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Percent Dissociated from Titration Curve
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Percent Dissociated from Titration Curve
The Next CEO of Stack OverflowVolume 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 3 hours ago
Mathew Mahindaratne
5,899623
asked 6 hours ago
JonJon
252
$endgroup$
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 3 hours ago
Mathew Mahindaratne
5,899623
asked 6 hours ago
JonJon
252
$endgroup$
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 3 hours ago
Mathew Mahindaratne
5,899623
asked 6 hours ago
JonJon
252
$endgroup$
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 3 hours ago
Mathew Mahindaratne
5,899623
asked 6 hours ago
JonJon
252
edited 3 hours ago
Mathew Mahindaratne
5,899623
asked 6 hours ago
JonJon
252
edited 3 hours ago
Mathew Mahindaratne
5,899623
edited 3 hours ago
Mathew Mahindaratne
5,899623
edited 3 hours ago
Mathew Mahindaratne
5,899623
5,899623
asked 6 hours ago
JonJon
252
asked 6 hours ago
JonJon
252
asked 6 hours ago
JonJon
252
252
add a comment |
add a comment |
1 Answer
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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 5 hours ago
andseliskandselisk
18.9k660124
$endgroup$
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
5 hours ago
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 5 hours ago
andseliskandselisk
18.9k660124
$endgroup$
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
5 hours ago
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 5 hours ago
andseliskandselisk
18.9k660124
$endgroup$
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
5 hours ago
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 5 hours ago
andseliskandselisk
18.9k660124
$endgroup$
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 5 hours ago
andseliskandselisk
18.9k660124
answered 5 hours ago
andseliskandselisk
18.9k660124
answered 5 hours ago
andseliskandselisk
18.9k660124
answered 5 hours ago
andseliskandselisk
18.9k660124
18.9k660124
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
5 hours ago
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
5 hours ago
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
5 hours ago
$begingroup$
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
5 hours ago
add a comment |
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