Buffer Problem PLEASE HELP!!!

Do you have an idea of what equations you need to use?

For part a, you need to use an ICE table (Initial-Change-Equilibrium) to find the hydrogen ion concentration of the solution. Then you find the negative log of that and you have the pH of the solution.

For part b, you need to first list all the ions that are present in the equlibrium chemical equation, also known as the major species. From that list, pick the ion that will drive this reaction (in this caseit would be the formatio of water from the ions H+ and OH-.). Write out the equlibrium reacton, and place it into the Ka expression (Ka=[H+][A-]/[HA]). You'll need to rearrange the equation to find the [H+] value, but once you find that, calculate the amount of the [H+] that would bond with the OH- in the solution. If there is an excess of [H+], the solution is acidic and you can directly calculate the pH by taking the negative log- but if there is an excess of [OH-], the solution is basic and when you take the neg. log, you'll get the pOH, not the pH. Just subtract the pOH from 14 and you'll get the pH.

For part c, again start off by listing the major species and find the reaction that will occur. Find the number of moles of the reactants both before and after the reaction occurs, and place the final concentrations of the major species into the Henderson-Hasselbalch equation (pH=pKa+log([base]/[acid])). Remember that H+ is your acid, and OH- is your base. The H+ comes from the acetic acid, so use the Ka of that in the H-H equation.

I hope I helped you instead of confusing you. Just let me kow if you need more of a clarification on some of the steps.

I'm glad I could help you. I don't have my old tetxtbook with me right now, but I'll update this later and try to clarify that a little for you.

Okay- the H-H equation is really just another form of the Ka expression. It says that the pH of a solution is equal to the negative log of the Ka value of a solution's acid (represented by pKa) subtracted from the log of the concentration of a weak acid (represented by [HA]) divided by the concentration of its conjugate base (represented by [A-]). The formula is pH=pKa-log([HA]/[A-]). A conjugate base is the part of an acid that is left over after it dissasociates in water (for example, the conjugate base of HCl is Cl-). It can be found by subtracting the [H+] value at the end of the reaction from the original molarity.

I hope this cleared things up for you...

It looks good...
I'm sorry, I just re-read your other post, and realized that you didn't ask me to clarify the H-H equation.:o I really hope I didn't confuse you too much...But it does look like I helped a bit after all:rolleyes:

There is really no need to learn the H-H equation. Learn the formula for dissociation of a weak acid and the corresponding equilibrium law:

Ka= [H+][A-]/[HA]

and derive the H-H by taking negative logs all through

-logKa = -log[H+] - log([A-]/[HA])

rearrange this to get

pKa = pH - log([A-]/[HA])

If you do it this way there is less chance of making a mistake with the signs of the terms involved in HH (provided that you are careful when mutiplying through by -log)

I still think the H-H equation is good to learn, though. It really comes in handy when you need to find certain concentrations and you're only given so much information. But then again, I just happed to like using it, too...