Can someone explain 2013 #56?
Is that 72 composite or a percentage?
Composite.
Can anyone explain the difference between weak acid- strong base and strong acid- weak base calculations?
Does anyone know how to do:
A .1 m solution of a weak monoprotic acid has a ph equal to 4. The ionization constant ka is:
1x10^-7 is the answer, but why? and why isn’t it 1x10^4?
Thanks!
It’s #45
for #3, (see edit)
i’m not sure why C can’t be right.
This site specifically states both C and D as the reason: http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/Periodic_Properties_of_the_Elements
But i think that the key word is “BEST justification”, and i guess that effective nuclear charge is more clear than an atom having a full valance shell.
for #11, you need to find out what is left after NaOH neutralizes HCl (what is in excess)
NaOH + HCl → NaCl + H2O
.200L of 2M NaOH = .4mol NaOH
.500L of 1M HCl = .5mol HCl
.1 mol HCl remains, and it ionizes completely (since it is a strong acid) into .1 mol H+.
[H+] = .1 / .7
Without using a calculator, you can tell that it’s not 3 or greater because a pH of 3 would mean that [H+] is .001, a lot smaller than .1 / .7.
EDIT: i thought about it some more and for #3, C is not a COMPLETE justification. saying that it has a “completely filled valance shell” does not fully explain it. By choosing C, you are implying that filled valance shell’s are more stable without explicitly stating it; therefore, C is not the best.
2nd Edit: i think D is better because “higher effective nuclear charge” is by definition a greater attraction between the valance electron and the nucleus, unlike with filled shell and stability.
@Mathman97 For #3, the College Board wants justifications regarding ionization energy and other periodic trends to always be based on the effective nuclear charge of an atom.
Also, could someone explain #17 on the 2014 practice test? That’s the one with the potential energy as a function of internuclear distance graph. I chose B, because I thought N2 would have a higher internuclear distance than O2, but the answer is A.
that makes sense about #3
for 11, why are you doing things with .7? i understand now its ph after neutralization but would you just subtract .5 and .4 to get .1 mol of HCl, then do -log[H] to get 1?
sleepdeprived i was just gonna ask for 17 for the same reason haha
@sleepdeprived4
you have to look at their structures.
The higher bond order the more stable the bond and shorter the bond length (shorter internuclear distance).
More stable = less energy.
I get O2 and N2 but not H2 for #17 on 2014. I guess that H2 is where it is because it is a few energy levels smaller than O2 or N2.
@Mathman97
pH is from the CONCENTRATION of H+, not the moles of H+.
.5L of HCl was mixed with to .2L of NaOH, and .7 is the total volume of solution.
@stemscholar lets take a look at the dissociation equation
HA --> H+ + A-
That would give us an equillibrium expression of:
Ka = [H+][A-]/[HA]
Remember when the weak acid dissociates, it will form equal concentrations of the H+ ions and and weak conjugate base. Since we know the pH is 4, the concentration of the H+ is 110^-4. Likewise, since the A- ions are in the same concentration, their concentration is also 110^-4. Plug in the values:
Ka = [H+][A-]/[HA]
Ka = (110^-4)(110^-4)/(0.1)
Ka = 1*10^-7
Can someone also explain 55. which of the following is true? Answer: equilibrium vapor pressure of a liquid in a closed system is independent of the volume of the vapor pressure.
I was thinking that if vapor pressure was greater, then the liquid would be pushed down and would also have greater pressure, but why is that wrong?
@taw1020 How does 0.1mol of Zn=6.54g?
So how many total points are on this test if we need 72 points for a 5?
There are 100 total points @alldaboston
A solution containing HCl and the weak acid HClO2 has a pH of 2.4. Enough KOH is added to the solution to increase the pH to 10.5. The amount of which of the following species increases as the KOH is added?
(A) Cl-(aq)
(B) H+(aq)
© ClO2-(aq)
(D) HClO2 (aq)
Can anyone help me with this problem? I suck at these kinds of questions. 
thanks taw… i guess we know who’s get a definite 5 
could i get help with 23 on 2014, why is it not 2.4
@stemscholar
Vapor pressure involves two functions that are in dynamic equilibrium (it even says equilbirium in the problem):
(1) Escape of molecules from liquid phase to gas phase
(2) Re-entry of gas molecules to the liquid phase.
Like with reactions, changing the volume just causes the system to restore equilibrium.
Decreasing volume would increase the pressure of the vapor initially, but the system will restore equilibrium by the re-entry of gas molecules into the liquid phase.
Increasing volume would decrease the pressure of the vapor initially, but the system will restore equilbirium by the escape of molecules from the liquid phase to the gas phase.
Thanks @APScholar18
@bm1999 both calculations are done with an ICE (or RICE) chart. weak base with strong acid, most likely you will have to find the Kb and weak acid with strong base, probably Ka.
However, the CURVES are different, in that for weak-base and strong acid, you start off at a pH above 7, not all the way at 14, and then end at somewhere below 3 or 4 pH. you will have to see the halfway EQUIVALENCE point, to find out the pKa (pKa=pH at this point remember) and you might be asked to find Ka. for weak acid, strong base, it is the same thing, except you start somewhere above 2 for the pH and go all the way up to a 12-14 pH (indicating a very strong base). here you also find the halfway equivalence point to find the pKa, and then go to Ka.
22:
H2 is consumed with a ratio of 2:1 compared to CO2. For any reduction in CO, the reduction is twice for H2.
It was easier for me to do this problem if I used dummy values (based on the Kc value, they are way off, but it demonstrates the concentrations relative to each other while the system establishes equilibrium).
If .2 mol H2 were used, [H2] would be .8, [CO] would be .9, and [CH3OH] would be .1
checking this with the choices, the only one that aligns with this is B.
23: (not sure about this explanation)
What I think is that you use P1V1 = P2V2 to find the INITIAL total pressure. The system is now at that new pressure without having established equilibrium yet.
Since pressure increases, equilibrium shifts to the right (least number of moles)
When the system moves towards equilibrium, 1 mol of CO and 2 mol of H2 are consumed for 1 mol of CH3OH, reducing the total number of moles (change in moles is -2). Based on this, it can not be C or D.
But, since the K value of the reaction was so high, the change that results from decreasing the partial pressures of reactants is so minimal that it does not bring it below 1.2 atm again. So, the answer is B.