Chemical Forums
Specialty Chemistry Forums => Biochemistry and Chemical Biology Forum => Topic started by: Simons2 on September 17, 2017, 12:26:35 PM
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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?
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You are moving into another buffer territory - H3PO4/H2PO4-.
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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.
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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 pKa1 to calculate pH (this approach won't give an exact answer, but finding a better one is probably beyond the scope of the exercise).
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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.
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Show your work, hard to comment on something we can't see.