Identify the conjugate bases of Al(H2O)63+ and Zn(H2O)62

[https://pubchem.ncbi.nlm.nih.gov/periodic-table/png/Periodic_Table_of_Elements_w_Chemical_Group_Block_PubChem.png ]
As before, while the behavior of all molecules is rooted in the orbital structure and electron dynamics, it is convenient to define broad categories of molecules that behave in similar ways. In the past and in the absence of any way to determine precise structures, molecular behavior was essentially all that previous chemists had to work with. Despite the psychological weight of all the work that has been performed making categorizations and trying to fit all known compounds into a few well-defined categories, the reality of chemistry is and always has been messy. With that in mind, let’s talk about ions, acids, and compounds.
First up, I should spell out the difference between an element, a molecule, a compound, a substance, a chemical, and however else you want to refer to things. An atom of an element has a distinct atomic number, a distinct number of protons, and in the neutral and ground state (uncharged and in the lowest possible energy electron configuration) has a distinct orbital and electronic structure. It is of course possible to have many atoms of the same element as well as different numbers of neutrons in/isotopes of the same element, but (ie) a boron atom is going to have fundamentally different bonding properties than a carbon atom or any other atom. Once you start bonding atoms together, you end up with molecules that have distinct structures, molecular orbitals, and chemical behaviors. There will of course be similarities between similar molecules, but a molecule of methane (CH4) is different than one of ethane (C2H6 or H3C-CH3) or one of propane (C3H8 or H3C-CH2-CH3). Continuing with the hydrocarbons, once you get to four carbons (C4H10) and above it is possible to have multiple configurations of the main carbon chain, so technically you would need to distinguish between butane (historically unbranched/normal/n-butane) and 2-methylpropane (historically isobutane/i-butane). However, we are now getting into the different isomers (butane, 2-methylpropane) of the same chemical formula (C4H10) instead of sticking with the idea of compounds. So, I would define a compound as a unique three dimensional arrangement of atoms, but trying to describe a three-dimensional and possibly quite complicated shape in words concisely gets difficult extremely quickly. As with atoms, you can have very many identical molecules in the same place, although in practice mixtures of similar molecules are more common unless extensive purification work has been conducted. By the time we are talking about a substance, we have progressed from a 3D picture of one unique molecule to whatever you are physically dealing with. This could be a mixture of different compounds, different compounds with trace contaminants, one compound with trace contaminants, any of the above with more than trace contaminants, etc. In practice, figuring out what is actually in the test tube/beakeflask/reaction glassware is quite difficult, and you may notice that we are at the practical level after having started in theory.
Generally speaking, I would suggest starting with IUPAC naming conventions, which are intended to include as much structural information in the name as possible, as seen with the butane and 2-methylpropane names in the preceding paragraph. In practice, nearly every specialization will have their own naming schemes that they will defend to the point of exhaustion regardless of whether or not retaining a separate naming scheme that people then have to learn makes sense. My position is that the benefits of a naming scheme that can be applied as universally as possible and be comprehended by as many people as possible is going to outweigh the inconvenience of the extra characters in 2-methylpropane versus “i-butane”. Speaking of unique naming schemes, I had to attempt to memorize several during the course of my undergraduate instruction, including dihydrogen monoxide (H2O)/carbon dioxide (CO2), the difference between nitrate and nitrite (one is NO3^-, the other is NO2^-), sodium chloride (NaCl)/disodium carbonate (Na2CO3), and a bunch of other stuff that I clearly have not retained very well. When dealing with covalently bonded compounds like water or carbon dioxide, the main way to figure out whether you have a valid structure or a false memory is to keep track of/draw out the valence electrons of each atom and make sure that they all “own” the correct number of valence electrons and are participating in an appropriate number of bonds. In charged molecules/ionic compounds/salts, there is the additional complexity of making sure that the charges end up on the correct counterions. As an example, sodium (Na) can either have a charge of 0 (neutral) or a charge of +1, while a chlorine atom (Cl) has a charge of 0 and the chloride ion (Cl-) has a charge of -1. If your structure requires chlorine to have a positive charge, you have almost certainly made a mistake. Also, the net charge on any molecule defaults to 0, so adding up all of the charges on all of the ions is another useful check. In practice, only a few ions that will be commonly encountered, and searching either the structure or the name quickly brings you to the missing information. Unless you’re in an exam, of course.
For our purposes, that is a sufficient introduction to the rules of relatively simple and mostly inorganic compounds and their naming. I have left plenty of information out, but the best way to actually learn about this is to encounter the concepts and naming schemes during the course of your normal work or exploration and doing enough internet searching to fill in most of the gaps in your knowledge rather than trying to memorize enough information to immediately identify every compound in the entire universe. We also need to get into acid-base chemistry, which suffers from a surplus of definitions. The most familiar definition is the Brønsted-Lowry definition of an acid as a proton (H+) donor, and a base a proton acceptor. In this definition, an acid’s ability to donate H+ depends on how closely the proton is bonded to the atom it detaches from, leading to a spectrum of strong and weak acids and their conjugate bases. It is important to note that all of this is taking place in water, with the protons actually being donated to water molecules to form hydronium/H3O+, which is then complexed to and stabilized by the negative (non-hydrogen) ends of adjacent water molecules. The positive charge on the hydronium complex is countered by the negative charge on the former acid that is caused by the departure of a proton without its electron. Or, an acid (HA) donates a proton to water to form hydronium (H3O+) and the deprotonated conjugate base (A-) of the acid. The net charge is still zero, the number of electrons remains the same, etc. The same thing can happen when a base (B or B-) accepts a proton from hydronium (H3O+) to form the conjugate acid of the base (HB+ or HB), although the existence of B- in solution will depend on the presence of a positively charged counterion. Complicating matters further, water will always contain some hydronium (H3O+) and some hydroxide (OH-) due to spontaneous disocciation even before any Brønsted-Lowry acids or bases are added. It logically follows that hydronium (H3O+) is the strongest acid that can exist in water and hydroxide (OH-) is the strongest base that can exist in water. The degree of acidity or basicity that can be ascribed to a substance added to water is due to the degree of dissociation – strong acids or bases will almost completely dissociate, very weak acids or bases will hardly dissociate, and everything in between.
So and in summary, Brønsted-Lowry acids or bases can only donate or receive protons in water or another suitable medium. If we instead use the Lewis definition of acids as accepting electron pairs and bases as donating electron pairs, we are well situated to get into the organic chemistry reactions. It must be said that most people will not have much contact with situations in which the concept of Lewis acids and bases adds anything to the Brønsted-Lowry definition. At the same time, if we view the protons detached from Brønsted-Lowry acids as the actually acidic component (because this is the case) and the HA “acid” as a delivery medium for the proton, we find that the Lewis definition squares perfectly well with the idea of H+ accepting an electron pair and being “donated” to another molecule, with the other molecule donating an electron pair and “accepting” the hydrogen. The Lewis definition is both more broadly applicable and focuses on the electrons as the most important part of any chemical interaction, which makes sense because chemistry is the movement or presence of electrons more than anything else. The general chemistry professor who was responsible for teaching me about acid-base chemistry was an inorganic specialist, dismissed the Lewis definition as being unimportant, and caused me several years of confusion and anxiety as a direct result.

Acid rain has a reaction where Nitric acid reacts with magnesium carbonate. The reaction I get use modified Arrhenius theory is H3O(+) + CO3(2-)->HCO3(1-)+H2O. Assuming there isn't enough nitric acid neutralize all the magnesium carbonate would the soil pH change much? I've got two varying opions, one is that it wouldn't change as all the acid is neutralized. However the conjugate acid still remains however hydrogen carbonate is also a base so I can't determine if its possible the pH would rise. I don't think the removal of magnesium carbonate would effect pH much as in its solid form it can't dissociate into hydroxide ion.

Hey Reddit,
I've never gone onto these forums to ask chem questions but I am writing my exam for grade 12 chem in a few days and I could really use some help with these practice questions and understanding the concept:
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1. Redox Reactions

H2O2+MnO4^(1-)+H^(1+) →Mn^(2+)+H2O+O2
So I understand how to find the oxidation numbers here -
H[2+]O[-2] + Mn[7-]O[-8]...
but I am getting confused about comparing the compounds.. which Oxygen gets compared with which on the product side? Which Hydrogen on the reactant side gets compared to the Hydrogen attached to the water compound?
H2O2+MnO4^(1-)+H^(1+) →Mn^(2+)+H2O+O2
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1. Equilibrium:

a 1.00L solution is made of 0.40 mol/L NaNO2 and 0.25 mol/L HNO2 (Ka= 4.5*10^-4). What is the pH of the solution?
b. to this solution, 20.0 ml of 0.15 mol/L HNO3 is added. What is the resulting pH?
for part a, the answer key states HNO2+H2O < H3O(+)+ NO2(-) so where does the Na go?
for part b, HNO3 + NO2(-) -> HNO2 +NO3(-) -> how do you know what the HNO3 pairs with? I guess my question has to do more of understanding how to identify conjugate base/acids.
Thank you!

Why is the conjugate base of sulfuric acid weaker (more stable) than water acting as a base? I know the negative charge of HSO4- is resonance stabilized across three oxygen atoms, and oxygen is very electronegative, but wouldn't H2O be more stable because it has no formal charge at all? I'm trying to understand why sulfuric acid (or hydrochloric acid, for example) is a stronger acid than the hydronium ion.
Thank you!

Hi guys, I’m in organic 1 and still really struggling with acids/bases. I get the whole ARIO concept but this problem is really confused me: Which is most basic out of NH3, CH3COCH3, CH3COOH, and H2O. I have no idea how to determine the stability of a conjugate acid but I know how to do conjugate base so that’s what I was doing here. I got NH2-, CH3COCH2-, CH3COO-, and OH-. I am really confused at how NH3 has a higher pka than CH3COCH3 because the conjugate base of acetone has a negative on the carbon and the conjugate base of NH3 has a negative on the nitrogen. So shouldn’t NH3 have a more stable conjugate base and thus be a stronger acid? Just really confused here. •also bonus question will the conjugate acid of acetic acid and acetone have the hydrogens bonded to the O atom or the C? Someone said the acetone conjugate base would have 5 bonds on the carbon and thus be an unstable base and idk why you wouldn’t just put the hydrogen on the oxygen?
Please please help if possible, I’m so confused

Through a targeted high‐throughput screening strategy, birefringence‐active functional modules and a series of crystals containing the C−O units with giant optical anisotropy spanning from 0.1 up to 1.35 were found. The compound with [C6O6]2− has a birefringence about 11 and 6 times that of α‐BaB2O4 and YVO4. The experimental measurement for the (NH4)2C2O4⋅H2O crystal confirms the feasibility of the screening framework.

Abstract

The main commercially used birefringent oxides α‐BaB2O4 and YVO4 have the birefringences of 0.12 and 0.22. We propose a targeted high‐throughput screening system to search birefringence‐active functional modules (FMs) and large birefringent materials. A series of π‐conjugated C−O units [C2O4]2−, [C2O6]2−, [C4O4]2−, and [C6O6]2− are discovered to be birefringence‐active FMs. Theoretical and experimental studies on the crystals with C−O units confirm the feasibility of strategy. Based on this, the C−O containing compounds ranging from deep‐ultraviolet to near‐infrared region with large birefringence from 0.1 to 1.35 are found, and most of them break through the birefringent limit of oxides. The (NH4)2C2O4⋅H2O crystal is grown and its experimental birefringence is 0.248 at 546 nm, which is identified as a promising UV birefringent crystal. The A‐site cations play significant roles in optical properties by influencing the density and arrangement of the C−O units.
https://ift.tt/3likCSe

HCl + NaOH —> NaCl + H2O
I know the Na and Cl are spectator ion. I know the reason for this is that both are effective as conjugates because HCl and NaOH are strong acids and bases respectively, so their conjugates are ineffective. However, can I also look at this in terms of net ionic? Because HCl is a strong acid, it’s mostly present as H+ and Cl- right? Same thing with the NaOH in that it’s mostly present as Na+ and OH-. So here’s the net ionic:
H+ + OH- —> H2O or if I’m being accurate, H3O+ + OH- —> 2 H2O
This is why the pH doesn’t change because the product side doesn’t produce any conjugate acids or bases that are effective.
My question: Is the net ionic representation ok to explain why Na+ and Cl- are spectators in a strong acid strong base reaction?

If I am given the following equation:
H2O + NH2 gives OH- + NH3
and asked to identify the acid base, conjugate acid, and conjugate base, I can.
But
If I’m given:
Complete the following reaction
CH3=O-OH + CH3O- gives?
Or
CH3CH2OH + NH2- gives?
I don’t know how to since I don’t know which of the starting compounds is the acid or base, and hence I don’t know which one should be turned into the conjugate acid or conjugate base.
Help!

So I got this question wrong because I completely blinded myself to the fact that a lactone is present and even though the first OH- attacks the carbamate, forming a acid COOH essentially, the next OH- should attack the C=O of the lactone.
That being said, it is being assumed that the -O-C=O that is formed acts as a leaving group and leaves when N grabs proton from H2O.
SO I guess my question is more about how do we know that thats a good leaving group? I thought good leaving groups are the weak conjugate bases of strong acids or weak bases in general?
https://preview.redd.it/0uvrpe3kfp651.png?width=926&format=png&auto=webp&s=99e31bc82b7365b54e3e0cbc4a65fd2aead6d343
https://preview.redd.it/nzp2jz9jfp651.png?width=1796&format=png&auto=webp&s=77766c4f49c50a547399dfab73b024f95f7e44c7

In the kaplan book, it says aldol condensation gives off an h2o and an alpha beta unsaturated carbonyl. For the sake of the mcat, do we just recognize this reaction as water being removed or should we understand that its more along the lines of a base attacking the alpha hydrogen resulting in a conjugate acid and then the next step being the OH- group leaving and C-C double bond forming? Because in my head its like yea a net loss of h2o occurs but its not actually h2o leaving.

This is a question I found online. There seems to be a pretty big step between going from starting concentrations to ending concentrations (italicized in text). I feel that they just solved it intuitively. My question is how do you solve this mathematically? Do you always need an ICE table with these type type of Henderson-Hasselbach equations?
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To solve this question you need to think about the chemical reaction occurring.
CH3COOH+NaOH⇌CH3COO−+H2O+Na+
We can ignore water and sodium ions for the sake of this question. The reactants exist in a 1:1 ratio, so that for every mol of NaOH we add, we lose one mol of acetic acid and gain one mol of acetate. We can determine the moles of acetic acid by using M = mol/L, which gives us mol = ML = (2M) * (0.5L) = 1mol acetic acid. If we use the Hendersen Hasselbach equation we can see that the pH equals the pKa when the concentration of conjugate base (acetate) equals the concentration of acid.
pH=pKa+log[base][acid]
If we have 1mol of acetic acid and add 0.5mol of NaOH, we will lose 0.5mol of acetic acid and gain 0.5mol of acetate. We will then be at a point where acetic acid equals acetate. This is summarized in the ICE table below. Now we know the moles of NaOH (0.5 moles) and the concentration (2M) so we can find the volume by doing M = mol/L.
L = mol/M = (0.5mol)/(2M) = 0.25L
Acetic acid
NaOH
Acetate
I 1 mol 0.5 mol 0 mol
C -0.5 mol -0.5 mol +0.5 mol
E 0.5 mol 0 mol 0.5 mol

For the dissociation of a weak acid, like acetic acid, I know it will dissociate with water into acetate ion and hydronium. I know the hydronium will leave the water slightly acidic...but here's my question.
I know that acetic acid is an acid...and thus will leave the solution somewhat acidic.
But since its a weak acid...doesn't that mean its conjugate base (CH3COO-) is a STRONG base...and thus will pull protons off of H2O in solution? thus creating OH- and making the solution basic?
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am I totally wrong here? because that would make the solution basic, yet i I know dissociation of an acid makes it acidic even if its not that acidic. Maybe the fact that it does this is why its not THAT acidic and only a little acidic?
​
thanks for your help

To find pH of a salt solution, we only consider the Anion which will hydrolyze. On the other hand, the cation does not hydrolyze and is considered to be spectator ion. Why?

The ions present are Na+ and OCl- as shown by the following reaction:
NaOCl(s)→Na+(aq)+OCl−(aq)NaOCl(s)→Na(aq)++OCl(aq)−
While Na+ will not hydrolyze, OCl- will (remember that it is the conjugate base of HOCl). It acts as a base, accepting a proton from water.
OCl−(aq)+H2O(l)⇌HOCl(aq)+OH−(aq)OCl(aq)−+H2O(l)⇌HOCl(aq)+OH(aq)−
Na+ is excluded from this reaction since it is a spectator ion.

Why doesn't Na⁺ react with OH^- to form NaOH?
SOURCE/Acids_and_Bases/Acids_and_Bases_in_Aqueous_Solutions/Aqueous_Solutions_Of_Salts)

Buffer is a solution of weak acid and its conjugate base which resists the change in pH when strong acid/base is added to the solution.
But how does it work?
Consider,
HA + H2O ⇄ H3O⁺ + A^-
When we add strong acid, the H+ from the strong acid reacts with the conjugate base A^- to for HA. Hence, the concentration of HA increases while the concentration of A^- decreases. According to Le Chatelier's principle the equilibrium reaction shifts to the right. So then won't the H3O⁺ concentration increase further? We want it to shift to the left right?
When we add strong base, the OH^- reacts with H⁺ from the weak acid to form H2O. So concentration of HA decreases while the concentration of A^- increases hence the reaction should shift towards the left, am I correct? It was given it shifts towards the right. How?
Thanks!

During the dehydration of either tertiary or secondary alcohols catalyzed by TsOH or H2SO4 it proceeds via an E1 mechanism after the creation of a better leaving group like H2O. My book says this is a rather clean method for producing alkenes as it does not produce any SN1 products. My question is why does it not have any SN1 products? The carbocation intermediate is formed so what’s stopping SN1 mechanism from occurring?
Is it because the conjugate base formed during the acid base reaction in the first step is not a good nucleophile due to lots of electron withdrawing groups? Thus it will only have the properties of a weak base allowing only E1 to happen?

So when a stong acid is put in water [H3O+] is gonna equal [strong acid]. Does that mean that the conjugated base will also equal this concentration?
For example HClO4 + H2O = H3O+ + ClO4-
Will the final concentration of the H3O+ be equal to the concentration of ClO4 and the initial concentration of HClO4?
Sorry if my terms are not really precise, english is not my first language.

I have a few questions that are on the worksheet I'm doing...
First of all, why is this redox reaction Mn2+ + Cr2O72- → Mn3+ + Cr3+ in an acidic solution? I know what redox reactions are but I don't understand why there's acid and base conditions
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Next, in this reaction, HPO42- + H2O ⇌ H2PO4- + OH- apparently water is the acid and OH- is the conjugate base... why not HPO42- as the conjugate base and H2PO4- as the acid?
​
My last question is, how do you solve this problem?
"A solution of a strong base with pH = 13 is mixed together with a solution of a strong acid with pH = 2. The resulting solution has a pH of 11. What is the ratio of the volumes of strong base solution and strong acid solution added?"
​
Thanks in advance :)

It all made sense to me (Acid donates H, becoming more electronegative, Base accepts Proton)
but then I encountered this problem: What is the conjugate base of H2PO4^1-?
OH- + H2(PO4)^1- > H(PO4)^2- + H2O
Isn't HPO4^2- the conjugate base? Or am I doing this wrong?

So I just wrote this for another comment, but I am willing to share it here as well. just a fair warning: I didn't really proof read it so please notify me if there are any errors.
A buffer is composed of an acid and it's conjugate base. It's function is to keep the pH relatively the same. HOWEVER, a buffer can run out. The Henderson Hosselbott (totally butchered that name) equation/process can help us figure out the pH of a buffer and when it is going to run out.
Equation: pH = pKa +log(B/A) to calculate pH
or
pOH = pKb + log(A/B) to calculate pOH.
yes, you can just do Kw=Ka * Kb and figure out the Ka from Kb and plug in that way.
An example of a strong buffer is one with high and similar concentrations of the acid and it's conjugate base.
For example, using Acetic acid (Ka = 1.8x10^-5 and its conjugate base, Sodium Acetate):
A buffer with 10M Base and 1M Acid is going to have a log(10/1) = 1. (pKa = 4.76 -- HAc) thus pH = 5.76.
pH = -log(1.8x10^-5 ) + log(10/1) = 5.76
Now change that to 9M base and 2M Acid. The equation becomes:
pH = -log(1.8x10^-5 ) + log(9/2) = 5.41. Not a very large movement, but still a pH change of .35.
Now watch this:
Let's make the concentrations equal.
10M base and 10M acid.
In this case, the pH = pKa because log(10/10) = 0. thus pH = pKa + 0.
therefore the buffer using equal concentrations of base and acid will always have a pH = pKa. (In this case pH = 4.76.)
Now let's take some Base away and add it to the acid.
9M base and 11M acid
pH = -log(1.8x10^-5 ) + log(9/11) = 4.67. Only a pH change of .09!
Really the thing you should take away from this is the closer the concentrations of acid and base, the less of a pH change there will be.
TITRATIONS (because they are so closely related)
Usually a buffer is used in titration questions because essentially a titration is what happens after a buffer runs out.
For this problem we are going to be working with the titration of 25mL .5M of HAc with .5M NaOH. HAc is in the beaker originally. Equation: **HAc + NaOH NaAc + H2O
The general formula for these problems is the following:
1. IS THIS A SB SA TITRATION?? If the answer is yes then you cannot use the Henderson hosslebott equation because both the Base and the Acid have a 100% dissociation. The pH = -log([SA]). However, to find the [SA] at that point you must follow the same steps.
2. Write the net ionic! For our case the net ionic is HAc + OH- - AC- + H2O This will help us later. (Note: SA and SB does not have a net ionic because it always forms a neutral salt and water, thus the equation is H + OH H2O.
2b. Calculate the Equivalence point and half equivalence point! Will be a huge time saver for later in the question. Equivalence point = MaVa=MbVb.
(.5M)(25mL) = (.5M)(xmL) . X = 25mL thus when 25mL of NaOH is added, we will be at equivalence point. I will solve for this pH later.
The 1/2 equivalence point is then 1/2 of the NaOH added at equivalence! Thus our half equivalence point is 12.5mL of NaOH added. Why is this helpful? Well, at 1/2 equivalence pH = pKa! Saves a couple minutes.
3. Calculate the pH BEFORE any base is added.
This is pretty easy. We only have HAc in the beaker right now. Set up the equilibrium expression: Ka = [H][Ac-]/[HAc] and plug in. (H and Ac- are unknown). 1.8x10⁵ = [x][x]/[.5]. X = .003. Thus the pH = 2.52
4. Calculate the pH after 1mL of NaOH is added. This is where it gets tricky. Pay close attention.

1. Find MOLES of HAc. (.025L)(.5M) = .0125mols HAc. <--- hold onto this. We are going to be using it a lot.
2. Find MOLES of Ac- (NaAc -- the conjugate base) (.001L)(.5M) = .0005mols Ac-.
3. Find how many moles of acid are left after the reaction and how many moles of the conjugate base are created.

.0125molsHAc - .0005molsNaOH = .012 moles HAc .0005 (NaOH) = .0005mols NaAc (AC-).
Now divide those concentrations by the total volume. (.025L + .001L = .026L).
.012HAc/ .026 = .46M .0005NaAc / .026 = .019M
Now lets plug into the Henderson hosslebott. pH = -log(1.8x10^-5 )+ log(.019/.46) = 3.35
This is the EXACT SAME process for any concentration of base added as long as it ISNT the equivilance point. (You can technically do this for the 1/2 eq point but it will turn out to be pH = pKa, so just save time by memorizing that shortcut.)
5. Calculate the pH after 12.5 mL NaOH added
Hey! That's half equivalence point, like we calculated earlier. Now we don't have to waste 5 minutes doing all the work from part 4.
pH = pKa .Thus pH = -log(1.8x10^-5 ) = 4.74. simple as that.
6. Calculate the pH after 25mL NaOH is added
Hey! That's equivilance point, like we calculated earlier. Now I know that I cannot use the Henderson hosslebott here.

1. Mols HAc = .0125 (still)
2. Moles NaOH = (.025L)(.5M) = .0125mols.
3. .0125 mols HAc - .0125 mols NaOH = 0 mols HAc left! See? this is why we can't use the hosselbott because we can't find the concentration without Acid!
4. .0125mols NaOH = .0125 mols NaAc. divide by total volume (.025L + .025L = .05L) .0125/.05 = .25M. *All the acid reacted with the base to create a salt. However, unlike in a SA SB titration, the salt is NOT neutral. Thus the salt will effect the pH. That is what we will calculate now!.
5. Earlier, we wrote the net ionic (HAc + OH- - AC- + H2O) now we can flip this to figure out the dissociation of the Salt, because all that is in the beaker now is H2O and NaAc,thus the salt will dissolve (partially depending on Kb).

the equation becomes. Ac- + H2O < OH- +HAc. From here, we now see OH- is being produced! (see how useful the net Ionics can be?) Thus we are technically finding pOH.
Now lets set up the equilibrium expression to find out how much OH is produced. Be careful to not use Ka, but Kb because OH is produced and not H.
Thus, lets find Kb using the equation Kw=KbKa. 10^-14 =(1.8x10^-5 )(Kb) Kb = 5.56x10^-10 . <--- very small, so the equilibrium most likely lies towards the solid salt.
Now the expression: Kb = [OH][HAc]/[Ac-] -- 5.56x10^-10 = [x][x]/[.25M] X = 1.18x10^-5M OH. Now, the pOH = -log(1.18x10^-5 ) = 4.93. Lastly, the pH = 14-pOH. Thus, pH = 14 - 4.93 = 9.07
7. Calculate pH after 50mL NaOH is added. - At this point, only OH ions are in the beaker because all the acid is the limiting reagent.
(.05LNaOH)(.5M) = .025mols NaOH .0125mols HAc - .025mols NaOH = -.0125 mols HAc. This really means there are .0125 mols NaOH left over. Thus, divide by total volume (.05L + .025L = .075L) .0125/.075 = .17M.
Since NaOH is a STRONG BASE it has 100% ionization thus .17M NaOH = .17M OH- and therefore the pOH = -log(.17) = .77. Finally, the pH = 14-.77 = 13.22. VERY basic. Which makes a lot of sense because its a lot of a strong base in a little of a weak acid.
Check my work
Note: the reason why I included this with buffers in because the region before the equivalence point is the buffer region, where the pH doesn't move that much.
I really hope this helped. I wrote this all just now because I too am studying for my acids-bases exam.
Please PM if you have any questions!

I understand self ionization of water. But I don't get why it's in equilibrium. Wouldn't the reaction heavily favor H2O isntead of the harsh strong acid and base counterparts? The pka would heavily favor the reactants, right? I understand H2O is amphoteric and all, but the whole stability of products vs reactants is bothering me. Why would a system allow this equilibrium when water is stable and the conjugates are nothing but strong and highly reactive? I'm probably overthinking it, but it bothers me

Hello everyone, I am trying to learn how to figure out how to figure out the most acidic molecule/substance when given a choice. My post has a question but also I'd like your wisdom in validating/refining my understanding of acids/bases as they relate to organic chemistry.
To my understanding so far, Water is more acidic than all alcohols except methanol. The relevant pKa values:
CH3OH 15.54
H2O 15.74
CH3CH2OH 15.9
Tert-Butanol around 19
While I appreciate that all have similar acid strength, there are subtle differences and I would like to understand why (relating back to the structure of the molecule) that causes it to be more acidic. Can anyone explain nice and cleanly why the order of acidity (from strongest to weakest) is methanol, water, ethanol then tert-butanol?
Possible reasons that I have seen on the internet (that I either don't understand or were poorly explained) mention: inductive effect, solvation effects, something about being a better nucleophile, CH3 being an "electron releasing group", ethoxide being a stronger base than hydroxide, something about linearity
My current understanding of deciding acid strength is: The stronger acid has the more stable conjugate base, due to:

• increased electronegativity e.g. HF stronger than NH3 because fluorine is more electronegative than nitrogen
  
• larger ion size/atomic size (e.g. HI is a stronger acid than HF because the larger iodine has more room to spread out and stabilise its negative charge)
  
• carbons hybridised with more S character (sp>sp2>sp3) since the s-orbital is closer to the nucleus and hence those electrons experience a stronger force of attraction, resulting in carbon being more electronegative and better at stabilising the conjugate base
  
• resonance stabilisation of conjugate base (more resonance structures = more stable)
  
• inductive effect of electron withdrawal groups (e.g. more electronegative halogen substituent can better pull electrons towards itself, resulting in a more stable conjugate base)
  
• distance between the acidic part and electron withdrawal group (e.g. in 2 constitutional isomers, the molecule with the acidic part and electron withdrawal group closer together will be more acidic)
  

I am also aware that lower pKa values mean stronger acidity and that the stronger the base, the weaker its conjugate acid.
I am trying to obtain a simplified (first year uni level) conceptual understanding so as to be in the best position to make educated guesses about what is the stronger acid, ultimately to help me predict reactants/products of complicated biochemistry reactions.
Please let me know if there is anything I wrote that I completely misunderstood. My apologies for the long post!

In general, what is the pKa of an acid's conjugate base? If HCl gives away its hydrogen, does Cl- have a pKa at all?
I understand that pKa is used to measure the acidity of a molecule, but theoretically do all molecules have a pKa, even if they don't have hydrogen? Or do they not? Or is it infinite?
I ask because I know both H2O and H3O+ have significant pKas, as well as NH3 and NH4+. But no one has ever explicitly taught me how to consideignore pKa of other conjugate pairs.