<p>Try 140s XD</p>
<p>I’ll end up at the 100-115 range haha.</p>
<p>@bleach: college board will probably be looking for some specific things, with essay questions like this, it’s always a crap shoot.</p>
<p>@ubermevin: who said I was a student ;)</p>
<p>@hypnotoad107: :o</p>
<p>hahahaa. So I guess you are a teacher(Idk) . 
. But i got to admit that you are really kind.<br>
@Can’tConcentrate- Lol that is nice. I am expecting somewhere in between 115-125. Hey guess what? We were in that study room with leehannah. xD</p>
<p>did anyone not get that form but take it on the same day…?</p>
<p>@uber: student told me about this site, thought I’d check it out to see what students thought and extrapolate a curve (the more difficult, the better the curve). Saw some questions and thought I’d help out. Did screw up on that rate constant though 
My students thought it was hard, but then they all ended up doing better than they thought when I went over the answers. End of 1 and middle of 2 just freaked them out. They made up a LOT of points in 4-6</p>
<p>how do u do part f of FRQ 1</p>
<p>@ hypnotoad. You sound exactly like my teacher. :D. My teacher did the same thing with rate constant as you did. I appreciate for working out all the problems boss. :D</p>
<p>Determine amount of HA left after reaction (.0025), and amount of A- formed (.0075). Buffer soln so use Henderson Hasselbach equation to solve for pH. then H+ from there.</p>
<p>@Adezar. Go back like one or two pages. Hypnotoad worked out all the problems.</p>
<p>@Uber: your teacher sounds awesome :)</p>
<p>Hahaha yes he does. Tomorrow, I will tell about you to my teacher. :D</p>
<p>for part F, i found the number of moles of HA and OH that were in the solution, then i did a ice table where all of the OH- disappeared so i got .016mol - .0075 mol. so the concentration of HA became .0085mol/.08 or .10625M. the concentration of A- then became .0075mol/ .08 or .09375M. then i just did the ICE table equation and did [A-][H3O+]/[HA] or [.09375][H3O+]/[.10625] = Ka which was 7.14 x 10^-5=[H3O+]</p>
<p>i thought it was similar to the Form B frq 1
<a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools;
<p>Initial HA was .01 mol. .01-.0075= .0025 mol of HA. Not sure where you got .016.</p>
<p>@adezar: yup, pretty much exactly the same.</p>
<p>@ec1026 I had a different form also. I had form E, most people have form O (the one online)</p>
<p>i got the .016 from the .2M (80mL of solution) solution times the amount so .2 x .08= .016mol</p>
<p>i think part e is a new situation and is independent of parts a-d</p>