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Percent Dissociated from Titration Curve

2

$begingroup$

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

share|improve this question

edited Apr 1 at 17:10

Mathew Mahindaratne

6,393725

asked Apr 1 at 14:31

JonJon

232

$endgroup$

add a comment | 

2

$begingroup$

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

share|improve this question

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

share|improve this question

edited Apr 1 at 17:10

Mathew Mahindaratne

6,393725

asked Apr 1 at 14:31

JonJon

232

$endgroup$

share|improve this question

edited Apr 1 at 17:10

Mathew Mahindaratne

6,393725

asked Apr 1 at 14:31

JonJon

232

share|improve this question

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

asked Apr 1 at 14:31

JonJon

232

asked Apr 1 at 14:31

JonJon

232

1 Answer
1

active

oldest

votes

6

$begingroup$

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, $[ceH+] approx pu1e-4 mol L-1$:

$$α = frac[ceH+]V_mathrmac_mathrmbV_mathrmb = fracpu1e-4 mol L-1cdotpu50 mLpu0.1 mol L-1cdotpu50 mL = pu1e-3~textor~0.1%$$

share|improve this answer

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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6

$begingroup$

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, $[ceH+] approx pu1e-4 mol L-1$:

$$α = frac[ceH+]V_mathrmac_mathrmbV_mathrmb = fracpu1e-4 mol L-1cdotpu50 mLpu0.1 mol L-1cdotpu50 mL = pu1e-3~textor~0.1%$$

share|improve this answer

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 | 

6

$begingroup$

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, $[ceH+] approx pu1e-4 mol L-1$:

$$α = frac[ceH+]V_mathrmac_mathrmbV_mathrmb = fracpu1e-4 mol L-1cdotpu50 mLpu0.1 mol L-1cdotpu50 mL = pu1e-3~textor~0.1%$$

share|improve this answer

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, $[ceH+] approx pu1e-4 mol L-1$:

$$α = frac[ceH+]V_mathrmac_mathrmbV_mathrmb = fracpu1e-4 mol L-1cdotpu50 mLpu0.1 mol L-1cdotpu50 mL = pu1e-3~textor~0.1%$$

share|improve this answer

answered Apr 1 at 14:50

andseliskandselisk

19.7k665128

$endgroup$

share|improve this answer

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