I have to calculate the pH of citric acid(0.1M) and sodium citrate(0.1M).I calculated the pH from ionic strength and the charge balance equation and found it to be 3.105. However, while doing the calculations I have assumed the activity coefficients to be equal to one. Now from the pH which I have calculated I get concentration of all species. Now I am guessing using these concentrations I recalculate the ionic strength and then the activity coefficients. After getting the new activity coefficients, do I again use them in charge balance equation, recalculate all concentrations and the pH and continue the iterations till the value of ionic strenghth and hence pH doesnt change. The value of this ph should be 5. However, I am unable to reach the solution. Is my approach correct or am I missing something.
Thanks,
ssn85
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Ph Of Citric Acid In Sodium Citrate Buffer
Started by ssn85, Dec 19 2009 06:54 PM
1 reply to this topic
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#1
Posted 19 December 2009 - 06:54 PM
#2
Posted 25 December 2009 - 04:58 PM
My understanding is that something is missing, although your procedure looks theoretically correct. Change of pH from 3.1 to 5 due to ionic strength may be too high, especially for 0.1 M solutions. Ionic strength may be neglected for simplicity (i.e. all activity coefficients=1); even so calculations are rather complex.
Probably calculation considered citric acid as monoprotic (K1=8.3x10-4) and assumed concentrations of citric acid, Na+, citrate- equal to 0.1 M, resulting in pH=3.08~3.105. But K1 is relatively high, so a monoprotic acid would give pH~1.60 in my opinion (equation x(x+0.1-(1E-14)/x)/(0.2-(x+0.1-(1E-14)/x)=8.3E-4, x=H3O+ concentration, x+0.1=(citrate-)+(OH-) due to charge balance).
Besides citric acid is polyprotic with K1=8.3E-4, K2=4.1E-5, K3=3.2E-6 (20 deg C), as one can see by googling "citric acid dissociation constants" (values may be slightly different). It is worthwhile looking at "acid dissociation constant" in wikipedea, where the behavior of citric acid is showed on diagram. Seeing K1, K2, K3 are close to each other, all three dissociations (or at least K1, K2, then checking dissociation due to K3) should be considered simultaneously. Supposing this was not taken into account in the calculation (?).
Considering ionic strength, it may not be an issue for 0.1 M solutions. Dissociation constants of acids are usually based on concentrations (just to simplify things) and College Chemistry by Schaum uses 0.001M-1M solutions without mention to activity coefficients (Ionic Equilibrium).
Well, calculation may not be so complex as it seems at first look. Hope this is of some help.
Probably calculation considered citric acid as monoprotic (K1=8.3x10-4) and assumed concentrations of citric acid, Na+, citrate- equal to 0.1 M, resulting in pH=3.08~3.105. But K1 is relatively high, so a monoprotic acid would give pH~1.60 in my opinion (equation x(x+0.1-(1E-14)/x)/(0.2-(x+0.1-(1E-14)/x)=8.3E-4, x=H3O+ concentration, x+0.1=(citrate-)+(OH-) due to charge balance).
Besides citric acid is polyprotic with K1=8.3E-4, K2=4.1E-5, K3=3.2E-6 (20 deg C), as one can see by googling "citric acid dissociation constants" (values may be slightly different). It is worthwhile looking at "acid dissociation constant" in wikipedea, where the behavior of citric acid is showed on diagram. Seeing K1, K2, K3 are close to each other, all three dissociations (or at least K1, K2, then checking dissociation due to K3) should be considered simultaneously. Supposing this was not taken into account in the calculation (?).
Considering ionic strength, it may not be an issue for 0.1 M solutions. Dissociation constants of acids are usually based on concentrations (just to simplify things) and College Chemistry by Schaum uses 0.001M-1M solutions without mention to activity coefficients (Ionic Equilibrium).
Well, calculation may not be so complex as it seems at first look. Hope this is of some help.
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