@skieurope I am not saying you are wrong but why would the college board release last years exam and say you needed a 72-100 for a 5
Okay, so determining pKa with a titration curve is easy because of Henderson Hasselbach, but… How would we find pKa with given concentrations of titrant and analyte? That seems like we’d be using ICE boxes, and I thought that wasn’t in the new curriculum anymore D:… I’m a bit fuzzy on those because I haven’t done those in ages.
Im pretty sure you still need to know rice tables
Although pka=ph at half way equivelence point
Also if it says 200 ml of .2M titrant and 200ml of titrated, solve for miles of titrant, do some stoichiometry and youll find concentration of titrated
@Mathman97 Thanks for the info!
Anybody wanna work through this problem with me?
20 mL of .2M HCN is titrated with .2M NaOH. The pH at equivalence is 11.1035. What is the Ka of HCN?
@APScholar18 The score is a composite. Total was 72%; FRQ was 31/46 or 67%. Regardless, that does not mean that the curve TY will be similar, although my guess is that it will.
How much time are we going to have and how many questions are we going to have on the frq section of this test ?
Not sure if this is right bc I didnt use molarity/volume, but if ph at equivelence is 11.1035 and ph=pka at equivelence, then pka is 11.1035
Pka=-logka so ka=10^-pka
Which equals 7.88e-12
105 minutes for 7 questions (3 long/4 short)
@Mathman97 pH=pKa at half equivalence though? Henderson Hasselbach says that pH=pKa+log(base/acid), so you can only get pH=pKa when log(base/acid)=1, which is at half equivalence.
Ah yes yes give me one second didnt read correctly
20 mL of .2M HCN is titrated with .2M NaOH. The pH at equivalence is 11.1035. What is the Ka of HCN?
HCN—> H+CN
[H]*[CN]/[HCN]
10^-11.1035=7.9E-12=[H]
CN should be the same as H+
HCN is .2 minus the concentration of H+, found using the rice tables
Does the 105 minutes include a 15 minute reading period or is it separate ?
Going along with the previous post.
Because they told you what the pH is, you don’t have to worry about the hydrolytic expression.
Simply treat it like a simple weak acid pH problem.
^ There is no separate reading period @Kitsyxoxo
@tau628 Please elaborate, pl@x.
Hey does anyone know how the 1999 MC test compares with the more recent AP chem tests in terms of difficulty, topics, etc? I’m planning to use the 1999 one (on college board site) to study if it is similar enough.
Thanks!!!
The topics are different, this test will be more general, will use more charts/graphs, also no nuclear chem apparently
@Newdle
HCN + NaOH -> NaCN + H2O
CN- + H2O -> HCN + OH-
NaOH neutralizes HCN to form CN- ions. From the pH you can calculate the [OH-] and then the Kb value for HCN. Then its just Ka = Kw / Kb.
I got 6.13 x 10^-10 for the Ka value.
Can someone explain number 38 from the 2014 exam? I don’t understand it at all.
@taw1020 Thanks! That was a lot of work ._.
Also, would Kb=[HCN][OH-]/[CN-] ? How would you find [HCN] ? I know you can get OH- with pH and CN- because of stoichiometry with the first equation… Does [HCN]=[CN-] ? Thanks!