Processes > Mixing

Base titration of acetic acid model result vs reality

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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: ---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
--- End code ---

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-

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