Processes > Mixing

Base titration of acetic acid model result vs reality

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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: ---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

--- End code ---

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

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