Percent Dissociated from Titration Curve Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Volume required to dilute solution for a pH changeHow to calculate the composition of a borate buffer with a defined pH using the Henderson-Hasselbalch equation?Titration of alpha-amino acid with strong baseWhy do these two calculations give me different answers for the same acid-base titration?Titration of NaOH with acetic acidUnderstanding how to calculate the pH of a buffer with ice tablesTitration with Ca(OH)2calculating ph of a mixture of acidsFinding the concentration of hydrochloric acid by titrationWhy does pH increase as a weak acid becomes more dissociated?
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Percent Dissociated from Titration Curve
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Volume required to dilute solution for a pH changeHow to calculate the composition of a borate buffer with a defined pH using the Henderson-Hasselbalch equation?Titration of alpha-amino acid with strong baseWhy do these two calculations give me different answers for the same acid-base titration?Titration of NaOH with acetic acidUnderstanding how to calculate the pH of a buffer with ice tablesTitration with Ca(OH)2calculating ph of a mixture of acidsFinding the concentration of hydrochloric acid by titrationWhy does pH increase as a weak acid becomes more dissociated?
$begingroup$
Question 818 references the titration curve. Answer is A because $ceH+$ conc $= 10^-4$. This is conc of dissociated acid. The conc of the undissociated acid is the original concentration minus this: $0.1 - 0.0001$, which is about $0.1$. So then its $frac0.00010.1times 100 = 0.1%$.
Where in the world are they getting the original concentration? how did they get $0.1$ as the concentration of undissociated acid?
acid-base aqueous-solution analytical-chemistry titration
edited Apr 1 at 17:10
Mathew Mahindaratne
6,393725
asked Apr 1 at 14:31
JonJon
232
$endgroup$
add a comment |
$begingroup$
Question 818 references the titration curve. Answer is A because $ceH+$ conc $= 10^-4$. This is conc of dissociated acid. The conc of the undissociated acid is the original concentration minus this: $0.1 - 0.0001$, which is about $0.1$. So then its $frac0.00010.1times 100 = 0.1%$.
Where in the world are they getting the original concentration? how did they get $0.1$ as the concentration of undissociated acid?
acid-base aqueous-solution analytical-chemistry titration
edited Apr 1 at 17:10
Mathew Mahindaratne
6,393725
asked Apr 1 at 14:31
JonJon
232
$endgroup$
add a comment |
$begingroup$
Question 818 references the titration curve. Answer is A because $ceH+$ conc $= 10^-4$. This is conc of dissociated acid. The conc of the undissociated acid is the original concentration minus this: $0.1 - 0.0001$, which is about $0.1$. So then its $frac0.00010.1times 100 = 0.1%$.
Where in the world are they getting the original concentration? how did they get $0.1$ as the concentration of undissociated acid?
acid-base aqueous-solution analytical-chemistry titration
edited Apr 1 at 17:10
Mathew Mahindaratne
6,393725
asked Apr 1 at 14:31
JonJon
232
$endgroup$
Question 818 references the titration curve. Answer is A because $ceH+$ conc $= 10^-4$. This is conc of dissociated acid. The conc of the undissociated acid is the original concentration minus this: $0.1 - 0.0001$, which is about $0.1$. So then its $frac0.00010.1times 100 = 0.1%$.
Where in the world are they getting the original concentration? how did they get $0.1$ as the concentration of undissociated acid?
acid-base aqueous-solution analytical-chemistry titration
acid-base aqueous-solution analytical-chemistry titration
edited Apr 1 at 17:10
Mathew Mahindaratne
6,393725
asked Apr 1 at 14:31
JonJon
232
edited Apr 1 at 17:10
Mathew Mahindaratne
6,393725
asked Apr 1 at 14:31
JonJon
232
edited Apr 1 at 17:10
Mathew Mahindaratne
6,393725
edited Apr 1 at 17:10
Mathew Mahindaratne
6,393725
edited Apr 1 at 17:10
Mathew Mahindaratne
6,393725
6,393725
asked Apr 1 at 14:31
JonJon
232
asked Apr 1 at 14:31
JonJon
232
asked Apr 1 at 14:31
JonJon
232
232
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm not sure I follow your logic.
For a monobasic acid S $(ceHA)$ dissociation degree $α$ is
$$α = frac[ceH+]c_mathrma,$$
where $c_mathrma$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_mathrmb$:
$$c_mathrmaV_mathrma = c_mathrmbV_mathrmb implies c_mathrma = fracc_mathrmbV_mathrmbV_mathrma$$
Finally, taking $V_mathrma = V_mathrmb = pu50 mL$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[ceH+] approx pu1e-4 mol L-1$:
$$α = frac[ceH+]V_mathrmac_mathrmbV_mathrmb = fracpu1e-4 mol L-1cdotpu50 mLpu0.1 mol L-1cdotpu50 mL = pu1e-3~textor~0.1%$$
answered Apr 1 at 14:50
andseliskandselisk
19.7k665128
$endgroup$
1
$begingroup$
Thank you! I appreciate your help. I seem to have missed the entire purpose of the titration... so the idea is that you are setting the molar amount of base to the molar amount of acid specifically at the equivalence point, then solving for the original concentration of acid. then you divide the hydrogen ion concentration by this value
$endgroup$
– Jon
Apr 1 at 15:00
add a comment |
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$begingroup$
I'm not sure I follow your logic.
For a monobasic acid S $(ceHA)$ dissociation degree $α$ is
$$α = frac[ceH+]c_mathrma,$$
where $c_mathrma$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_mathrmb$:
$$c_mathrmaV_mathrma = c_mathrmbV_mathrmb implies c_mathrma = fracc_mathrmbV_mathrmbV_mathrma$$
Finally, taking $V_mathrma = V_mathrmb = pu50 mL$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[ceH+] approx pu1e-4 mol L-1$:
$$α = frac[ceH+]V_mathrmac_mathrmbV_mathrmb = fracpu1e-4 mol L-1cdotpu50 mLpu0.1 mol L-1cdotpu50 mL = pu1e-3~textor~0.1%$$
answered Apr 1 at 14:50
andseliskandselisk
19.7k665128
$endgroup$
1
$begingroup$
Thank you! I appreciate your help. I seem to have missed the entire purpose of the titration... so the idea is that you are setting the molar amount of base to the molar amount of acid specifically at the equivalence point, then solving for the original concentration of acid. then you divide the hydrogen ion concentration by this value
$endgroup$
– Jon
Apr 1 at 15:00
add a comment |
$begingroup$
I'm not sure I follow your logic.
For a monobasic acid S $(ceHA)$ dissociation degree $α$ is
$$α = frac[ceH+]c_mathrma,$$
where $c_mathrma$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_mathrmb$:
$$c_mathrmaV_mathrma = c_mathrmbV_mathrmb implies c_mathrma = fracc_mathrmbV_mathrmbV_mathrma$$
Finally, taking $V_mathrma = V_mathrmb = pu50 mL$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[ceH+] approx pu1e-4 mol L-1$:
$$α = frac[ceH+]V_mathrmac_mathrmbV_mathrmb = fracpu1e-4 mol L-1cdotpu50 mLpu0.1 mol L-1cdotpu50 mL = pu1e-3~textor~0.1%$$
answered Apr 1 at 14:50
andseliskandselisk
19.7k665128
$endgroup$
1
$begingroup$
Thank you! I appreciate your help. I seem to have missed the entire purpose of the titration... so the idea is that you are setting the molar amount of base to the molar amount of acid specifically at the equivalence point, then solving for the original concentration of acid. then you divide the hydrogen ion concentration by this value
$endgroup$
– Jon
Apr 1 at 15:00
add a comment |
$begingroup$
I'm not sure I follow your logic.
For a monobasic acid S $(ceHA)$ dissociation degree $α$ is
$$α = frac[ceH+]c_mathrma,$$
where $c_mathrma$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_mathrmb$:
$$c_mathrmaV_mathrma = c_mathrmbV_mathrmb implies c_mathrma = fracc_mathrmbV_mathrmbV_mathrma$$
Finally, taking $V_mathrma = V_mathrmb = pu50 mL$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[ceH+] approx pu1e-4 mol L-1$:
$$α = frac[ceH+]V_mathrmac_mathrmbV_mathrmb = fracpu1e-4 mol L-1cdotpu50 mLpu0.1 mol L-1cdotpu50 mL = pu1e-3~textor~0.1%$$
answered Apr 1 at 14:50
andseliskandselisk
19.7k665128
$endgroup$
I'm not sure I follow your logic.
For a monobasic acid S $(ceHA)$ dissociation degree $α$ is
$$α = frac[ceH+]c_mathrma,$$
where $c_mathrma$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_mathrmb$:
$$c_mathrmaV_mathrma = c_mathrmbV_mathrmb implies c_mathrma = fracc_mathrmbV_mathrmbV_mathrma$$
Finally, taking $V_mathrma = V_mathrmb = pu50 mL$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[ceH+] approx pu1e-4 mol L-1$:
$$α = frac[ceH+]V_mathrmac_mathrmbV_mathrmb = fracpu1e-4 mol L-1cdotpu50 mLpu0.1 mol L-1cdotpu50 mL = pu1e-3~textor~0.1%$$
answered Apr 1 at 14:50
andseliskandselisk
19.7k665128
answered Apr 1 at 14:50
andseliskandselisk
19.7k665128
answered Apr 1 at 14:50
andseliskandselisk
19.7k665128
answered Apr 1 at 14:50
andseliskandselisk
19.7k665128
19.7k665128
1
$begingroup$
Thank you! I appreciate your help. I seem to have missed the entire purpose of the titration... so the idea is that you are setting the molar amount of base to the molar amount of acid specifically at the equivalence point, then solving for the original concentration of acid. then you divide the hydrogen ion concentration by this value
$endgroup$
– Jon
Apr 1 at 15:00
add a comment |
1
$begingroup$
Thank you! I appreciate your help. I seem to have missed the entire purpose of the titration... so the idea is that you are setting the molar amount of base to the molar amount of acid specifically at the equivalence point, then solving for the original concentration of acid. then you divide the hydrogen ion concentration by this value
$endgroup$
– Jon
Apr 1 at 15:00
1
1
$begingroup$
Thank you! I appreciate your help. I seem to have missed the entire purpose of the titration... so the idea is that you are setting the molar amount of base to the molar amount of acid specifically at the equivalence point, then solving for the original concentration of acid. then you divide the hydrogen ion concentration by this value
$endgroup$
– Jon
Apr 1 at 15:00
$begingroup$
Thank you! I appreciate your help. I seem to have missed the entire purpose of the titration... so the idea is that you are setting the molar amount of base to the molar amount of acid specifically at the equivalence point, then solving for the original concentration of acid. then you divide the hydrogen ion concentration by this value
$endgroup$
– Jon
Apr 1 at 15:00
add a comment |
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