Percent Dissociated from Titration Curve
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Question 818 references the titration curve. Answer is A because $ce{H+}$ 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 $frac{0.0001}{0.1}times 100 = 0.1%$.
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
6,022725
asked Apr 1 at 14:31
JonJon
232
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add a comment |
$begingroup$
Question 818 references the titration curve. Answer is A because $ce{H+}$ 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 $frac{0.0001}{0.1}times 100 = 0.1%$.
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
6,022725
asked Apr 1 at 14:31
JonJon
232
$endgroup$
add a comment |
$begingroup$
Question 818 references the titration curve. Answer is A because $ce{H+}$ 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 $frac{0.0001}{0.1}times 100 = 0.1%$.
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
6,022725
asked Apr 1 at 14:31
JonJon
232
$endgroup$
Question 818 references the titration curve. Answer is A because $ce{H+}$ 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 $frac{0.0001}{0.1}times 100 = 0.1%$.
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
6,022725
asked Apr 1 at 14:31
JonJon
232
edited Apr 1 at 17:10
Mathew Mahindaratne
6,022725
asked Apr 1 at 14:31
JonJon
232
edited Apr 1 at 17:10
Mathew Mahindaratne
6,022725
edited Apr 1 at 17:10
Mathew Mahindaratne
6,022725
edited Apr 1 at 17:10
Mathew Mahindaratne
6,022725
6,022725
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
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I'm not sure I follow your logic.
For a monobasic acid S $(ce{HA})$ dissociation degree $α$ is
$$α = frac{[ce{H+}]}{c_mathrm{a}},$$
where $c_mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_mathrm{b}$:
$$c_mathrm{a}V_mathrm{a} = c_mathrm{b}V_mathrm{b} implies c_mathrm{a} = frac{c_mathrm{b}V_mathrm{b}}{V_mathrm{a}}$$
Finally, taking $V_mathrm{a} = V_mathrm{b} = pu{50 mL}$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[ce{H+}] approx pu{1e-4 mol L-1}$:
$$α = frac{[ce{H+}]V_mathrm{a}}{c_mathrm{b}V_mathrm{b}} = frac{pu{1e-4 mol L-1}cdotpu{50 mL}}{pu{0.1 mol L-1}cdotpu{50 mL}} = pu{1e-3}~text{or}~0.1%$$
answered Apr 1 at 14:50
andseliskandselisk
19k660125
$endgroup$
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 $(ce{HA})$ dissociation degree $α$ is
$$α = frac{[ce{H+}]}{c_mathrm{a}},$$
where $c_mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_mathrm{b}$:
$$c_mathrm{a}V_mathrm{a} = c_mathrm{b}V_mathrm{b} implies c_mathrm{a} = frac{c_mathrm{b}V_mathrm{b}}{V_mathrm{a}}$$
Finally, taking $V_mathrm{a} = V_mathrm{b} = pu{50 mL}$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[ce{H+}] approx pu{1e-4 mol L-1}$:
$$α = frac{[ce{H+}]V_mathrm{a}}{c_mathrm{b}V_mathrm{b}} = frac{pu{1e-4 mol L-1}cdotpu{50 mL}}{pu{0.1 mol L-1}cdotpu{50 mL}} = pu{1e-3}~text{or}~0.1%$$
answered Apr 1 at 14:50
andseliskandselisk
19k660125
$endgroup$
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 $(ce{HA})$ dissociation degree $α$ is
$$α = frac{[ce{H+}]}{c_mathrm{a}},$$
where $c_mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_mathrm{b}$:
$$c_mathrm{a}V_mathrm{a} = c_mathrm{b}V_mathrm{b} implies c_mathrm{a} = frac{c_mathrm{b}V_mathrm{b}}{V_mathrm{a}}$$
Finally, taking $V_mathrm{a} = V_mathrm{b} = pu{50 mL}$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[ce{H+}] approx pu{1e-4 mol L-1}$:
$$α = frac{[ce{H+}]V_mathrm{a}}{c_mathrm{b}V_mathrm{b}} = frac{pu{1e-4 mol L-1}cdotpu{50 mL}}{pu{0.1 mol L-1}cdotpu{50 mL}} = pu{1e-3}~text{or}~0.1%$$
answered Apr 1 at 14:50
andseliskandselisk
19k660125
$endgroup$
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 $(ce{HA})$ dissociation degree $α$ is
$$α = frac{[ce{H+}]}{c_mathrm{a}},$$
where $c_mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_mathrm{b}$:
$$c_mathrm{a}V_mathrm{a} = c_mathrm{b}V_mathrm{b} implies c_mathrm{a} = frac{c_mathrm{b}V_mathrm{b}}{V_mathrm{a}}$$
Finally, taking $V_mathrm{a} = V_mathrm{b} = pu{50 mL}$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[ce{H+}] approx pu{1e-4 mol L-1}$:
$$α = frac{[ce{H+}]V_mathrm{a}}{c_mathrm{b}V_mathrm{b}} = frac{pu{1e-4 mol L-1}cdotpu{50 mL}}{pu{0.1 mol L-1}cdotpu{50 mL}} = pu{1e-3}~text{or}~0.1%$$
answered Apr 1 at 14:50
andseliskandselisk
19k660125
$endgroup$
I'm not sure I follow your logic.
For a monobasic acid S $(ce{HA})$ dissociation degree $α$ is
$$α = frac{[ce{H+}]}{c_mathrm{a}},$$
where $c_mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_mathrm{b}$:
$$c_mathrm{a}V_mathrm{a} = c_mathrm{b}V_mathrm{b} implies c_mathrm{a} = frac{c_mathrm{b}V_mathrm{b}}{V_mathrm{a}}$$
Finally, taking $V_mathrm{a} = V_mathrm{b} = pu{50 mL}$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[ce{H+}] approx pu{1e-4 mol L-1}$:
$$α = frac{[ce{H+}]V_mathrm{a}}{c_mathrm{b}V_mathrm{b}} = frac{pu{1e-4 mol L-1}cdotpu{50 mL}}{pu{0.1 mol L-1}cdotpu{50 mL}} = pu{1e-3}~text{or}~0.1%$$
answered Apr 1 at 14:50
andseliskandselisk
19k660125
answered Apr 1 at 14:50
andseliskandselisk
19k660125
answered Apr 1 at 14:50
andseliskandselisk
19k660125
answered Apr 1 at 14:50
andseliskandselisk
19k660125
19k660125
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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