Acids & Bases .

Acids & Bases

Warm Up Give you 5 minutes to discuss with your peers before checking your answers.

Conceptual-Use it until you can’t!
Arrhenius Acid - anything that can give off a H+ ion. Arrhenius Base - anything that can give off an OH- ion.

Conceptual Bronsted-Lowry Acid (1923) - anything that will donate a proton (H+) Bronsted-Lowry Base - anything that will accept a proton

ACID-BASE THEORIES Lewis Definition Acid – accepts an electron pair
Base – donates an electron pair This terminology is not as widely used as it used to be because it is too general. Instead, we focus on coordination complexes, which are products of Lewis Acid-Base reactions.

Conjugate Pairs Acids LOSE an H+ and have conjugate bases
Bases GAIN an H+ and have conjugate acids

Conjugate Pairs HC2H3O2 + H2O  CH3CH3NH2 + H2O 

Bronsted-Lowry NH3 + H2O ==> NH4+ + OH-
base acid conjugate conjugate acid base Know how to label!

Warm Up-Have article on desk
For questions 1 and 2 identify the acid, base, conjugate acid, and conjugate base. 1. OH- + HPO42- → H2O + PO43- 2. HF + NH3 → NH4+ + F- For questions 3-7 Name the acid or base. H3PO HNO2 HBr LiOH Mg(OH)2 Given a pH of 9, find the [H+], pOH, and [OH-]. Is the solution acidic or basic? (show all work) If [H+] = 2.5 x 10-5, find the pH, pOH, and [OH-]. Is the solution acidic or basic? (show all work) Warm Up-Have article on desk

Motivation

CONCENTRATED VS. STRONG
“Concentrated” – refers to the amount dissolved in solution. “Strong” – refers to the fraction of molecules that ionize. For example, if you put a lot of ammonia into a little water, you will create a highly concentrated solution. However, since only 0.5% of ammonia molecules ionize in water, this basic solution will not be very strong.

HA(aq) + H2O(l) ⇄ H3O+(aq) + A– (aq)
ACID STRENGTH A strong acid completely ionizes in solution A weak acid only partially ionizes. The strength of an acid depends on the equilibrium: HA(aq) + H2O(l) ⇄ H3O+(aq) + A– (aq) If the equilibrium lies far to the left, the acid is weak (only a small percentage of the acid molecules ionize) If the equilibrium lies far to the right, the acid is strong (completely ionizes)

Strong Acids Ionize(dissociate) completely
HCl, HBr, HI, HClO3, HClO4, HNO3, H2SO4 Only do it one at a time.

Strong Bases Ionize (dissociate) completely
Alkali metal hydroxides and heavy alkaline earth hydroxide. (Ca, Sr & Ba) Facts: 1) No rule about breaking apart 2) Why can we eat some strong bases?

Bronsted-Lowry NH3 + H2O ==> NH4+ + OH-
base acid conjugate conjugate acid base Understand: It is a competition between bases for the proton H+

Finish the statement! NH3 + H2O ==> NH4+ + OH-
base acid conjugate conjugate acid base The more readily a substance gives up a proton……. The more acidic the substance is

base acid conjugate conjugate acid base The more readily a substance accepts a proton….. The more basic it is

Finish the statement! HCl + H2O ==> Cl- + H3O+
acid base conjugate conjugate base acid Strong acids have _______conjugate bases. weak

base acid conjugate conjugate acid base Weak acids always have _________ conjugate bases. strong

Conjugates and Strength

Finish the statement! Water can act as an acid or a base. Water is called __________. Amphoteric

Warm Up Define concentrated vs. strong
Clarification: Why is HI stronger then HCl? Which statement is correct? Why? HClO2 is stronger then HClO3 HI is a weaker acid then HCl CH3COOH is a stronger acid then CH2BrCOOH HNO3 is stronger then HNO2

STRONG ACIDS An HCl solution contains virtually no intact HCl.
Single arrow indicates complete ionization. HCl(aq) + H2O(l) → H3O+(aq) + Cl– (aq) An HCl solution contains virtually no intact HCl. The HCl has essentially all ionized to form H3O+(aq) and Cl–(aq).

Calculating pH of Strong Acid Solutions
Because a strong acid completely ionizes, or dissociates ions, [HA] = [H+] for the first hydrogen ion released by the acid. Monoprotic acids only release one hydrogen ion. Polyprotic acids (di- and tri-protic) completely dissociate from their first hydrogen ion, but the second hydrogen ion is much less likely to dissociate.

Finding the pH of a Strong Acid
Strong acids Ionize completely HA  H+ + A- Do it only one at a time So?? The concentration of the strong acid is EXACTLY the same as the [H+]

Example – Find the pH of 1.0 M HCl. What contributes to H+? HCl is a strong acid pH = - log [H+] = - log [1.0] = 0.00

Example – Find the pH of 0.10 M HNO3. What contributes to H+? HNO3 is a strong acid [HNO3] = 0.10 M  [H+] = 0.10 M pH = - log [H+] = - log [0.10] = 1.00

Sometimes an acid is so dilute that water is actually better Example – Find the pH of 1.0 x M HBr. What contributes to H+? HBr is an acid, water could be an acid [HBr] = 1.0 x M  [H+] = 1.0 x M [H+] in water = 1.0 x 10-7 M So WATER is actually a better acid pH = - log [H+] = - log [1.0 x 10-7] = 7.00

Strong Bases Strong bases (soluble in water): NaOH, KOH, LiOH, RbOH, CsOH Strong bases (not very soluble in water): Ca(OH)2, Ba(OH)2, Sr(OH)2 Strong bases are OH- combined with group I or II cations. (NEED TO KNOW!!) When a strong base completely ionizes, or dissociates ions, [MOH] = [OH-] for the hydroxide ion released by the base. (Note: M is for metal)

Work the same way =] Strength of bases Strong = group 1 hydroxide
group 2 (Ca and below) hydroxide Weak  based on Kb Bigger the Kb, the better the base Some bases may not be on there because they are conjugates of weak acids! Remember Ka x Kb = 1 x 10-14

Work the same way =] Strong Base Find the pH of 5.0 x 10-2 M NaOH.
[H+] = 1.0 x log [5.0 x 10-2] 5.0 x pOH = 1.30 -log [2.0 x 10-13] pH = pH = 12.70

Remember! There’s no one at a time rule for bases!
pH of 3.0 M Ca(OH)2? [OH-] = 6.0 M pOH = pH = 14.78

WEAK ACIDS Double arrow indicates partial ionization. HF(aq) + H2O(l) ⇄ H3O+(aq) + F – (aq) An HF solution contains a large number of intact (or un-ionized) HF molecules. It also contains some H3O+(aq) and F –(aq). Slides 38 (Weak acids) through 53 (K calculations) correspond to p. 7-8 in notes packet. Then complete questions on slide 54 as a class.

ACID IONIZATION CONSTANT (Ka)
A way to quantify the relative strength of a weak acid The equilibrium constant of the weak acid HA(aq) ⇄ H+(aq) + A–(aq)

Ka Hydrocyanic acid (HCN) Ka = 4.9 x 10-10
Hydrocyanic acid (HCN) Ka = 4.9 x 10-10 Nitrous acid (HNO2) Ka = 4.6 x 10-2 Acetic acid (HC2H3O2) Ka = 1.8 x 10-5 Hydrofluoric acid (HF) Ka = 3.5 x 10-4 Chlorous acid (HClO2) Ka = 1.1 x 10-2

Practice…. If mol of HF is diluted with distilled water to a volume of 500 mL, what is the pH of the solution? Ka = 6.6x10-4

Work….. Write the reaction
HF + H2O ↔ F- + H3O+ Write your equilibrium expression (should we include water in our equilibrium expression?) Set up your ICE box

HF H2O ↔ H+ F- I C E

Write your Ka expression and solve for x
HF H2O ↔ H+ F- I 0.200M - C -x x E .2 -x Write your Ka expression and solve for x Ka = [H[[F]/[HF] x2 / .2-x = 6.6x x=0.0115 [H] = M pH = -log[H] pH = 1.94

BASE IONIZATION CONSTANT
The equilibrium expression for a base is known as Kb. This expression is used to quantify the relative strength of a weak base. B + H2O → HB+ + OH- Kb = [HB+] [OH-] [B]

What is the pH of a 0.100 M solution of ammonia? Kb = 1.77 x 10-5
NH3 + H2O = NH4 + OH (remember because it’s a weak base, bases accept proton) ICE pOH = 2.88 pH =11.12

Something better? H2O (l) + H2O (l) <==> H3O+(aq) + OH-(aq)
Auto-ionization of water=one molecule donates an electron pair while another accepts it. Kw =

Something better? H2O(l) + H2O(l) <==> H3O+(aq) + OH-(aq)
Kw = 1 x at 25⁰C Kw = 1 x at 95⁰C Temperature affects K – so only a scale of 14 at standard lab conditions

What about the [ ] of H+’s and OH-’s
Developed by a brewmaster(Sorenson) to test acidity of beer. pH scale pH = -log[ H+ ] pOH scale pOH = -log [OH-]

What about the [ ] of H+’s and OH-’s
Problem. What is the pH and pOH of water at 25⁰C? H2O(l) + H2O(l) <==> H3O+(aq) + OH-(aq) Kw = [H+] [OH-] 1 x = [x][x] 1 x 10-7 = x pH = -log1 x pOH = -log 1 x 10-7 pH = 7 pOH = 7

Acidic/Basic/Neutral
Can look at pH Less than 7 = ACIDIC More than 7 = BASIC Neutral = 25⁰C

Acidic/Basic Neutral Can also compare [H+] and [OH-]
If [H+] > [OH-] then it is ACIDIC If [H+] < [OH-] then it is BASIC Neutral is [H+] = [OH-]  at ANY temp!!

Problems? 1) What is the pH if the [H+] = 4.5 x 10-6?

Problems? 2) What is the pOH if the [H+] = 1.5 x 10-3?

Problems? 3) What is the [H+] if the pH = 4.63

Problems? 4) What is the pH of 12 M HCl?

Finding the pH of a Weak Acid
Don’t ionize completely (only partially) So they make an EQUILIBRIUM Write the equation with water using the Bronsted-Lowry definition They’ll have a Ka value associated with them

A Couple Things about Ka
The relative strength of acids compared to each other can be found by comparing the Ka values for the acids Strong acids have a Ka of infinity Weak acids have known Ka values (look them up) You can also find the Kb of its conjugate base Ka x Kb = 1.00 x 10-14 same as [H+][OH-] = 1.00 x 10-14

Example – find the pH of 1.0 M HF. Look up the Ka for a weak acid (6.6 x10-. HF (aq) + H2O (l)  F- (aq) + H3O+ (aq) I C - x x x E 1.0 – x x x

7.4 x 10-4 = [x][x] [1.0-x] very small x = 2.7 x  5% rule? What’s x? So [H3O+] = 2.7 x 10-2 pH = - log [0.027] = 1.57

For the quiz… HCN Ka = 6.2 x 10-10 HOC6H5 =

A Little Different Weak Acid calculation
Find the molarity of acetic acid is the pH of the solution is 4.15. HC2H3O2 (aq) + H2O (l)  C2H3O2- (aq) + H3O+ (aq) I ??? C - x x x E ??? – x x x

A Little Different Weak Acid calculation
pH = 4.15 So [H3O+] = = 7.1 x 10-5 M So x = 7.1 x 10-5 M 1.8 x 10-5 = [7.1 x 10-5] [7.1 x 10-5] [??? – 7.1 x 10-5] Solve for ??? 2.8 x 10-4 M

Last thing they could do
Give you the initial concentration of an unknown acid and its pH, find Ka M HQ, pH of 5.31

Mixtures of Acids When two or more acids are mixed together, you must calculate them in order from best “strongest” to weakest Based on Ka

Mixtures of Acids If both are strong….
Basically finding the new molarity of H+ and taking the pH Ex – calculate the pH of a mixture if 50.0 mL of 1.00 M HCl and mL of 3.00 M HNO3 are mixed.

Mixtures of Acids Calculate the pH of a solution that contains 1.00 M HCN (Ka=6.2 x 10-10) and 5.00 M HNO3. Also calculate the [CN-] in this solution.

Mixtures of Acids Start with the better acid
HNO3 (aq) + H2O (l)  NO3- (aq) + H3O+ (aq) I C x x x E – x x x 4.0 x 10-4 = [x][x]  x = = [H3O+] [5.00-x]

Mixtures of Acids Then do the next acid (BUT YOU ALREADY HAVE A CONCENTRATION OF H3O+ FROM THE FIRST ACID) HCN (aq) + H2O (l)  CN- (aq) + H3O+ (aq) I C x x x E – x x x 6.2 x = [ x][x]  x = 1.4 x 10-8 = [CN-] [1.00-x] [H3O+] =  pH = - log [0.045] = 1.35

Problem Find the pH of 1.0 M sol of methylamine. Kb = 4.38 x 10-4

Problem CH3NH2 <==> CH3NH3+ + OH- I 1.0 M 0 0 C -x +x +x
E 1.0-x +x +x 4.38 x 10-4 = x2 [1.0-x] x is very small

Problem CH3NH2 <==> CH3NH3+ + OH- 4.38 x 10-4 = x2
[1.0-x] x is very small [OH-] = 2.1 x % rule = 2.1% pOH = 1.68 pH = = 12.32

Using the pH A solution of triethylamine has a pH of Find its initial concentration (Kb = 4.0 x 10-4) The [ ] = M

Example – find the pH of 1.0 M HF (Hint look up the Ka for HF) Ka = 6.8x10-4 HF (aq) + H2O (l)  F- (aq) + H3O+ (aq) I C - x x x E 1.0 – x x x

6.8 x 10-4 = [x][x] [1.0-x] very small x = 2.6 x  5% rule? What’s x? So [H3O+] = 2.6 x 10-2 pH = - log [0.027] = 1.58

Ka values for HCN and HOC6H5
HOC6H5 = 1.3 x 10-10 HCN = 6.2 x 10-10

PERCENT IONIZATION OF A WEAK ACID
Concentration of ionized acid (at equilibrium) Concentration of acid (initially) Determine equilibrium concentrations for all species in the ionization reaction (Use ICE table if needed). Use Ka and the equilibrium constant expression to find [H+]. Use [H+]eq and the initial acid concentration, [HA]0, to solve for percent ionization.

EXAMPLES OF PERCENT IONIZATION
What is the percent ionization in a M solution nitrous acid if the [H3O+] is M? If a 2.5 M solution of nitrous acid is 1.4% ionized, what is the hydronium concentration in the solution? What is the pH of the solution? 4.8% = [H+], pH = 1.46

[HNO2]eq = __________________ [H+]eq = _________ [NO2-]eq = _________
Determine equilibrium concentrations for all species in the ionization reaction (Use ICE table if needed). Use Ka and the equilibrium constant expression to find [H+]. Use [H+]eq and the initial acid concentration, [HA]0, to solve for percent ionization. Find the percent ionization of a 2.5 M HNO2 solution. The Ka of nitrous acid is 4.6 x 10-4. HNO2 ⇄ H+ + NO2- [HNO2]eq = __________________ [H+]eq = _________ [NO2-]eq = _________ x x

Determine equilibrium concentrations for all species in the ionization reaction (Use ICE table if needed). Use Ka and the equilibrium constant expression to find [H+]. Use [H+]eq and the initial acid concentration, [HA]0, to solve for percent ionization. Find the percent ionization of a 2.5 M HNO2 solution. The Ka of nitrous acid is 4.6 x 10-4. HNO2 ⇄ H+ + NO2- (x is small) Solve for x, which equals [H+]

Polyprotic Acids Acids that furnish more than one H+
Ex: *H2SO4, H2CO3, H3PO4 *****It is always one proton at a time**** H2CO3  H+ + HCO Ka1 HCO3-  H CO32- Ka2

Polyprotic Acids Typically: Ka1 > Ka2 > Ka3
Usually by at least a factor of 100 x Why? The conjugate base of a weak acid is a pretty “good” base – so it would rather act like a base than an acid in most cases

Polyprotic Acid Calculation
Find the pH of 5.0 M H3PO4 and the equilibrium [ ] of each conjugate base. (Ka1 = 7.7 x 10-3) (Ka2= 6.2x10-8) What are we trying to find? [H2PO4-] [HPO42-] [PO43-] [H3O+]  pH

Start with the first H+ (Ka1) H3PO4 (aq) + H2O (l)  H2PO4- (aq) + H3O+ (aq) I C x x x E – x x x 7.7 x 10-3 = [x][x]  x = = [H3O+] = [H2PO4-] [5.0-x]

Then do the next hydrogen (Ka2)– use the values for the things you just calculated in this one too!! H2PO4- (aq) + H2O (l)  HPO42- (aq) + H3O+ (aq) I C x x x E – x x x 6.2 x 10-8 = [ x][x]  x = 6.2 x 10-8 = [HPO42-] [0.196-x] [H3O+] now = x 10-8 = 0.038

Then do the last hydrogen (Ka3) – use the values for the things you just calculated in this one too!! HPO42- (aq) + H2O (l)  PO43- (aq) + H3O+ (aq) I x C x x x E 6.2 x 10-8 – x x x 4.8 x = [ x][x]  x = 7.8 x = [PO43-] [6.2 x x] [H3O+] still M  pH = 1.42

Problem Find the pH of 1.0 M sol of methylamine. Kb = 4.38 x 10-4

Warm Up Calculate the percent dissociation of a weak acid in a M HA solution. (Ka = 1.60 x 10-5)

ACID-BASE PROPERTIES OF SALTS
Salts are ionic compounds (cation + anion) When dissolved in water, some salts are pH- neutral, some are acidic, and some are basic IN GENERAL, anions tend to form basic or neutral solutions. IN GENERAL, cations tend to form acidic or neutral solutions.

Acid-Base properties of Salts: Salt Hydrolysis
Salts are ionic compounds (cation + anion) When dissolved in water, some salts are pH-neutral, some are acidic, and some are basic In order to determine acidity, take the salt and add water. The water will dissociate into the H+ and OH-; this process is hydrolysis.

Salt Hydrolysis Compare the ions separately.
Add H+ to the anion (to make its conjugate acid) and OH- to the cation (to make its conjugate base). Strong acids/bases completely dissociate. The weak acid/base does NOT complete dissociate. Ex. NaNO Na|NO3 NaOH HNO3 SA + SB = NEUTRAL

Salt Hydrolysis cont. Determine if the following salts will produce acidic, basic, or neutral solutions. KNO2 NH4Br CaCl2 Basic Acidic Neutral

Determining the pH of a salt solution
The same equilibrium expression is used Add in the hydrolysis of water in order to complete the expression Remember: solve for Kb if basic and solve for ka for acidic solutions. Example: Calculate the pH of a 1.0 M solution of sodium acetate. The ka for acetic acid is 1.8 x 10-5. Zuhmdahl book - p. 55 ex. 2.14 pH = 9.37

Polyprotic Acids Acids that furnish more than one H+
Ex: H2SO4, H2CO3, H3PO4 *****It is always one proton at a time**** H2CO3  H+ + HCO Ka1 HCO3-  H CO32- Ka2

Polyprotic Acids Typically: Ka1 > Ka2 > Ka3
Usually by at least a factor of 100 x Why? The conjugate base of a weak acid is a pretty “good” base – so it would rather act like a base than an acid in most cases

Warm Up Find the pH of 5.0 M H3PO4 and the equilibrium [ ] of each conjugate base. (Ka1 = 7.7 x 10-3) (Ka2= 6.2x10-8) **Hint this is a polyprotic acid  **

Find the pH of 5.0 M H3PO4 and the equilibrium [ ] of each conjugate base. (Ka1 = 7.7 x 10-3) (Ka2= 6.2x10-8) What are we trying to find? [H2PO4-] [HPO42-] [PO43-] [H3O+]  pH

Start with the first H+ (Ka1) H3PO4 (aq) + H2O (l)  H2PO4- (aq) + H3O+ (aq) I C x x x E – x x x 7.5 x 10-3 = [x][x]  x = = [H3O+] = [H2PO4-] [5.0-x]

Then do the next hydrogen (Ka2)– use the values for the things you just calculated in this one too!! H2PO4- (aq) + H2O (l)  HPO42- (aq) + H3O+ (aq) I C x x x E – x x x 6.2 x 10-8 = [ x][x]  x = 6.2 x 10-8 = [HPO42-] [0.038-x] [H3O+] now = x 10-8 = 0.038

Then do the last hydrogen (Ka3) – use the values for the things you just calculated in this one too!! HPO42- (aq) + H2O (l)  PO43- (aq) + H3O+ (aq) I x C x x x E 6.2 x 10-8 – x x x 4.8 x = [ x][x]  x = 7.8 x = [PO43-] [6.2 x x] [H3O+] still M  pH = 1.42

Titrations Standard solution - Solution in
which the molarity is known. (Typically this solution is placed in the buret) Analyte - solution whose concentration is unknown.

Titrations Equivalence point - The point during the titration when the moles of standard solution equal the moles of titrant. (moles acid = moles base) Endpoint - Indicates when the solution has reached the equivalence point. Usually by turning a different color. KNOW THE DIFFERENCE BETWEEN THE EQUIVALENCE POINT AND THE ENDPOINT!

Titrations - Strong Acid Strong Base
Consider the titration of mL of 1.00 M HCl with M NaOH. Calculate the [H+] in the solution after 50.0 mL of 0.50 M NaOH has been added. write the equation. HCl + NaOH → NaCl + H2O 2. Determine the number of moles in each of the solutions 1 = mol/ = mol/0.50 0.1 mol HCl = mol mol NaOH = mol

Titrations - SA/SB cont.
HCl + NaOH → NaCl + H2O 3. Since the reaction goes to completion, all of the NaOH is consumed. Therefore, the amount of moles of HCl left are: = moles HCl 4. Find the resulting concentration of [H+]. **Be sure to use total volume in the solution = 0.50 M HCl 5. Solve for pH pH = -log[0.50] = 0.3

SALTS 1) Another name for an ionic compound
2) When dissolved they break apart into ions The Catch?

SALTS The Catch? Under certain conditions these ions behave like acids and bases.

Hydrolysis The ability of ions to react with water to generate H+ (or H3O+) or OH-

SALTS Condition 1: Neutral Solutions
Salts that consist of anions of Strong Acids and cations from Strong Bases have no effect on pH

SA  H A- the conjugate base has no affinity for H+ SB  B+ + OH- the conjugate acid can’t produce H+ and has no affinity for them

SALTS Condition 1: Neutral Solutions Example: NaCl

SALTS Condition 1: Neutral Solutions Example: NaCl
Na+ is a cation of a strong base - NaOH

SALTS Condition 1: Neutral Solutions Example: NaCl
Cl- is an anion of a strong acid - HCl

Therefore adding NaCl to water has no effect on the pH of the water.

SALTS Condition 2: Basic Solutions
It will come from a weak acid; The conjugate base will be stronger than water(base)

SALTS Condition 2: Basic Solutions Example: NaC2H3O H2O

SALTS Condition 2: Basic Solutions Example: NaC2H3O2 + H2O 

SALTS Condition 2: Basic Solutions Example: NaC2H3O2 + H2O 
Neither A or B

SALTS Condition 2: Basic Solutions Example: NaC2H302 + H20 
Na C2H3O H2O conjugate base of weak acid Neither A or B

SALTS Condition 2: Basic Solutions Example: NaC2H302 + H20 
Na C2H3O H2O conjugate base amphoteric of weak acid (weak) Neither A or B

SALTS C2H3O2- (aq) + H2O (l)  How do we figure it out?

SALTS C2H3O2- (aq) + H2O (l)  HC2H3O2 (aq) + OH- (aq)
HC2H3O2 + H2O<====> C2H3O2- + H3O Ka= 1.8 x 10-5 but this isn’t what is happening. C2H3O2- + H2O <====> HC2H3O2 + OH- it is the reverse so remember Ka x Kb = Kw calculate Kb = Kw/Ka

SALTS C2H3O2- (aq) + H2O (l)  HC2H3O2 (aq) + OH- (aq) Kb = Kw Ka
Ka for HC2H3O2 (aq) is 1.8 x so Kb = =

So if you had a 0.50 M solution, what would be the pH? Set up ICE I C E

So if you had .50 M solution, what would be the pH? Set up ICE and solve:

SALTS Calculate the pH of a 0.30 M NaF solution.
The Ka value for HF is 7.2 x 10-4.

SALTS Bottom Line: For any salt whose cation has neutral properties and whose anion is the conjugate base of a weak acid, the solution will be basic!

SALTS Condition # 3. Acid solutions……a couple of ways.

SALTS Condition # 3. Acid solutions……a couple of ways.
a) Salts in which the anion is not a base and the cation is the conjugate acid of a weak base, produce an acidic solution.

SALTS Condition # 3. Acid solutions……a couple of ways.
Salts in which the anion is not a base and the cation is the conjugate acid of a weak base, produce an acidic solution. Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is 1.8 x 10-5. NH4+ + H2O <====> NH3 + H3O+

SALTS Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is 1.8 x 10-5. Write equation and find K: Set up ICE

SALTS Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is 1.8 x 10-5. Set up ICE NH4+ + H2O <====> NH3 + H3O+ I C E

SALTS Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is 1.8 x 10-5. NH4+ + H2O <====> NH3 + H3O+ I C x x x E x x x Solve:

SALTS If both are conjugates of weak acids and bases, then must look at which one is stronger (based on K values) If Ka > Kb the salt is acidic NH4F (Ka NH4+ = 5.6 x 10-10; Kb F- = 1.4 x 10-11) If Kb > Ka the salt is basic NH4ClO(Ka NH4+ = 5.6 x 10-10; Kb ClO- = 3.6 x 10-7)

SALTS ANOTHER WAY to get an acidic solution:
If you have a highly charged metal cation => it rips away the OH- from one of the waters producing H+ Al(H2O)63+(aq) <====> Al(OH)(H2O)52+(aq) + H+(aq) Generally speaking, the higher the charge on the cation, the greater the acidity! Yellow Pit Stains-Yikes

SALTS What if it’s amphoteric?? Na2HPO4 -- acid or base?

SALTS Master Problem! Na2HPO4 so you have HPO42- therefore it could
Equation 1: Ka 3 = 4.2 x 10-13 or Equation 2: Ka 2 = 6.2 x 10-8 Which one?

SALTS Use the better one and ICE to solve for 0.75 M solution

Acid Base Equilibrium Part 2

Solutions of Acids or Bases and Their Salts
We have a weak acid and its salt HF <==> H F- Ka = 7.2 x 10-4 NaF <==> Na+ + F- (dissolves into ions)

HF <==> H+ and F- Ka 7.2 x 10-4 NaF <==> Na+ + F- (dissolves into ions) Have both HF and F- initially!

HF + H2O <==> H3O+ + F- LeChatelier-> shift to left causing it to be less acidic.

HF + H2O <==> H3O+ + F- Problem: The equilibrium [ ] of H+ in a 1.0 M HF solution is 2.7 x 10-2 M and the % dissociation is 2.7 % Calculate the [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF

Acid Base Pt 2 HF + H2O <==> H3O+ + F- I C -x +x +x E 1.0-x x x

Acid Base Pt 2 HF + H2O <==> H3O+ + F- E 1.0-x x x 7.2 x 10-4 = [x][1.0+x] [1.0-x] x = 7.2 x 10-4 % = 0.072

Acid Base Pt 2 Compare Before common ion=2.7% Common ion = % Greatly inhibits dissociation!

Acid Base Pt 2 So what is the point of a buffer?? It contains two species that are conjugates of one another. One is an acid, the other is a base. With both an acid AND a base, you have the ability to neutralize anything added and maintain a fairly constant pH.

Problems Calculate the pH of a solution by adding 0.30 mol of HC2H3O2 and 0.30 mol of NaC2H3O2 to enough water to make 1.0 L. (Ka = 1.8 x 10-5)

Problems 2) Calculate the pH of a solution containing M of HNO2 and 0.10 M KNO2. (Ka = 4.5 x 10-4)

Problems 3) Calculate [F-] and the pH of a soln that is 0.20 M HF and 0.10 M HCl.

Warm Up

Buffers-Acid Rain

Buffers-resist the change in pH
FUN FACT: During exercise, the muscles use up oxygen as they convert chemical energy in glucose to mechanical energy. -This O2 comes from hemoglobin in the blood. CO2 and H+ are produced during the breakdown of glucose, and are removed from the muscle via the blood. T -The production and removal of CO2 and H+, together with the use and transport of O2, cause chemical changes in the blood. These chemical changes, unless offset by other physiological functions, cause the pH of the blood to drop. If the pH of the body gets too low (below 7.4), a condition known as acidosis results. This can be very serious, because many of the chemical reactions that occur in the body, especially those involving proteins, are pH-dependent. Ideally, the pH of the blood should be maintained at 7.4. -If the pH drops below 6.8 or rises above 7.8, death may occur. Fortunately, we have buffers in the blood to protect against large changes in pH.+ *

Awesome site for more info…

What is a “Buffer”? A buffer is a solution that resists changes in pH
when small amounts of acid or base are added to it.

What is it made of? Buffers can be placed into two categories: acidic buffers and basic buffers Acidic buffers Consists of a weak acid and a salt of its conjugate base e.g. CH3COOH and CH3COONa pH less than 7 Basic buffers Consists of a weak base and a salt of its conjugate acid e.g. NH3 and NH4Cl pH greater than 7 Our blood is a buffer, as cells can only survive within a narrow pH range The pH of a buffer can be changed by altering the ratio of acid and its conjugate base or base and its conjugate acid. The amount of acid/conjugate base or base/conjugate acid determine its buffer capacity, or amount of acid or base that can be added for it to still maintain constant pH

How does it work? Buffers follow one of the following equilibriums:
There are both products and reactants present in both cases Buffers use Le Chatelier’s Principle, which states that a system will relieve stress put on a system by “shifting” to the left or right side of the equation accordingly. In other words, if you add reactant to the equilibrium, the system would respond by removing reactant (and thus creating product), and vise versa

If acid is added to the system…
How does it work? If acid is added to the system… + + + + + The H3O+ from the acid will cause the system to shift left Or react with OH- to form H2O

If base is added to the system…
How does it work? If base is added to the system… + + + + + The OH- from the base will cause the system to shift to the left Or react with H3O+ to form H2O

Summary Buffers are solutions that resist changes in pH with the addition of small amounts of acid or base They consist of a weak base and its conjugate acid salt or a weak acid and its conjugate base salt They work through Le Chatelier’s Principle, shifting to the left or right to counterbalance the addition of acid or base

A Better Way! (or at least a different way….)
Can always use ICE or……… Henderson-Hasselbach Equation: shows relationship of variables in a buffer solution.

Henderson-Hasselbach
pH = pKa + log([base]/[acid]) or pOH = pKb + log([acid]/[base]) Log ([Base/[Acid]) = log[A-]/[HA] **see reference sheet**

What is the pH of a buffered soln that is 0.12 M lactic acid(HC3H5O3) and 0.10 M sodium lactate? Ka = 1.4 x 10-4

What is the pH of a buffered soln that is 0.12 M lactic acid(HC3H5O3) and 0.10 M sodium lactate? Ka = 1.4 x 10-4 1.4 x 10-4 = [.10+x][x] x = 1.68 x 10-4 [.12-x] pH = 3.77

What is the pH of a buffered soln that is 0.12 M lactic acid(HC3H5O3) and 0.10 M sodium lactate? Ka = 1.4 x 10-4 pH = pKa + log([base]/[acid]) pH = -log(1.4 x 10-4) + log(.10/.12) pH = 3.77

Buffers Problem: A buffered solution contains 0.40 M acetic acid and 0.60 M sodium acetate. Calculate the pH. Ka = 1.8 x 10-5

Buffers pH = pKa + log [B] log [A] = 4.74 + log (0.60/0.40) = 4.92
HC2H3O2 <==> H C2H3O2- pH = pKa + log [B] log [A] = log (0.60/0.40) = 4.92

Buffers Option 1: Can manipulate Ka expression
You can adjust the [ ] of each to make a buffer solution at any pH. Option 1: Can manipulate Ka expression Ka = [H+][X-] manipulate H+ = Ka[HX] [HX] [X-]

Buffers Option 2: Can use H-H equation and solve for ratio.
You can adjust the [ ] of each to make a buffer solution at any pH. Option 2: Can use H-H equation and solve for ratio. pH = - log Ka + log (B/A)  B / A = 10(pH + log Ka)

Buffer Capacity Amount of acid or base it can neutralize before pH begins to change! 1 M HC2H3O2 and 1M NaC2H3O2 versus 0.1 M will have the same pH but different capacity to be a buffer.

Buffers [H+] or pH is determined by 2 factors:
Ka expression and value Ratio of [ ] CA or CB By adding OH- it will react and dec[acid] and inc [base]. By adding H+ it will react and dec [base] and inc [acid]

Optimal Buffering Problem: A chemist needs a solution buffered at pH 4.30 and can choose from the following acids(and their salts). Chloroacetic acid Ka 1.35 x 10-3 Proponoic acid Ka 1.3 x 10-5 Benzoic acid Ka 6.4 x 10-5 Hypochlorous acid Ka 3.5 x 10-8 Find the Ka closest to the pH you are looking for…..10^ = 5.01x10-5 ; Benzoic Acid

Acid-Base Titration The process of adding a strong acid or a strong base to a titrate of acid or base with the purpose of neutralizing the substance in the beaker Looking to reach equivalence point  mol H+ = mol OH-

Acid-Base Titration Generally done with a pH probe, which is used to create a graph of pH vs. volume added pH Volume added

Determination of the Equivalence Point
Ways to find the equivalence point Can find it stoichiometrically using stoichiometry for the neutralization reaction mol H+ = mol OH- stoichiometrically

Determination of the Equivalence Point
Ways to find the equivalence point Can find it graphically based on pH vs. volume graph When the titrate is neutralized completely by the titrant, the pH changes dramatically The midpoint of this dramatic change is the equivalence point – the middle of the area where neutralization is happening

Use of an Acid-Base Indicator
If the pH at the equivalence point is known, an indicator can be used Want the pKa of the indicator to match the pH at the equivalence point Indicator will then change color as pH changes dramatically, giving visual end to the titration

Indicators They are weak acids and weak bases themselves!
Given Ka of possible indicators RANGE of indicators = pKa ± 1 (10/1 or 1/10 ratio of A/B) Ex – bromothymol blue has Ka = 1.0 x 10-7 HA side of indicator = yellow, A- side of indicator = blue

Indicator Chart

Things to KNOW about Titrations
Need to be able to EXPLAIN the shape of the curve at ALL points through the titration What happens initially after a small amount of acid/base added When excess of the titrant still remains What happens near the equivalence point What happens beyond the equivalence point – when excess of the strong acid or strong base used as a titrate is added

Things to CALCULATE about Titrations
Calculate the initial pH of acid/base in the beaker Normal acid/base calculation for strong or weak acid/base Calculate the pH at the halfway point of the titration Or use pH at halfway point to find Ka or Kb of substance titrated Calculation of the pH at the equivalence point OR using the equivalence point to find the concentration of unknown acid/base you are titrating Calculation of the pH at a point beyond the equivalence point

Five Types of Titrations
Strong acid titrated with a strong base Strong base titrated with a strong acid Weak acid titrated with a strong base Weak base titrated with a strong acid Polyprotic acid titrated with a strong base

Strong Acid titrated with Strong Base
Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find initial pH

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH pH Volume added Starts at LOW pH (strong acid in beaker)

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH at halfway point

Upon adding strong base, there is a very SMALL change as long as there is still excess acid. pH Volume added A very small change with excess strong acid still present

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH at equivalence point

As the equivalence point is reached, there begins to be a dramatic jump in pH as moles acid = moles base pH Volume added Dramatic change

Even just beyond the equivalence point, the addition of excess base leads to a basic pH pH Volume added Excess base now = jump to high pH Midpoint = equivalence point = 7

Example – you have 50.0 mL of M HNO3 and look to titrate it with M NaOH Find pH after mL of NaOH has been added

As more excess strong base is added, there is NOT a dramatic change in pH pH Volume added Same excess base = no dramatic change in pH

Strong Base titrated with Strong Acid
OPPOSITE!!!!! To find initial pH – To find pH at halfway point – To find pH at equivalence point – To find pH beyond equivalence point -

Strong Base titrated with Strong Acid
OPPOSITE!!!!!

WEAK Acid titrated with Strong Base
Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find initial pH

Still starts at a low pH (though not quite as low) with weak acid in beaker pH Volume added Starts at LOW pH (weak acid in beaker)

The REACTION that occurs is now slightly different: HC2H3O2 + OH-  C2H3O H2O Makes a GOOD conjugate base now that also influences the pH and

The REACTION that occurs is now slightly different: HC2H3O2 + OH-  C2H3O H2O With not all weak acid reacted and good conjugate base also formed, it creates a BUFFER

Initial spike in pH due to formation of the conjugate base. Then mostly levels off as with both weak acid and conjugate base you have a BUFFER. pH Volume added Jump = conjugate base made, then buffer region

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH at halfway point

HALFWAY POINT Ratio of weak acid to conjugate base is the same
Half of weak acid used up, equal amount of conjugate base formed pH = pKa!!!

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH at equivalence point

As the equivalence point is reached, there again is the point where the base neutralizes all of the acid and there is a significant jump in pH. The pH at the equivalence point is GREATER than 7 since the GOOD conjugate base remains. pH Volume added Dramatic jump up as equivalence point is reached again Halfway point: pH = pKa

Even just beyond the equivalence point, the addition of excess base leads to a basic pH pH Volume added Excess base now = jump to high pH Midpoint = equivalence point = GREATER THAN 7

As more excess strong base is added, there is NOT a dramatic change in pH pH Volume added Same excess base = no dramatic change in pH

Example – you have 50.0 mL of M HC2H3O2 with 0.100 M NaOH (Ka of acetic acid = 1.8 x 10-5) Find pH after mL NaOH added

WEAK Base titrated with Strong Acid
OPPOSITE!!!!! To find initial pH – To find pH at halfway point – To find pH at equivalence point – To find pH after equivalence point -

WEAK Base titrated with Strong Acid
OPPOSITE!!!!!

Polyprotic Acids Acids that have more than one H+
That means there will be more than one equivalence point!! Each H+ reacted must be neutralized with a stoichiometric amount of base Example – H2CO3 with NaOH

Acids & Bases .

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