02-04-2018, 06:31 PM

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?

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?