<p>On PR and Sparknotes SAT chem, they use the Henderson equation for Buffer problems. Will the real test expect us to calculate logs of random values in our head?</p>
<p>I’m 99.9% sure you’re not expected to know the Henderson-Hasselbach equation for the chem SAT II. It’s not in Barrons, on another site a chem teacher said you don’t need to know it for the SAT II, and I didn’t learn it in Honors Chem (my Honors Chem class taught the SAT II curriculum, unlike a lot of Honors Chem classes; a lot of people in my Honors Chem class got a 790-800 on the chem SAT II). I’m taking the chem SAT II October 9th, and I don’t know the equation. I didn’t see it on any College Board test I’ve taken or any practice test I’ve taken.</p>
<p>Yeah it’s not in Barrons (which usually goes OD with info) but I found it weird that it was in PR and Sparknotes. Thanks.</p>
<p>In PR, they have the equation screwed up…</p>
<p>I wish CB would just publish what the hell they want us to know.</p>
<p>Comming from someone who has not taken the Chem SAT II test (though I have taken the AP Chem exam), there are several things college board will do:</p>
<p>-You don’t have to know it
-You will be given simple concentrations
-You will be given harder concentrations (How its tested on the AP Chem exam) </p>
<p>Not much to say about the first option besides don’t expect it. The second option is the most likely: the [H+] or [OH-] concentration will be 1 to some exponent and the exponent is the pH or pOH. The third option is not much harder than #2 but still trip up some students. The problem will give you a [H+] concentration like 2.0^-4 and will give you decimals that fall into the following categories: >3 but <4, 4, >4 but <5, and then nonsensical answers. When the concentration is not given with a base of 1, the answer will always be less than the exponent but never by a margin greater than 1. For instance, in the example I just gave the concentration is exactly 3.70, the pH of a solution with [H+]=9.90^-4 is 3.004, etc. They will never give you a non-whole number exponent with a non-1 base, so this is as hard as pH problems will get.</p>
<p>EDIT: Whoops, wrote examples for wrong scenario (finding pH). The principles of logs still apply, just not the problem solving.</p>
<p>The equation in the 2009-2010 PR book is pH = pKa + log[conj acid/conj base] or</p>
<p>pH = pKa + log[A/HA]</p>
<p>What I’m confused about it: How is [A] the conjugate acid? and [HA] the conjugate base? I thought something with the “H” was usually the acid.</p>