[newageanubis]

Hey guys,

I'm currently working my way through some basic buffer solution problems. I understand the concepts and everything, but my answers are always 0.03 to 0.04 pH points off of the answer in the solutions manual. Normally, I wouldn't care, but given that these are buffer solution problems, I believe that kind of deviation is significant since 0.04 is on par with the pH change exhibited by a buffer solution resisting a significant pH change.

Here is the one of the problems from the set I solved:

"Calculate the pH of 0.100 L of a buffer solution that is 0.25 M in HF and 0.50 M in NaF. What is the change in pH on addition of the following?

a) 0.002 mol of HNO₃
b) 0.004 mol of KOH

I got 3.76 as the pH of the buffer solution, as did the textbook. However, I got 3.74 as the pH for a), while the textbook's answer is 3.71, and I got 3.83 as pH for b), while the textbook's answer is 3.87.

The only difference I can find between my solutions and those given in the solutions manual are that the solutions manual uses initial concentrations for the weak acid and its conjugate base as equilibrium concentrations because the change in their concentrations due to the dissociation of the weak acid is negligible. While this is frequently the case, I have been considering the change to be non-negligible in my calculations so I can practise the longer method of solving equilibrium problems that involves the quadratic formula (which is easy to make careless errors with). Anyway, if the change in concentration is truly negligible compared to the initial concentrations of the weak acid and its conjugate base, my answer should not be affected. Moreover, all of my concentration values satisfy the respective acid-dissociation constant equations, and all of my calculated pH and base and acid concentration values satisfy the Henderson-Hasselbalch equation.

Does this mean that my answers are still correct?

[Cavillus]

I repeated your calculations and I found where the mistake happens.

The thing you have to consider in this kind of problems is that, adding a small amount of strong acid to a buffer solution, we cause the acid-base reaction to be complete, for example:

                         NaF + H⁺ HF + Na⁺   

So, in the first calculation, we have, using the Henderson-Hasselbach equation:

M_HNO3 = C / V = 0,02 M

pH = pK_a + log ^{(C_NaF - C_HNO3)} / _{(CHF + CHNO3)}.

With this equation, you obtain 3,71 as the final pH, as the textbook says.

[newageanubis]

Thanks! I knew the something was wrong; this just goes to show that one should never trust the people over at Yahoo! Answers...xD

I went over my solutions, and saw that I indeed forgot to account for the production of HF when HNO₃ is added, as well as the production of F^- when KOH is added.

Once again, thanks so much!