +2NaCı
[H*1?
(F – [H*])
(3.31x10-4 M )²
Ka1
1.64 x10-5
0.00703M – 3.31x10-4M
pKa1 = – log(Ka1) = – log(1.64x10-5) = 4.79
pH = 1/2 (pKa1 + pKa2)
pKa2 = 2 * pH – pKa2 = 2 * 7.62 – 4.79 = 10.45
Unknown 243
27.1%
Percent Recovery Correction
% CO32¯ Experiment
27.1%
% CO32 Corrected
100 = 46.5%
* 100 =
% Recovery
58.3%](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20a2f328-bced-4431-a632-3ebcfa0e6361%2F96da7ed1-b3b1-4c6e-9403-ed82ace1d07d%2F7aurq0o.png&w=3840&q=75)
Transcribed Image Text:Initial concentration of Carbonate
mol CO-
L solution
0.000844 mol co
[CO} ]; =
= 0.00844 M co?-
0.100 L solution
Concentration of Carbonate at the Equivalence Point
[CO} ]; * Vi
Veg
0.00844 M CO * 100 mL
[CO? leg
0.00703 M CO?-
120.00 mL
Finding the Concentration of Hydronium at the Equivalence Point
[H*1 = 10 pH = 10-3.48 = 3.31x10»M [H*]
Finding the pKai
2 HC1 + Na,CO3 +
H2CO3 +2NaCı
[H*1?
(F – [H*])
(3.31x10-4 M )²
Ka1
1.64 x10-5
0.00703M – 3.31x10-4M
pKa1 = – log(Ka1) = – log(1.64x10-5) = 4.79
pH = 1/2 (pKa1 + pKa2)
pKa2 = 2 * pH – pKa2 = 2 * 7.62 – 4.79 = 10.45
Unknown 243
27.1%
Percent Recovery Correction
% CO32¯ Experiment
27.1%
% CO32 Corrected
100 = 46.5%
* 100 =
% Recovery
58.3%
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