Base titration of acetic acid model result vs reality

[waynehwm]

Hi David,

I am trying to model a base titration curve of acetic acid to match the published titration curve (https://users.humboldt.edu/rpaselk/C109.S11/C109_Notes/C109_lec41.htm).

As we know the acetic acid has a buffering effect around pH 4.5 that delays the rise of pH. But the output curve from my model does not have that same trend as the published data. I wonder if it is because my model is set up incorrectly, or if you think the model is correct what do you think could be the cause of the difference between model and actual data?

As always, thank you David.

Code: [Select]

SOLUTION_MASTER_SPECIES
Ac     Ac-         0.0 60.00 60.00

SOLUTION_SPECIES
#Acetic acid
Ac- = Ac-
        log_k   0.0;    -gamma  1e7   0.0
Ac- + H2O = HAc + OH-
        log_k   4.75;    -gamma  1e7   0.0

SOLUTION 1
    temp      20
    pH 2.5
    redox     pe
    units     mol/l
    density   1
    water    0.050 # kg
            Ac 1
SELECTED_OUTPUT; -file case2.prn; -reset false; -high_p true
USER_PUNCH
-headings pH
 -start
 10 PUNCH -LA("H+")
 -end

USE SOLUTION 1
REACTION 2
    H2O        1
    NaOH       0.00180
0.2 moles in 40 steps
END

[dlparkhurst]

You have the wrong K for the reaction, or you have the wrong equation for the K:

Ac- + H+ = HAc
        log_k   4.75; 

[waynehwm]

Hi David,

Yes, you are correct, now the titration curve fits data perfectly.

Thank you very much!

[pawanagra0@gmail.com]

Hello,

My name is pawan. I have recently started learning PHREEQC. I was looking at this titration example and did not understand how he has defined these reaction steps

REACTION 2
    H2O        1
    NaOH       0.00180
0.2 moles in 40 step

and second, if log k =4.75 is not the correct value than what is the right value?

Thank you
pawan

[dlparkhurst]

The REACTION is 0.0018 mol NaOH and 1 mole--0.018 kg--of water; so, the reaction represents about 0.018 L of 0.1 N sodium hydroxide. A total of 0.2 mol of this reaction (or about 3.6 mL of 0.1 N NaOH) is added to the SOLUTION incrementally.

log_k of 4.75 is the correct value for the following equation:

Ac- + H+ = HAc

It is not the correct value for the original equation. log_k = -14 for the reaction H2O = H+ + OH-. So, 4.75 + (-14.0) = -9.25 would be appropriate for this equation:

Ac- + H2O = HAc + OH-

[pawanagra0@gmail.com]

Hello,

Thank you so much for your quick reply. It was really helpful. I asked you this coz I am working on the same thing. I have titrated 0.05M acetic acid with 0.1M NaOH but I want to calculate only some pH values from PHREEQC as I got it from the experiment and I am not getting them. Hereby, I am attaching the google drive link for the excel file that has the experimental titration results and I want to calculate the three pH values which are in the black boxes in the excel file. The link to that is given below:

https://docs.google.com/spreadsheets/d/16ksYdMJyPi14yngB7w7Fkaabys46J4ee/edit?usp=sharing&ouid=114700544890023608870&rtpof=true&sd=true


Phreeqc code is also given below. Please help me. I will be very thankful to you.

SELECTED_OUTPUT 1
    -file                 D:\output files\selected_output_1.txt
    -solution             true
    -pH                   true
    -totals               NaOH
    -molalities           NaOH
SOLUTION_MASTER_SPECIES
Cl            Cl-              0     Cl              35.4527
Cl(-1)        Cl-              0     Cl             
Cl(1)         ClO-             0     Cl             
Cl(3)         ClO2-            0     Cl             
Cl(5)         ClO3-            0     Cl             
Cl(7)         ClO4-            0     Cl
Ac            Ac-              0     60              60
Na       Na+            0.0     Na              22.9898
SOLUTION_SPECIES
Ac- = Ac-
    log_k     0
    -gamma    10000000 0
Ac- + H+ = HAc
        log_k   4.75;
Cl- + 0.5O2 = ClO-
    log_k     -15.1014
    delta_h   -66.0361 kJ
Cl- + O2 = ClO2-
    log_k     -23.108
    delta_h   -112.688 kJ
Cl- + 1.5O2 = ClO3-
    log_k     -17.2608
    delta_h   -81.3077 kJ
Cl- + 2O2 = ClO4-
    log_k     100
    delta_h   -62.0194 kJ
ClO- + H+ = HClO
    log_k     7.5692
ClO2- + H+ = HClO2
    log_k     3.1698
SOLUTION 1
    temp      25
    pH        3.69
    pe        4
    redox     pe
    units     mol/l
    density   1
    Ac        0.05
    Cl(7)     0.01
    Na        0.01
    -water    0.032 # 32ml
USER_PUNCH 1
    -headings z
    -start
10 PUNCH pH = -LA("H+")
    -end
USE solution 1
REACTION 1
    NaOH      7.38e-05
   # 0.1 moles in 42 steps
END
USE solution 1
REACTION 1
    NaOH      7.88e-05
   # 0.1 moles in 42 steps
END
USE solution 1
REACTION 1
    NaOH      8.12e-05
   # 0.1 moles in 42 steps
END

[dlparkhurst]

I calculate that the concentration of Ac in your solution is 0.002 L * 0.05 mol/L / 0.032 L = 0.003125 mol/L.

Using that concentration, the experimental and calculated curves are similar.

Code: [Select]

SOLUTION_MASTER_SPECIES
Cl(-1)        Cl-              0     Cl             
Cl(7)         ClO4-            0     Cl
Ac            Ac-              1     60              60
SOLUTION_SPECIES
Ac- = Ac-
    log_k     0
    -gamma    10000000 0
Ac- + H+ = HAc
        log_k   4.75;
Cl- + 2O2 = ClO4-
    log_k     100
    delta_h   -62.0194 kJ
END
SOLUTION 1
-units mol/L
pH  4 charge
Na 0.01
Cl(7) 0.01
Ac 0.003125  # 0.002 L * 0.05 mol/L / 0.032 L = 0.003125 mol/L
-water 0.032
END
USER_GRAPH 1
    -headings               pH Calculated Experimental
    -axis_titles            "NaOH added, mL" "pH" ""
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
 10 DATA   \
0 ,3.67   ,  \
0.006 ,3.67   ,  \
0.012 ,3.68   ,  \
0.018 ,3.7    ,  \
0.034 ,3.75   ,  \
0.098 ,3.93   ,  \
0.154 ,4.09   ,  \
0.232 ,4.27   ,  \
0.32 ,4.45   ,  \
0.42 ,4.64   ,  \
0.528 ,4.84   ,  \
0.636 ,5.05   ,  \
0.738 ,5.29   ,  \
0.788 ,5.45   ,  \
0.812 ,5.54   ,  \
0.838 ,5.64   ,  \
0.882 ,5.9    ,  \
0.892 ,5.98   ,  \
0.9 ,6.04   ,  \
0.92 ,6.25   ,  \
0.928 ,6.37   ,  \
0.934 ,6.48   ,  \
0.942 ,6.6    ,  \
0.95 ,6.74   ,  \
0.96 ,6.99   ,  \
0.966 ,7.25   ,  \
0.972 ,7.62   ,  \
0.978 ,8.06   ,  \
0.984 ,8.56   ,  \
0.99 ,9.02   ,  \
0.996 ,9.35   ,  \
1.006 ,9.62   ,  \
1.02 ,9.83   ,  \
1.04 ,10.06  ,  \
1.064 ,10.24  ,  \
1.098 ,10.45  ,  \
1.138 ,10.6   ,  \
1.198 ,10.77  ,  \
1.274 ,10.91  ,  \
1.378 ,11.06  ,  \
1.512 ,11.2   ,  \
1.686 ,11.33
20 DIM mL(42), pH(42)
30 RESTORE 10
40 FOR i = 1 TO 42
45 PRINT i
50   READ mL(i), pH(i)
60   IF ABS(mL(i) - RXN*1000) < 1e-5 THEN GOTO 110
70 NEXT I
110 GRAPH_X rxn*1000
120 GRAPH_Y -LA("H+")
130 GRAPH_Y pH(i)
  -end
    -active                 true
USE solution 1
REACTION # reaction definition is 1 L of 0.1 M NaOH
NaOH 0.1    # 0.1 mol
H2O  55.506 # ~ 1 L
# additions of 0.1 M NaOH in liters
0.0000e-3
0.0060e-3
0.0120e-3
0.0180e-3
0.0340e-3
0.0980e-3
0.1540e-3
0.2320e-3
0.3200e-3
0.4200e-3
0.5280e-3
0.6360e-3
0.7380e-3
0.7880e-3
0.8120e-3
0.8380e-3
0.8820e-3
0.8920e-3
0.9000e-3
0.9200e-3
0.9280e-3
0.9340e-3
0.9420e-3
0.9500e-3
0.9600e-3
0.9660e-3
0.9720e-3
0.9780e-3
0.9840e-3
0.9900e-3
0.9960e-3
1.0060e-3
1.0200e-3
1.0400e-3
1.0640e-3
1.0980e-3
1.1380e-3
1.1980e-3
1.2740e-3
1.3780e-3
1.5120e-3
1.6860e-3


[pawanagra0@gmail.com]

Hello,

Thank you so much for your help. It means a lot to me. This help has given me more clarification and understanding about how phreeqc thinks and what I need to be careful about writing programs in this software.

I am running out of words. I do not know what else to say but thank you so much, Sir.

Have a perfect day ahead.

Pawan