Calculate pH Titration

  • Thread starter nautica
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  • #1
nautica

Main Question or Discussion Point

50.00 mL of 0.1000 M citric acid (Ka = 7.41 x 10^-4).

titrated 50.00 mL NaOH added.

Calculate pH.

I have equal moles of Acid and equal moles of Base. But, I am titrating a strong base into a weak acid. So, instead of my pH being 7, I assume it should be just above 7. But, I am not sure how to go about this.

I would not think that Henderson-Hass would work b/c we will be dealing with a log of undefined, which is undefined. The zero concentrations of each have me confused.

Thanks
Nautica
 

Answers and Replies

  • #2
chem_tr
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Citric acid is a triprotic acid as you know, and as a result trisodium citrate will be formed. This may be a cue.
 
  • #3
nautica
But, will that effect the pH at these concentrations???
 
  • #4
chem_tr
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Well, start with millimole amounts to find the first equivalence point; you have 50*.1=5 mmol of acid and 5 mmol of base is needed. How many mililiters will you use?

Also, you should know the starting concentration of the base, without knowing this property, it would be useless to consider anything.
 
Last edited:
  • #5
nautica
Sorry, I see the confusion now. I left out that the NaOH concentration is 0.1000 M. This is what is confusing me. At these concentrations they cancel out so Henderson Hass can not be used.

I have also not worked with triprotics, but I wouldnt think at these concentrations it would matter. I was planning on using the above Ka.

Thanks
Nautica
 
  • #6
chem_tr
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Well, since only one Ka value was given, we can assume that the first ionization is under consideration. If you want to study the other stages of deprotonization, you'll have to give us the other Ka values along with this one.

When you start the titration, you add, say, 1 mL of NaOH onto 50 mL of HA (I will regard this acid as monoprotic here). Here, the total volume will be 51 mL, and you can recalculate the concentration after each addition step, and find the relevant concentrations (base concentration will be practically zero, since it will immediately reacted by the virtual pool of acid).
 
  • #7
nautica
Yes, only one Ka was given, so I am considering it a monoprotic

Nautica
 

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