<p>how would you do this problem:
if an acid is 15% dissociated in a 2.00M solution, what is the Ka?</p>
<p>theradtomato person, ur choice c and d are the same.</p>
<p>yo orangefoot i am not sure but basically its asking for the concentration after 15 %of the solution dissociates(leaves/breaksapart). what i would do is take 15% of 2.00M solution but i could be really wrong .</p>
<p>what do u think of this reasoning?</p>
<p>@tomato molarity = mol/V like vadata said, but density changes with temperature. Thus volume changes too. BUT molaLity - mole/kg. Mass doesnt change with temp…if D) was actually supposed to be molality :)</p>
<p>%diss= [H+]/[HA] therefore [H+]=.3 and K = .3^2/2 = .045
I think</p>
<p>^^yah i meant molality. My teacher didn’t really teach vapor pressure, are there any key things to know about it?</p>
<p>Vadata, I think you’re right.
So does anybody know if we need to know Beer’s law?</p>
<p>@orangefoot - Ka = ([H][A])/[HA]
15% of 2 is the value of [H] and [A].
2 - (15% of 2) is the value of [HA].
Solve to get Ka, which I got to be 0.0529</p>
<p>Ka = ((0.3)(0.3))/(2-0.3)</p>
<p>what is the short form of ICE during MCQs? its like x^2/(something) = Ka?</p>
<p>Vapor pressure occurs when particles in a liquid or solid “break free” from its IMFs (and the rest of the liquid/solid) and become gaseous. When vapor pressure = atmospheric pressure, the liquid boils.</p>
<p>ChickenonRice, ICE is x^2/[concentration of WA/WB/Salt]= ka/b</p>
<p>@rva2012 I have taken almost most of the MC tests since 1989 and 5 FRQ tests (teacher gave us a full length test every week for the past 5 weeks, I’ve gotten 5’s each time) and I don’t recall every seeing a Beer’s Law question in the MC or the FRQ. Regardless it is on the formula sheet so just plug and chug.</p>
<p>@ChickenOnRice - yeah, that’s usually how it is: x^2 over given molarity is usually Ka.</p>
<p>@4khoas You’re right, I assumed it was a weak acid which isn’t stated.</p>
<p>Okay, because 2005’s Free Response requires you to draw the calibration curve of Beer’s Law. I got worried.</p>
<p>Well according to my AP chem teacher, we’ll never need to do a calculation where we DON’T ignore the concentration of [H+] or [OH-], so I wouldn’t worry about it</p>
<p>how bout osmotic pressure?</p>
<p>thanks everyone for your explainations.
also for 2009 FRQ #3 why do we need to divide by the avogadro constant?</p>
<p>alright thanks guys. and normally what is the quickest way to find x?</p>
<p>I think it’s pretty unlikely that we’re going to get a question that requires an ICE box (or even the short form) on the MCQ’s. Unless they’re really simple, and I mean really simple, calculations, considering we don’t have a calculator…</p>
<p>For question no. 2 @ <a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board; , how is the cathode on the right? I thought that we had to look at the half equations and Fe clearly is being reduced to Fe2+ from Fe3+. But Mg also changes from +7 to +2. So how exactly do we figure out the cathode and anode?</p>