Chapter 14: Acids and Bases

Bases

Substances that react w/ water to form hydroxide ion (OH-) - Receive H+ in water (donates OH- in H2O?) Why is NH3 a base even tho has no OH-? NH3 + H2O <--> NH4+ + OH- NH3 FORMS OH- in water - It reacts w/ water to FORM hydroxide (so doesn't necessarily need to have OH- in formula to be an acid, just needs to FORM OH-)

Electrolytes

Substances whose aq solution conducts electricity --> electrolytes - Acids, bases, salts

Discovery of Acids

Svante Arrehenius

Base Properties

Taste *bitter* *Caustic* --> Dissolves biological tissue (will make fingers feel slippery, thats your skin dissolving) Do not react w/ metals to form H2 Change color of acid/base indicators

Elutant (moving phase) -- The liquid moving up paper

The (l) or (g) that flows thru a chromatography system, moving the materials to be separated @ different rates OVER the stationary phase

Carboxylic Acid

The C=O bond -- O pulls e- away from C Now C e- deficient so pulls e- from the O in the OH bond, OH bond now polarized, H can be donated Any C-H -- The H cannot be donated bc C-H is too strong bond

Here's an example of where you def can't use negligible rule: ka = 1.2x10^-2 and [HA] = .50

The difference in powers of 10 is less than 3, so the amount dissociated is significant and will ∆ the [HA] once equilibrium is reached

Salt derived from a weak base and a strong acid

Cation ion is a relatively strong conjugate acid Cation hydrolyzes to form H+ Solution has a pH < 7 (Acidic salt) FeCl2, Zn(NO3)2 FeCl2 - We don't care about the Cl- bc it came from HCl which is a strong acid, so neg conjugate - We do care about Fe2+ bc came from Fe(OH)2 which is weak base - This means that Fe2+, as relatively strong CA, can hydrolyze water to produce acidic solution: Fe2+ + H2O --> Fe(OH)2 + H+

Fe3+ is better Lewis acid bc more + charge and smaller (Coloumb's law applies when talking about which ion is better as Lewis acid)

Coloumb -- Smaller + bigger charge = stronger bonds/better as Lewis acid

Complex ions

Complex ions will normally have 2x as many things attached to it as the charge on the metallic ion - So Zn2+ will have 4 things attached to it - So Zn(OH)2 + 2H+ --> Zn2+ + 2H2O? Ok here are some complex ion equations to memorize Al(OH)3 as bronsted base (donates OH-): Al(OH)*3* + *3*H+ --> Al*3+* + *3*H2O Al(OH)3 as Lewis acid (accepts e- pair): Al(OH)*3* + OH*-* --> Al(OH)*4^-* Zn(OH)2 as bronsted base (donates OH-): Zn(OH)*2* + *2*H+ --> Zn*2+* + *2*H2O Zn(OH)2 as Lewis acid (accepts e- pair): Zn(OH)*2* + OH*-* --> Zn(OH)*3^-*

Absorbant (stationary phase)

For example, H2O molecules in chromatography paper - The solid or (l) phase of a chromatography system on which materials will be separated or selectivity adsorbed

H-O-Cl, H-O-Cl-O, H-O-Cl-O2, H-O-Cl-O3

We know that last compound is HClO4 aka a strong acid The oxidation #s for the Cl is +1, +3, +5, +7 As oxidation number increases (gets more positive), acid strength increases - As ox # increases, the Cl becomes increasingly e- deficient - Basically, the Os "on the right" make Cl e- deficient, so Cl pulls e- from the left O, which makes the -OH bond polarized and therefore H is donatable - The more oxygens attached to the central atom, the stronger the acid

Example 2 -- A .200 M has [H+] = 9.86x10^-4 Find ka

ka = (9.86x10^-4)/(.2 *- 9.86x10^-4*) When problem says weak acid - A .200 M solution of weak acid has [H+] of x - The .200 M is saying the [weak acid] before dissociation (aka the .200M = [HA] = concentration of weak acid, by weak acid it means the [HA]) - So when finding ka, remember you are using the equilibrium concnetrations/the [ ]s after dissociation - So have to subtract the amount dissociated (which = [H+]/[A-]) from original [HA]

HA (aq) + H2O (l) <--> H3O+ (aq) + A- (aq)

keq = [H3O+][A-]/[HA][H2O] - Bc H2O is (l), its [ ] stays constant SO keq*[H2O] = [H3O+][A-]/[HA] - keq*[H2O] is a constant times a constant so = a new constant = ka *ka = [H3O+][A-]/[HA]* ka = acid dissociation constant Can write ka for any acid If have strong acid - [HA] is (very) small --> ka is (very) big If have weak acid - [HA] = big --> ka = small

NH3 weak base --> CA = NH4+ is relatively strong (not strong acid, but stronger acid than NH3 is base) HF --> F- = relatively stronger (not strong base, but stronger base than HF is acid)

kw = kakb for acid/base conjugate pairs pkw = pka + pkb F- and NH4+ can hydrolyze water NH4+ can hydrolyze water to create acidic solution NH4+ + H2O <--> NH3 + H3O + F- can hydrolyze water to create basic solution F- + H2O <--> HF + OH- Basically, the CA of a weak base can hydrolyze water and create an acidic solution And the CB of a weak acid can hydrolyze water to create a basic solution So you can have acidic salts and basic salts

pH/pOH

pH = -log[H+] = log(1/[H+]) As [H+] increases --> More acidic --> *pH goes down* - The more acidic the lower the pH pOH = -log[OH-] pkw = -log(kw) - If @ 25 C --> -log(1x10^-14) - pkw = 14 @ 25 C pkw = pH + pOH 14 = pH + pOH kw = 14 only @ 25 C pH scale 0 (acidic) --> 14 (basic) ex. [HCl] = 12.1 M pH = *-*1.083 - Having - value for pH or pOH is OK - You typically don't get acidic solutions more acidic than pH of 0 (<0 means SUPER DUPER acidic) - However it IS possible to get -pH values So can have -pH values, -pOH values OR pH values > 14, pOH values > 14

How does [ ] affect the degree of ionization of weak acid?

% dissociation = ratio of the concentration of the dissociation hydrogen ion [H+] to the concentration of the undissociated species [HA] = [H+]/[HA] * 100 a) .10M HF solution -- 7.9% ka = 6.8x10^-4 ka = [H+][F-]/[HF] = 6.8x10^-4 = x^2/.1 - x Solve for x, x = [H+], then find % dissociation b) .01M HF solution -- 23% dissociation [ ] affects % dissociation As [ ] decreases, % dissociation increases, pH increases (aka gets more basic), [H+] decreases As [ ] increases, % dissociation decreases, [H+] increases, pH decreases

Log review

-log(1.00x10^-14) If the input of the log = 1.00 x10^x (where the 1.00 is exactly 1) The -log of the number = the opposite of the #s exponent (aka -x, or 14 for the first example) SF -log ( ) The s.f. of the input into log should = the # of decimals of your answer (aka the pH should have the same number of decimals as sf in the concentration) THe # of deci places in your pH should = the # of sf in your [ ] Remember log(base)x = n --> base^n = x So -log[H+] = pH log[H+] = -pH 10^-pH = [H+] SO do 10^-pH or 10^-pOH to get [H+] or [OH-]

1. Calc the pH of a .4 M solution of NH4ClO4

1. Dissociate salt (check w/ solubility rules, but unless told otherwise will fully dissociate *for now*) NH4ClO4 --> NH4+ + ClO4- .40 M 0 0 0 .40 .40 2. Determine if one or both ions is important - Does the ion come from strong or weak acid? - If from strong, the ion is neg and can be ignored - We care about ions that come from weak 3. After determining which ion(s) is/are important, write the water hydrolization equation for that ion NH4+ + H2O <--> NH3 + H+ 4. Use ICE chart to solve for [H+] - Note: ka of NH4+ is not in the book, so use its conjugate's, NH3's, kb to calc the ka of NH4+ 5. Then find pH Remember HClo4 is strong, so ClO4- is neg conjugate so can't hydrolyze

The strength of an acid has nothing to do w/ its concentration --> the strength of an acid has to do with whether the acid fully or partially dissociaties/how much it dissociates

Strong acid does not necessarily have lower pH than weak acid - Bc its [H+] that determines pH, if the strong acid has very low concentration, then going to have low [H+] --> in same vein, a very high [weak acid], while it doesn't fully diss --> it will still have higher [H+] than the minute amount of the strong acid (bc [strong acid] was so small) - Equal [strong acid] and [weak] tho does mean pH of strong solution will be higher

Bronsted Lowry

Acid -- Donates proton Base -- Accepts a proton

Arrehenius

Acid -- Reacts H2O to form H3O+/donates H+ in water Base -- Reacts w/ H2O to form OH-/donates OH- in water

HCl (acid) + H2O (base) --> H3O+ (conjugate acid) + Cl- (conjugate base) NH3 (base) + H2O (acid) --> NH4+ (conjugate acid of base [NH3], aka the acid for the reverse rxn) + OH- (conjugate base of acid aka the base for the reverse rxn) H2O was base in one rxn but acid in another -- *Amphoteric*

Substance that can act as acid and base

Acids

Substances that react w/ water to form hydronium ion (H3O+) - Donate H+ in water H+ + H2O --> H3O+ (H3O+ is H2O that's gained a proton) You can use H+ and H3O+ interchangeably

2. An unknown salt is NaCN, NaC2H3O2, NaF, NaCl, or NaOCl. If the pH of .100 mol of the salt in 1.00 L of solution is 8.07 determine the identity of the salt

8.07 means the solution is basic, so the hydrolization of water created OH- A CB of a weak acid can produce OH- when hydrolyzing water So B- + H2O <--> HB + OH- Find kb kb = [HB][OH-]/[B-] = (1.2x10^-6)^2/.100 - 1.2x10^-6 - Note: Where did the .100 come from? - The salt NaB fully dissociates, so the .100 M of salt becomes .100 M of Na+ and .100 M of B- - Then also you gotta -amount dissociated from original concentration Once find kb, find ka - We want to identify which acid the B came from - So want to find ka of HB from kb of its conjugate, B-: HB <--> H+ + B- which can identify HB - Once find ka, look @ table to find which ka value is closest to this - In this case, its HF, so B = F-, and the salt is NaF

Weak Acid

Acid that does not fully dissociate HA + H2O <--> H3O+ + A- There's competition b/w bases (H2O and A-) for who's gonna get the proton - A- is the stronger base, so equilibrium lies to the left (when A- has the proton and H2O does not) - *Weak acids* have *relatively strong CBs* (*stronger than H2O*) so rxn lies to left ka = [H3O+][A-]/[HA] = [H+][A-]/[HA] *As ka decreases, acid is weaker* Big ka means stronger acid (HUGE ka means the acid is strong and rxn goes to completion, so no equilibrium) Also, weak DOES NOT mean acid is less reactive - Weak means does not fully dissociate - Weak or strong does not affect reactivity - HF = weak, but very reactive Weak acid problems are completed in the same way as equilibrium problems (Use ICE chart...etc.)

Polyprotic Acid

Acid w/ more than one donatable proton (H+) - CH3COOH = MONOPROTIC (only that one H at end of -COOH [carboxyl group] is protonizable); ITS NOT POLY, only 1 donatable proton - If have carboxyl group --> Organic acid - Oxyacid -- contains oxyanion (See digram, notice how donatable protons are usually attached to oxygens)

Polyprotic Acids

Acids with more than one mole of ionizable hydrogen atoms per mole of acid - Many protons - Acids that donate more than 1 proton - Look @ dissociation in 2 or more steps - You donate one H+ at a time H2SO3 <--> H+ + HSO3- ka1 = 1.7x10ˆ-2 HSO3- <--> H+ + SO32- ka2 = 6.8x10^-8 ka1 > ka2 meaning that first dissociation is stronger than 2nd - If the difference in powers of 10 b/w ka1 and ka2 or ka2 and ka3...etc. is 3 or more, the 2nd dissociation (aka the one w/ the smaller ka value) doesn't affect the pH/only the first step affects pH/the [H+] from the 2nd step is negligible compared to the [H+] from the 1st step - So if ka1 > ka2 by 3 or more powers of 10, 2nd step can be ignored or 3rd or 4th or whatever - Why is step 1 stronger than step 2? - The CB produced in step 1 has a - charge, so (2nd?) H+ is attracted to the - charge and that makes 2nd H+ harder to donate bc H+ wants to stay w/ - charged HSO3-/CB - This is typically the case that ka1 > ka2 bc CB usually formed is - charge --> so harder to donate H+ - Bigger ka value = more dissociation, so the smaller ka2 means that HSO3- doesn't dissociate much bc H+ attracted to - charge HSO3- and won't leave the molecule (more attracted to - charge of HSO3- than to unbonded e- pairs on H2O)

NaHCO3

Adding an acid to solution of NaHCO3: The H added by the acid donates an H+ to HCO3- to create H2CO3. H2CO3 is very unstable and decomposes immediately to H2O and CO2. The Co2 (g) is what causes bubbling

Why do complex ions do their thang?

Al3+ + 6H2O --> Al(H2O)6^3+ Al has positive charge, so is able to accept e- pairs (Lewis acid) H2O, the O has unbonded e- pairs to donate (Lewis base)

Acidic/Basic Anhydrides

Anhydride = w/o water - These are acids/bases w/o water Nonmetal will create acid in water CO2 + H2O --> H2CO3 NOx, SOx will do same Metal oxide will create base in water CaO + H2O --> Ca(OH)2

Salt with a highly charged metallic ratio and the anion is from a strong acid (Complex Ions)

Anion does not hydrolyze Hydrated cation acts as acid High charge in the metal polarizes the O-H bonds in water These water molecules have an increased acidity Al(H2O)6Cl3 Al(H2O)6Cl3 Cl- came from SA, so CL- as base is neg base (so ignore it) Al(H2O)63+ --> Al(H2O)5OH2+ + H+ Al --> Al3+ (lost e-) + H2O (which has unbonded e- pairs) Al3+ accepts e- pair from H2O so Al3+ gets surrounded by 6 H2O molecules (don't worry about this, later chapter??)

Salt derived from a strong base and a weak acid

Anion is a relatively strong conjugate base Anion hydrolyzes to produce OH- Solution has pH > 7 (Basic Salt Ca(NO2)2, NaC2H3O2 Ca(NO2)2 --> Ca2+ + 2NO2- Ca2+ comes form Ca(OH)2 which is strong base, so Ca2+ is negliglabe so can't hydrolyze water (we don't care about it bc came from strong base) NO2- comes from HNO2 --> a weak acid - Conjugates of weak acids/bases form relatively strong conjugates which can hydrolyze water H2O + NO2- <--> HNO2 + OH- which produces basic solution So Ca(NO2)2 is a basic salt

Thing ka Cl-O-H 3x10^-8 Br-O-H. 2x10^-9 I-O-H. 2x10^-11 CH2-O-H. 0

As move down this column, acid strength is decreasing, and electronegativity is decreasing Cl has highest electro neg, C has lowest So as electronegativity increases, the stronger the e- pull away from O is, the more O then needs to pull e- from H, the stronger the acid (the more donatable the proton)

3c Calculate the [OH-] and pH for ea of the following solutions: .10 M NaNO2 and .20 M in Ca(NO2)2

Bc these are both salts, the dissociate fully, so can find the total [NO2-] that comes from both salts BE CAREFUL OF COEFFICIENTS: Ca(NO2)2 --> Ca2+ + 2NO2- The mole ratio isn't 1:1 so M Ca(NO2)2 ≠ [NO2-] Ca(NO2)2 --> Ca2+ + 2NO2- .2 0 0 -x=.2 +x +2x 0 .4 Then after seeing how much NO2-dissociates from both salts, add the [ ]s of the NO2- from NaNO2 and NO2- from Ca(NO2)2 - Then proceed as normal - Write hydrolization rxn w/ NO2- - Determine kb using ka of HNO2 - IN General terms - Write hydrolization rxn w/ CA/CB - Determine ka/kb of hydrolization rxn using kb/ka of acid that the CA/CB came from - Then use ICE to solve for [H+] or [OH-] the find pH/pOH

Litmus

Blue in base, red in acid

Salt derived from a weak acid and a weak base

Both ions hydrolyze pH depends on the extent to which each ion hydrolyzes Compare the respective ka and kb, whichever is greater indicates the ion that hydrolyzes to the greater degree Cu(C2H3O2)2 Cu2+ + 2H2O <--> Cu(OH)2 + 2H+ ka C2H3O2- + H2O <--> HC2H3O2 + OH- kb Both of these happen, so to find pH you'll need to figure out which process dominates by comparing the ka and kb value - Whichever value dominates indicates pH To calc ka and kb values - There is ka value for HC2H3O2 and kb value for Cu(OH)2 but there is not kb value for C2H3O2- and ka value for Cu2+ - So you have to use kakb = kw to find the kb and ka value - So use known ka value of HC2H3O2 to find the kb value of its conjugate, C2H3O2- If ka = kb Neutral salt ka < kb Basic salt ka > kb Acidic salt

1. Calculate the pH of 5.0M H3PO4 solution and the equilibrium concentrations of the species H3PO4, H2PO4-, HPO42-, PO43- ka1 = 7.5x10^-3 ka2 = 6.2x10^-8 ka3 = 4.8x10^-13

Do ea dissociation step separately (will have 3 different ICE charts) If the problem only asked for pH, you could use the 3 or more powers of 10 rule to see that the [H+] produced in the 2nd step is negligible and therefore you could skip the extra math required to calculate the [H+] of the second step But bc this problem wants the cconetration of ea acid/base/CA/CB produced, you gotta do all three steps even tho steps 2/3 don't affect pH But remember, sometimes 2nd step's [H+] is NOT neg so don't just assume But bc in this problem we are doing all 3 steps, there's something you must know - In step 1, you produce a [H+] and [H2PO4-] - Step 2 -- H2PO4- --> H+ + HPO42- - The *I*nitial concentration of H2PO4- is = to the [H2PO4-] calculated in step 1 - AND the *I*nitial concentration of H+ is = to the [H+] calculated in step 1 - Basically, bc H+ is produced by step 1, that H+ will still be floating around in solution while step 2 is occurring, so that [H+] becomes the initial [ ] for step 2 - Remember, after 1st step, you already have a [H+] and [H2PO4-] --> these transfer down into step 2 - The [H+] doesn't just go away, its still in the solution - If do 3rd step, you'll have to transfer down [HPO42-] and [H+] calculated in step 2 down to step 3 Step 1 H3PO4 <--> H+ + H2PO4- I 5.0 M 0 0 C -x = .194 E .194 .194 So now transfer down your new [H+] and [H2PO4-] to step 2 as your initial concentrations for both Step 2 H2PO4- <--> H+ + HPO42- I .194 .194 0 C E Then continue and solve for the rest of the values it asks for, just remember to transfer down any [ ]s of things that appear in the previous step and the next step And also remember, that bc ka2 < ka1 by more than 3 powers of 10, the [H+] from step 2/3 have no affect on pH bc their concnetrations are neg - If only wants pH --> only need to do 1st step bc powers of 10 rule - But if need [PO43-], then need to do all steps! Always will need to write out multiple steps for polypro tic acids (H2SO4 <--> H+ + HSO4- AND HSO4- <--> H+ + SO42-) - Can't just write overall rxn - Also remember, 1st dissociation of H2SO4 is STRONG, but 2nd step is not If wanted to find overall ka by adding all steps of polyprotic acid: H3PO4 <--> 3H+ + PO43- ka overall = ka1xka2xka3 (multiply kas from ea step)

2. NaClO

Don't have time to fully write out, basically, just work backwards

In Bronsted Lowry

Every *acid* has a *conjugate base* and every *base* has a *conjugate acid* - Everything occurs in conjugate pairs

Autoionization of water

H2O (l) + H2O (l) <--> H3O+ + OH- = H2O <--> H+ (aq) + OH- (aq) keq = [H+][OH-]/[H2O] The [H2O] is so big that it doesn't change much over the course of the rxn --> aka its constant What is [H2O] in 1 liter of H2O 1 kg = 1000g H2O x 1 mol/H2O = 55.5 mol 55.5 mol/1 L = 55.5 M = [H2O] (in H2O?) So this is so big that its constant So lets move all constants to one side keq[H2O] = [H+][OH-] keq[H2O] = new constant = kw = ion product constant of water - *@ 25 C*, kw = 1.00 x10^-14 - Unless told otherwise --> assume you are @ 25 C kw = 1.00 x10^-14 = [H+][OH-] - As [H+] or [OH-] goes up, the other will go down (indirectly related) - Also you assume that kw is constant bc will only ∆ w/ temp and assuming temp is constant during rxn [H+] = [OH-] -- Neutral [H+] > [OH-] -- Acidic [H+] < [OH-] -- Basic @ 25 C, neutral solution [H+] = [OH-] = x 1x10^-14 = x^2 = [H+ = 1x10^-7][H+ = 1x10^-7] This is an awkward way to measure pH --> so we use pH scale

HA (aq, strong acid) + H2O (l, base) <--> (means in equilibrium, so can use law of mass action) H3O+ (aq, acid) + A- (aq, base)

H2O and A- are competing for the proton - (Acid in ionic form) If H2O > A- (stronger base = H2O) --> Rxn shifts right [mostly A- and H3O+, see H2O won the H+] - (Acid in molecular form) If H2O < A- (A- = stronger) --> Rxn shifts left [mostly HA and H2O, see the A- won the H+] So in this case, if there is more H3O+ and A- in the solution, then A- is a weaker base than H2O, so rxn shifts right

Oxyacids and Acidity

H2SO4 -- Oxyacid - The S pulls e- towards it from the Os, which pull e- towards it from the H, making the O-H bond very polarized and allowing H to be donatable Why does H not leave OH bond in NaOH - O is e- deficient, but bc in ionic bond w/ Na, the Na donates e- to O to make O not e- defiecient - Bc O no longer e- deficient, the O does not need to pull e- from H+, meaning the OH bond is not polar so H stays w/ O If the OH bond is not polarized, the H can't be donated So the more polarized the -O-H, the more donatable the proton - OH = polarized if O is e- deficient - O is e- deficient if centre atom is very electroneg (I think)/centre atom takes e- away from O --> So O now needs e- so takes e- form H, making the OH bond polarized So H2SO4 can donate proton bc its OH bond is polarized

The strong acids

HI, HBr, HCl, HNO3, HClO*4*, *H2SO4 (only strong for FIRST dissociation)* - If given ka --> weak, if not given ka --> assume strong acid (aka w/ HClO3, bc sometimes HClO3 is strong, sometimes not, depends on what problem gives you) - Sometimes HClO3 and HMnO4 are considered strong

Acid Properties

Have *sour* taste (when milk goes bad, sour, lactic acid formed from milk sugars being oxidized) Acids are very *corrosive* (like burn[ing]) and/or *posoinous* Acid contain *hydrogen* and some liberate H2 in runs w/ certain metals - If metal is above hydrogen on activity series --> will react w/ acid to create H2 -- Zn + HCl --> ZnCl2 + H2 - If metal is below H on activity series --> NO RXN Acids affect the colors of acid/base indicators - Can ∆ color of some plants w/ acid/base?

2. H2SO4 Problem, basically H2SO4 is a special snowflake

If [H2SO4] > 1.0M --> Can ignore 2nd dissociation If [H2SO4] < 1.0M --> CANNOT ignore 2nd dissociation H2SO4 is only acid where you have to worry about this - All other acids can follow ka1/ka2 rules w/ powers of 10

Remember, when doing Hess' law w/ ks

If add two equations, multiply the ks of those 2 equations If double the coefficients of equation, square the k If 1/2 the coefficients of equation, root the k If flip the equation, take 1/k

Various Ways to Describe Acid Strength -- Property: Equilibrium concentration of H+ compared w/ the initial concentration of HA

Strong Acid -- [H+] ~ [HA]0 Weak Acid -- [H+] << [HA]0

Various Ways to Describe Acid Strength -- Property: Ka

Strong Acid -- ka is LARGE Weak acid -- ka is small

NH4+ <--> NH3 + H+ ka = [H+][NH3]/[NH4+] NH3 + H2O <--> NH4+ + OH- kb = [OH-][NH4+]/[NH3] This is conjugate acid base pair rxn for the conjugate acid base pair NH4+ and NH3

If combine these runs, the net rxn = H2O <--> OH- + H+ If you have two runs that add together, to get the keq of the net rxn, overall keq/net keq = k1*k2 - If you have 5 runs that add up to one net rxn, keq net = k1xk2xk3xk4xk5 So for this keq net = ka*kb = kw - [H+][NH3]/[NH4+]*[OH-][NH4+]/[NH3] = [H+][OH-] So for conjugate acid base pairs, the ka of the acid and the kb of the conjugate base ka*kb = kw = 1.00x10^-14 (so if given kb of NH3, but need ka for NH4+ rxn, can do 1x10^-14/kb = ka) - This only works for conjugate acid base pairs (can't just do this w/ 2 random acids and bases lol) - Strong Acid = Weak CB - ka = Huge, kb = Small --> indirectly related, as one goes up the other goes down, meaning that when multiplied together, they = a constant = kw - Strong Base = Weak CA - kb = Huge, ka = small So ka*kb of acid base pairs = kw pkw = 14 = pka + pkb

How do you calculate [H+] or pH w/o calc (aka on MC section of AP)

If given pH = 4.5, what is [H+]? Solution If [H+] = 1x10^-4 --> pH = 4 If [H+] = 1x10^-5 --> pH = 5 (remember that the -log of something that is exactly one to a power of 10 is going to = the opposite of the exponent of the power of 10) So [H+] is somewhere b/w 1.0x10^-4 and 1.0x10^-5 - Usually, in MC questions, there will be 3 answers that are outside this range, so the actual answer will be the only one that falls b/w this range - If this does not happen, and you are left w/ 2 answers still, well, at least you have a better probability of guessing right So for this AP question a) 1.0 x10-6 b) 1.0x10^-3 c) 3.5x10^-5 d) 4.5 x10^-4 a is too small, b is too big, d is too big, c falls b/w the range!

ALSO, always how to know which is acid/which is base

If substance gains proton --> base NH3 --> NH4+ - So look at substance on reactant side, then look at substance on product side --> If product version of substance has extra proton, then the original substance was a base If substance loses proton --> acid HCl --> Cl-

General Rules about Acid-Base Properties of a Salt Solution

If the ion comes from a strong acid/base, the ion is a negligible conjugate --> neg conjugates cannot hydrolyze water and therefore do not affect pH - Na+ from NaOH, Cl- from HCl If the ion comes from a weak acid/base, the ion is a relatively strong conjugate --> can hydrolyze water and affects pH - Fe2+ from Fe(OH)2, NH4+ from NH3, F- from HF Basically, determine which base or acid ea ion comes from, then determine if said acid/base is strong or weak then determine if the ion is neg or strong, then determine type of salt

2. A solution of sulfuric acid has a pH of 4.62. What is the concentration of the acid?

In these problems, we assume that these = strong --> so fully diss But remember, only the first dissociation/the first H+ to be donate is strong So the equation is H2SO4 --> H+ + HSO4- - You cannot use H2SO4 --> 2H+ + SO42- bc the second dissociation is NOT strong - Second dissociation of HSO4- --> SO42- is NOT strong - We only do first dissociation bc thats the only strong part (and all the problem wants is the concentration of H2SO4, so you don't need to calculate any concentrations from the 2nd dissociation)

Lewis Theory

Lewis acid -- Electron pair acceptor Lewis base -- Electron pair donor The BH3 is Lewis acid and NH3 is Lewis base The BH3 does not have octet (has 6 e- surrounding it), so has room to accept e- pair NH3 has unbonded e- pair that can be donated

Strong Bases

LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂ W regards to ones like Ca(OH)2, Sr(OH)2, Ba(OH)2 - Remember the dissociation for these are Ca(OH)2 --> Ca2+ + 2OH- So if you calc the concentration of OH- (from pH) and you want to get the concentration of the Ca(OH)2, Sr(OH)2, or Ba(OH)2, remember to take into account the mole ratio of OH- to the full base - So for CaSrBa bases, the mole ratio is 2OH- to 1CaSrBa base, so the [OH-]≠[CaSrBa base] bc the mole ratio is not 1:1

If weak + weak

Need to compare the two ions' ka value to kb value (calc using kb/ka of the base/acid that the ion came from, use conjugate acid/base pairs to calc ka/kb for given ion)

Salt derived from a strong acid and a strong base

Neither ratio nor anion hydrolyzes Solution has pH = 7, Neutral Salt Examples: NaCl, CaBr2 NaCl Remember a salt = acid + base SO to get NaCl NaOH + HCl --> NaCl + H2O To determine if the salt is basic, acidic, or basic look at the ions that make up the salt - Think of what acid/base the ion comes from - If the ion comes from a strong acid/base, the ion is a negligible CB/CA (PATHETIC CONJUGATE) - Negligable acids/bases do nothing to pH - Two negligible conjugates are produced from strong acid + strong base so the solution is neutral Na+ = neg acid bc comes from strong NaOH Cl- = neg base bc comes from strong HCl Since they are neg acids/bases, they cannot hydrolyze/interact w/ H2O --> so the salts they can form are neutral salts

If you add acid (to water?) @ lower [H+] than [H+] of neutral water (1.0x10^-7), then you can no longer ignore the [H+] of neutral water

Normally you ignore [H+] of neutral water bc its too small/is negligible - So when [H+ added acid] is way bigger than [H+ neutral water] then [H+ neutral water] is neg bc too small - But when [H+ neutral water] is bigger than [H+ acid], then [acid] becomes negligible compared to [H+ neutral] *Basically if you get [H+ acid] < [H+ neutral water], add [H+ acid] and [H+ neutral water] = 1.00x10^-7, then calc pH w that new [H+]total* - Normally you don't need to do this bc [H+ neutral water] is negligible compared to [H+ acid] - You only need to do this when [H+ neutral water] and [H+ acid] are of similar [ ]s/[H+ acid] is smaller

Phenolphthalein

Pink in base, clear in acid

NH3 -- Weak base

Reacts w/ H2O NH3 aq + H2O <--> NH4+ + OH- keq = [NH4+][OH-]/[NH3][H2O] [H2O] = constant @ 55.5 M keq[H2O] = [NH4+][OH-]/[NH3] = kb

If trying to find pOH by doing pkw - pH

Remember pkw = 14 @ 25 So even if doing pkw - pH means you get -pOH value, that's ok, you can have - pOH value - Normally, everyday things only fall b/w pH of 0 and 14, but it is possible to have things outside this range (aka -pH/-pOH or pH/pOH > 14)

If ka > 1

Rxn lies to right Ka should approach (theoretically) infinite for strong acids (aka no HA left if strong acid)

1. A .02 M of an unknown acid (HX) has a pH of 3.26. What is the ka of the acid?

Set up like equilibrium problem HX <--> H+ + X- I .02M 0 0 C -x +x +x E .02 -x x x x = 10^-3.26 = 5.5x10^-4 M For this problem it doesn't really matter, but just remember the negligible rule - If x/[og] * 100 < 5%, then the -x can be ignored - It doesn't really matter for this problem bc there's no quadratic, but when there's a possibility for a quadratic to form, you'll really enjoy remember the negligible rule bc you can avoid quadratics w/ this rule Then just solve for ka (ka = [H+][X-]/[HX] = (5.5x10^-4)^2/.02) You end up w/ ka = 1.5x10^-5 - This is a pretty small ka, meaning equilibrium lies to the left - So we are justified to leave out the -x bc rxn lies to left so [HX] stays relatively constant/barely dissociates - Remember the other part of the negligible rule, if the ka is 3 or more powers of 10 less than the [og], the -x can be ignored - Tho I would be careful w/ differences in powers of 10 of exactly three bc these can go both ways (I would keep -x just in case in this case) - Also if worried that you weren't justified, just do the 5% rule after calculating x to prove that you were justified

IF given % dissociation , remember % dissociation = [H+]/[HA] * 100

So % dissociation/100 * [HA] = [H+] (and often [A-] as well) Or % dissociation/100 ÷ [H+] = [HA] You can use this information to find ka % diss = amount dissociated/[initial] * 100

The relationship b/w temperature change, endothermic/exothermic, and k is still true for kw

So if given kw value (or you're asked to find kw value @ temp other than 25, so you'll have two known kw values [one at new temp, one @ 25]) Compare the two known kw values to temperature If endothermic T increases, kw increases If exothermic T decreases, kw increases May be asked to determine if endo or exo Look at the two different Ts and kws, see the relationship (does kw increase as T increases or does it decrease when T increases --> If first example, then endo, if 2nd, then exo) Why does kw increase w/ increase T for endothermic? - We know that we added heat bc T increased H2O + heat <-> H+ + OH- If add heat, we want to go away from what was added (bc heat is reactant, we added reactant, so rxn will shift right to use up excess heat) - This means that [products] increases and [reactants] decreases, aka numerator increases, denominator decreases, so kw gets bigger

Various Ways to Describe Acid Strength -- Property: Strength of conjugate base compared w/ that of water

Strong Acid -- A- much weaker base than H2O Weak Acid -- A- much stronger base than H2O

Various Ways to Describe Acid Strength -- Property: Position of the dissociation equilibrium

Strong Acid -- Far to right Weak Acid -- Far to left

1. Will the salt K2HC6H5O7 form an acidic or basic solution in water? ka1 = 7.4x10^-4 ka2 = 1.7x10^-5 ka3 = 4.0x10^-7

The main idea: Polyprotic acids/ions that come from polyprotic acids can either act as acid or base --> So we gotta figure out which process dominates, the ion acting as an acid or acting as a base - You can identify polyprotic or not by the charge/by the salt - If the salt is made up of a cation + an anion that contains a donatable proton, the anion comes from a polyprotic acid - If the salt is made up of multiple of the cations + the anion, the anion comes from polyprotic acid Basically, to know if this amphoteric problem solving applies --> If the ion has an H to donate AND has a negative charge that indicates that it could accept a + proton, the ion is amphoteric and can act as both acid and base K2HC6H5O7 --> 2K+ + HC6H5O7 2- The HC6H5O72- has 2 options It could donate H+ --> Act as acid 1. HC6H5O72- <--> H+ + C6H5O73- ka3 It could accept H+ --> Act as base 2. HC6H5O72- + H2O <--> H2C6H5O7- + OH- kb comes from ka2 Which dominates? The second possibility has a kb value that we must find - HC6H5O72- is the CB of the 2nd dissociation, so do kw/ka2 = kb ka3 > kb --> Acidic solution Remember, if you have a polyprotic acid problem like this and you are not sure is the acidic step is ka1/2/3 / you don't know which ka to use to calculate a kb value 1. If the polyprotic acid is losing its 1st H, or gaining its last H --> Ka1 (meaning ka1 is talking about the acidic way, or use ka1 to calc kb) 2. If the polyptortic acid is losing its 2nd H, or gaining its 2nd H --> Ka2 3. If the polyprotic acid is losing its 3rd H, or gaining its 1st H --> ka3

Amino Acids

The reason amino acids have basic and acidic side: There is amine side w/ N --> Ns usually have unbonded e- pairs, so its basic bc has unbonded pair to donate Other side is acidic side and it has carboxyl group so thats why acidic

The stronger the acid/base

The weaker its conjugate - HCl's conjugate = Cl- - Cl- is a PATHETIC conjugate base (aka a negligible base) The stronger the acid, the weaker the conjugate base The weaker the acid, the stronger the conjugate base And same w/ bases SEE DIAGRAM The stronger the acid/base, the more it dissociates/the more ions in solution The weaker, the less dissociation/less ions in solution

2. Determine the pH of a solution that is 5.00 M in HNO2. ka = 4.0x10^-4

This is example where difference in powers of 10 b/w ka and [og] concentration is greater than 3, so you can pretty safely ignore the -x (but could still check w/ 5% rule --> x/[HA]*100 <5% then you're good ) Basically what I want you to understand w/ weak acid problems is that weak acid problems are basically equilibrium problems w/ maybe 1 extra step (calc pH, pOH...etc.)

When something is bonded to H+, it may or may not act as acid

X--H will it act as acid? There are 2 things that will determine if it will 1. Strength of Bond ex. C-H bond is really strong, so H in C-H bond is not donatble and won't act as acid 2. Polarity of Bond - The more polarized the bond b/w X and H is, the more likely that compound/bond will donate the proton (aka act as acid) - ex NaH - This is ionic bond, won't act as acid bc its a neutral compound/the bond is not polar - Ionic bonds can't be polar - Also ionic bonds are super strong H-F, H-Cl, H-Br, H-I - We know that out of these, H-F is only weak acid - But based on polarity (as F is most polar), we would predict that H-F would be strongest acid - But its weak - Based on Coulomb's law, bc F is so SMALL, the bond is very STRONG --> so can't donate the H+ - So based on size and therefore bond strength, HF is weak acid As move down period, bond strength decreases as atom size (radius) increases - As you go down column, you'd expect acidity to increase bc size of atoms increase, so bond strength b/w the atom and H decreases --> low bond strength = high acidity

Calc pH of solution w/ .10 M HOCl + .10M HNO2 ka HOCl = 3.5x10^-8 kaHNO2 = 4.0x10^-4

You need to calc [H+] in HOCl solution (using ICE chart bc weak) and [H+] from HNO2 solution (using ICE chart) Then add those separate [H+] to get [H+]total Then use [H+]total to find pH of total solution pH = 2.21 Ok, not entirely sure yet *BUT IM PRETTY SURE THIS IS INCORRECT* I think what you are supposed to do is first, do the dissociation of the weak acid that has the higher ka Then after you do that, do the dissociation of the second weak acid but make sure to bring down the [H+] from the first weak acid dissociation (bc remember the H+ doesn't just go away, it dissociated from the first acid and is still in solution) SO I'm not sure which approach is correct, so gotta check