<p>do u guys want to go through the free response and consolidate</p>
<p>For the buffer one, I put that it wasn’t a good buffer because in order to be a good buffer, the pH of the solution has to be in the range of +1 or -1 of the pH of the buffer. I think that was it.</p>
<p>That would be nice.</p>
<p>1)</p>
<p>a) PH of HCl is 1.</p>
<p>b) Kb = [NH4+][OH-] / [NH3]
[OH-] = 1.3e-3 M</p>
<p>I kinda guessed 6.d)
Is the final pressure 0.80 atm? Because ethanol decomposes into 2 particles, so the pressure is doubled?
I’m probably wrong, because that just seems too simple.
Anyone else get a different answer?</p>
<p>That’s exactly what I did, roobear!</p>
<p>Good to know I got 1.a) right . . .
I don’t have my sheet with my so I don’t know what answers I got for most of the other questions.</p>
<p>what did u guys get for 1c?</p>
<p>1(b).</p>
<p>i: Ka * Kb = 1 * 10^-14</p>
<p>Ka * (1.8 * 10-5) = 1 * 10^-14</p>
<p>Ka = 5.6 * 10^-10</p>
<p>Not 100% on ii so I won’t post it.</p>
<p>wait, why did you the kW constant? I didn’t think that was a hydrolysis reaction.</p>
<p>When will college board post up the answers?</p>
<p>Don’t expect them up until after AP exam score results are released.</p>
<p>BTW, did anyone get .100 for 1c i?</p>
<p>^ I don’t think so. I don’t remember.</p>
<p>What was 1.c) ii?
Is it supposed to be 7?
I got 6-something, but I"m pretty sure I did it wrong.</p>
<p>Kw is used to convert between Ka and Kb.</p>
<p>Yeah, I know. But in the equation, Kb is defined. The formula also fits with the kB description. That’s how I got my answer.</p>
<p>Whoops I accidentally put B instead of C. That’s the answer to (c)i not (b)i.</p>
<p>The question asked whether the new solution, after combining with the buffer would be an effective buffer.</p>
<p>First of all, i didn’t know how to calculat the ph, so i just bsed it. Here is what i wrote.</p>
<p>The new solution wouldn’t be an effective buffer because the ph doesn’t correspond to the pKa.</p>
<p>i said it was an effective buffer.</p>
<p>Since people seemed to have the most problems with 6, I’ll post my answers and people can check them.
6. (a) equal to; the amount of liquid ethanol is constant at equilibrium
(b) Reason one: greater temp => greater kinetic energy; more intermolecular bonds broken, molecules can escape as gases & exert vapor pressure
Reason two: pressure is directly proportional to temperature
(c) (i) 0 order: time vs. [ethanol] is linear
(ii) Rate = k
(iii) Because rate = k, we need to find the rate, which is change in [ethanol] over change in time. k = 4.0 * 10^-6 mol/L*s
(d) 0.40 atm * (2 mol products/1mol reactant) = 0.80 atm</p>
<p>did anyone get 0.8303 moles for 2b?</p>
<p>would you get double jeopardy if you get the wrong order and answered ii and iii according to the order you got?</p>
<p>Nope, they’ll give you points as long as you’re consistent.</p>