[Simons2]

I have to predict the change in pH when 0.50ml of 1M HCl is added to 40mL of .10M buffer at pH 6. The buffer was made with Na₂HPO₄ and NaH₂PO₄.

I did this successfully for a pH of 7.5; pH=pKa+log[b/a]; 7.5=7.2+log[b/a]; b=2a. Now I kept everything in milmols instead of changing it to moles so a+b=4mmols a=1.33 b=2.7.
HPO₄ + H+ ==> H₂PO₄
2.7.. 0 ... 1.3 -The addition of the .5mL of HCl ends up with the final being: 2.2mmol of HPO₄ and 1.8mmol of H₂PO₄. Plugging those numbers into the HH equation: 7.2+log[2.2/1.8]=7.29 so the change in pH is 7.5-7.29 = .21. I understood how to do all this.

Now if I were to do this with pH 6, I would end up with a negative concentration for HPO₄ and you can't plug a negative number into a log. Following the steps above; b=.0631a; a+b=4mmol a=3.76 b=.237. Add the HCl and you get a negative for b: (.237-.5=-.263). I understand when you get a number that is negative you have excess HCl but I can't work with a negative number in pH=pKa+log[b/a]. I have no idea what I'm doing wrong. I also have to do this for 40mL of .01M buffer with the addition of .5mL of 1M HCl as well, and all those numbers will also be negative. Do I just take the absolute value of it or is my math wrong?

[Borek]

You are moving into another buffer territory - H₃PO₄/H₂PO₄^-.

[Simons2]

You are moving into another buffer territory - H₃PO₄/H₂PO₄^-.

I don't understand how that helps when solving for pH change. I understand it hypothetically because of all the excess HCl in solution, but mathematically I don't see the relevance/how to use it for solving the problem.   

[Borek]

Follow the stoichiometry - once all HPO₄^- gets protonated into H₂PO₄^- excess HCl (the one giving you negative concentrations) will start to protonate H₂PO₄^-  into H₃PO₄. You need to calculate the concentrations of H₂PO₄^- and H₃PO₄ and use pK_a1 to calculate pH (this approach won't give an exact answer, but finding a better one is probably beyond the scope of the exercise).

[Simons2]

I've figured it out, but thank you! I do have a different question tho;

I'm asked to calculate the pH using phosphate correction factors when 0.50ml of 1M HCl is added to 40mL of .10M buffer at expected pH 6 and pH 6.5. So same numbers as above. I know the correction factors: Correction factors for .10M: HPO4-0.45 H2PO4-0.74 as well as the equation pH=pKa +log[b/a]+log[cf of b/cf of a]. I've been getting very messed up answers that don't come close to what I've measured in the lab. When I calculated it for pH's 7, 7.5,8 they all came close to my measured pH. pH 6 and 6.5 didn't. I'm unsure what to do.

[Borek]

Show your work, hard to comment on something we can't see.