I dont know what s/he did, but hcn is not equal to cn, the concentration of the products are the same though
@alecdtatum
At the half-equivalence point for a titration, pH = pKa of the thing being titrated.
Halve the volume of titrant added at the eq point to find the half-eq point. The pH at that point on the curve is equal to pKa.
To find the molarity of the titrant, moles of acid = moles of titrant at eq-point.
Calculate moles of acid from the given concentration and volume. Use the volume of titrant at eq-point and the moles of titrant (which is equal to moles of acid) to calculate molarity.
@Newdle
You’re right with the Kb expression.
You know that 4mmol of HCN reacts with 4mmol of NaOH*. This forms 4mmol of NaCN. To calculate [CN-], you need to add the volumes of both HCN and NaOH together (40mL) to get .1M.
Since the equation is CN- + H2O -> HCN + OH-…
[HCN] = [OH-]
It’s basically an ICE problem for this part.
*.20 M HCN = x / 20mL --> x = 4mmol HCN
Since it is at equivalence, there are 4mmol of NaOH. The volume of NaOH added:
.2M = 4mmol / y --> y = 20mL.
So how do you use naoh in the equilibrium expression?
Why is oxygens first ionization energy greater than that of Phosphorus?
Because ionization increases the further right it is and the higher it is on the pt
This is bc there are less protons to attract the electrons btw
@Mathman97
You don’t use NaOH in the equilibrium expression. All of the NaOH was consumed in the neutralization reaction to produce NaCN and H2O.
You need the NaOH to find the concentration of CN- (I edited it in my earlier post).
So you only use it to find number of moles of hcn?
No.
The problem tells you that the concentration of [NaOH] is .2M. It also tells you that the acid being titrated have a volume of 20mL and a concentration of .2M. From this, you can calculate 4mmol of HCN present. At the equivalence point, mole acid = mole base, so 4mmol of NaOH is added. Since the concentration of the NaOH is .2M, you can calculate the volume of NaOH added to be 20mL.
4mmol of HCN react with 4mmol of NaOH to form 4mmol of NaCN, which dissociates into 4mmol CN- ions. Calculate [CN-] to be .1M.
The problem tells you that the pH is 11.1035. This is the pH at equilibrium conditions.
pOH = 14 - pH = 2.8965
[OH-] = antilog(-2.8965) = .00127 M
The CN- acts as a base when it reacts with water
CN- + H2O -> HCN + OH-
The equilibrium expression for this equation is the Kb value for HCN, since CN- is the conjugate base of HCN.
At equilibrium, [HCN] = [OH-] because the CN- hydrolyzes to a 1:1 ratio of HCN and OH-.
Initially, [CN-] = .1M, [HCN] = [OH-] = 0
At equilbrium, [OH-] is .00127M (calculated above), so [CN-] at equilibrium is .1 - .00127 = .09873M and [HCN] = .00127M
Kb = ([HCN][OH-])/[CN-] = 1.63 x 10^-5
Ka = Kw / Kb = 6.13 x 10^-10.
Keep in mind that I didn’t bother with sig figs.
Thank you so much,I get all the calculations you did, my only questions is how do you know whether to calculate ka or kb
In a conjugate acid/base pair, Ka corresponds to the acid’s reaction with water, and Kb corresponds to the base’s reaction with water.
E.g:
NH3 and NH4+
For NH3 + H2O -> NH4+ + OH-, you would use Kb (NH3 is the base).
For NH4+ + H2O -> NH3 + H3O+, you would use Ka (NH4+ is the acid.
HC2H3O2 and C2H3O2-
For HC2H3O2 + H2O -> C2H3O2- + H3O+, you would use Ka.
For C2H3O2- + H2O -> HC2H3O2 + OH-, you would use Kb.
If you want to the the other from one, you would use the equation Kw = Ka * Kb, where Kw = 1x10^-14 at 25C.
Ahhhh ok, how are you so knowledgeable btw?lol are you in college
Nah, I just forced myself to do a bunch of these problems after I barely passed the test.
I’m also pissing myself in fear, trying to cram for the exam.
Ah haha yep just started studying this morning myself
@mathman97 your explanation doesn’t make sense oxygen has more protons not less?
Thats the general trend for ionization, lemme explain for this problem.
Phosporous is a row below oxygen meaning it has an extra shell of electrons. These electrons are further from the nucleus, making them easiwr to pull off, meaning the ionization energy is lower. The proton part only applies to elements in the same row.
@mathman97 ok but why does Nitrogen have a higher ionization energy than oxygen?
Does anyone have a list of the specific labs you need to know for the ap test ?
Oxygen has one sub shell with two electrons in the p orbitals while nitrogen symmetrically has 1 electron in each subshell, since oxygen wants to be like nitrogen, it more easily gives an electron
@Mathman97 doesn’t this defy the periodic trend?
Hey Guys! I’m getting ready for the AP Chem exam on Monday, but my teacher didn’t give us any information on the Lab Investigations. Do any of you know where I can find these? Thanks!