Murka

You didn't use the Ka of the acetic acid to find X, you used the pH=-log[H+] of the acid to find x=[H+]=[A-], then you use this X and [HA]=0.1M (from your data) to find Ka. Then you're being asked for the experimental data. your acid was neutralized at 12 ml of NaOH, right? This means that it was half-neutralized at 6. You know the pH at 6, what do you know about the half-titration point? What is special about it? leading you there... So, assuming 8 is correct, what do you know about 9?

You're coming at this from the wrong angle. From question #1 you are given pH and from there you can calculate [H+] and [CH3COO-] which are both X. Then, notice that in question #2 you're not given the Ka, but expected to calculate it with the X from question #1.

Read my scan, I've explained it, but it seems like you have missing data since you need the concentration of the unionized acid (CH3COOH) to calculate it.

Vinegar is essentialy diluted Acetic Acid (CH3COOH) which is a weak acid, Sodium Hydroxide is NaOH which is a strong base. The reaction would be CH3COOH + NaOH <-equilibrium-> (CH3COO-) + (Na+) + (H2O) but pay attention that since this is an equilibrium, not all the (H+) and (OH-) will be H2O, (H+) would be neutralized by (OH-) to form H2O but there will be more (H+) formed by the equilibrium reaction etc' etc' essentially you're looking at the simplified reaction for the acid HA <--equlilibrium--> (H+) + (A-)

I don't understand what would you want me to do, question 2 depends on the concentration of the acetic acid, and all the other answered are gained experimentally from your experiment's results i.e. half titration of the acetic acid from the naoh etc'. Also pay attention that from the Henderson-Hasselbalch equation (http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation) we see that when the concentrations of Acid and Base are equal the pH=pKa, thus the Ka of the acid can be calculated very easily.

Sorry, in the middle of a lecture, check this out see if this helps and we'll continue

Thank you everyone, timo I actually realized that cobstants commute with operators just before you posted it, thank you very much, as with many things, I have a feeling that the question is too simple thus I have overthought and overcomplicated things myself... Thus I present my proof H|psi>=a|psi> /*P() PH|psi>=Pa|psi> Use PH=HP as per given data Use Pa=aP, can be moved because a is a constant H(P|psi>)=a(P|psi>) Q.E.D.

Anyone?

That's the thing, I'm not sure I know how to proceed, maybe my definitions are wrong? From the definition of eigenvectors and eigenvalues: If Av = av, v is the eigenvecor, a is the eigenvalue of T. So if |phi> is the eigenvector of H with the same value as P|phi> as the eigenvector then... H|phi>=a|phi> HP|phi>=aP|phi> ? sigh...I realize those are really basic firsttimer questions, but I'm at my 3rd year at the university, and the last time I saw the word "operator" even mentioned was at my linear algebra course, which was at the 2nd semester of the 1st year...I've done all my math courses a long time ago

Hey guys, I have trouble grasping Quantum Mechanics...not such a rare condition, I guess I just need a nudge in the right direction, I seem to know the basics but somehow when I try and solve the homework we are given I get stuck and unsure on where to begin. Let P be an operator that commutates with the Hamiltonian HP=PH show that if |phi> is an eigenvector of H, P|phi> is also an eigenvector of H with the same eigenvalue. Thanks in advance!