Acids and Bases …and Airman 1st class HCl flew back over the front lines, confident that he had neutralized the enemy’s strongest base.

NaOH into HCl

NaOH into HC2H3O2

NaOH into NH4Cl

Arrhenius acids and bases
Substances that ionize in water to form H+ ions are acids. Substances that ionize in water to form OH- ions are bases.

Substances that ionize in water to form H+ ions are acids. Substances that ionize in water to form OH- ions are bases. Remember strong and weak electrolytes?

Substances that ionize in water to form H+ ions are acids. Substances that ionize in water to form OH- ions are bases. Remember strong and weak electrolytes? Strong acids and bases ionize completely Weak acids and bases ionize partially.

Examples? Acids Bases

Properties of Acids and Bases

Properties of Acids and Bases
Are electrolyte solutions Make ions in solution Affect indicators Have high pH Taste bitter Neutralize acids Can cause serious burns Corrode aluminum only Have more OH- than H+ (in solution) Acids Are electrolyte solutions Make ions in solution Affect indicators Have low pH Taste sour Neutralize bases Can cause serious burns Corrode reactive metals Have more H+ than OH- (in solution)

Properties of Both Acids Are electrolyte solutions
Make ions in solution Affect indicators Have low pH Taste sour Neutralize bases Can cause serious burns Corrode reactive metals Have more H+ than OH- (in solution) Bases Are electrolyte solutions Make ions in solution Affect indicators Have high pH Taste bitter Neutralize acids Can cause serious burns Corrode aluminum only Have more OH- than H+ (in solution)

BrØnsted-Lowry Definition
Substances that donate a proton (H+ ion) in a reaction are acids. Substances that accept a proton (H+ ion) are bases.

Lewis Definition Substances that accept an electron pair in a reaction are acids. Substances that donate an electron pair are bases.

Conjugates After an acid has donated a proton, the rest of the species is the conjugate base. HAA- + H+ After a base has accepted a proton, the resulting species is the conjugate acid. B- + H+ HB

What is the conjugate base of…
HCl CH3COOH H2SO4 HSO4- H2O NH4+ NH3

ACID (loses H+ to form its) Conjugate base HCl CH3COOH H2SO4 HSO4- H2O NH4+ NH3

ACID (loses H+ to form its) Conjugate base HCl (  H+ and) Cl- CH3COOH H2SO4 HSO4- H2O NH4+ NH3

ACID (loses H+ to form its) Conjugate base HCl (  H+ and) Cl- CH3COOH(  H+ and) CH3COO- H2SO4 (  H+ and) HSO4- HSO4- (  H+ and) SO4-2 H2O (  H+ and) OH- NH4+ (  H+ and) NH3 NH3 (  H+ and) NH2-

What is the conjugate acid of…
NO3- C2O4-2 HPO4-2 HSO4- H2O F-

Base (gains H+ to form its) Conjugate acid NO3- C2O4-2 HPO4-2 HSO4- H2O F-

Base (gains H+ to form its) Conjugate acid NO3- (+H+  ) HNO3 C2O4-2 HPO4-2 HSO4- H2O F-

Base (gains H+ to form its) Conjugate acid NO3- (+H+  ) HNO3 C2O4-2 (+H+  ) HC2O4- HPO4-2 (+H+  ) H2PO4- HSO4- (+H+  ) H2SO4 H2O (+H+  ) H3O+ F- (+H+  ) HF

Nomenclature If the anion name then the acid name ends in… is…

Fill in the blanks HCl is _____________acid HClO4 is _____________acid

Fill in the blanks Hydrogen chloride Hydrogen perchlorate
HCl is _____________acid HClO4 is _____________acid HClO3 is _____________acid HClO2 is _____________acid HClO is _____________acid Hydrogen chlorate Hydrogen chlorite Hydrogen hypochlorite

Nomenclature If the anion name then the acid name ends in…. is…
--ide Hydro___ic acid (hypo--) --ite Hypo___ous acid --ite ___ous acid --ate ___ic acid (per--) –ate Per ___ic acid

Fill in the blanks HNO3 is _____________acid HIO4 is _____________acid
H2CO3 is _____________acid H3PO3 is _____________acid HBrO is _____________acid

Fill in the blanks _____________is hydrocyanic acid
_____________ is perbromic acid _____________ is phosphoric acid _____________ is sulfurous acid _____________ is hypoiodous acid

Show the conjugate acid/base pairs in the following reactions.
C2O4-2 + H3O+  HC2O H2O CH3COOH + NH  NH3 + CH3COO-

C2O4-2 + H3O+  HC2O H2O CH3COOH + NH  NH3 + CH3COO- ConjugateBase Conjugate Acid Base

C2O4-2 + H3O+  HC2O H2O CH3COOH + NH  NH3 + CH3COO- ConjugateBase Conjugate Acid Base Conjugate Acid Base Acid ConjugateBase

Water dissociates! H2O  H+ + OH-
This makes an equilibrium for water where: Kw=[H+][OH-]=1 x 10-14 (at 25oC)

This makes an equilibrium for water where: Kw=[H+][OH-]=1 x 10-14 (at 25oC) (endothermic or exothermic?) (Does Kw increase or decrease at higher T?)

This makes an equilibrium for water where: Kw=[H+][OH-]=1 x 10-14 (at 25oC) (endothermic or exothermic?) (Does Kw increase or decrease at higherT?)

[H+] is inversely related to [OH-]
When [H+] increases, [OH-] decreases in a water solution, and vice versa.

[H+] is inversely related to [OH-]
When [H+] decreases, [OH-] increases in a water solution, and vice versa. (Why?)

Kw=[H+][OH-]=1 x 10 -14 (at 25oC)
pH The basic (and acidic) definitions are: pH= -log [H+] [H+]= 10-pH pOH= -log [OH-] [OH-]=10 -pOH Kw=[H+][OH-]=1 x (at 25oC) pH + pOH = 14 (at 25oC)

pH practice If pH is 3.38…. What is the pOH?

pH practice If pH is 3.38…. What is the pOH? 14-pH= 10.62

pH practice If pH is 3.38…. What is the pOH? 14-pH= 10.62
What is [H+]?

What is [H+]? = 4.17 x 10-4M

What is [H+]? = 4.17 x 10-4M What is [OH-]?

What is [H+]? = 4.17 x 10-4M What is [OH-]? =2.40x10-11M and Kw/4.17x10-4M=2.40x10-11 M!

pH practice If [OH-]= 4.8 x 10-6 M…

pH practice If [OH-]= 4.8 x 10-6 M… What is pOH?

pH practice If [OH-]= 4.8 x 10-6 M…
What is pOH? -log (4.8 x 10-6 )= 5.32

What is pOH? -log (4.8 x 10-6 )= 5.32 What is pH?

What is pOH? -log (4.8 x 10-6 )= 5.32 What is pH? = 8.68

What is pOH? -log (4.8 x 10-6 )= 5.32 What is pH? = 8.68 What is [H+]?

What is pOH? -log (4.8 x 10-6 )= 5.32 What is pH? = 8.68 What is [H+]? = 2.08 x 10-9 M and Kw/ 4.8 x 10-6 = 2.08 x 10-9 M !

Homework pH problems Back of the forecast (3/30-4/10)

Please recall: Strong acids and bases dissociate completely in a water environment. Weak acids and bases do not. Strong acids= nitric, hydrochloric, sulfuric, hydrobromic, hydroiodic, perchloric Strong bases-Group 1 & 2 hydroxides—(group 2’s might not dissolve well)

Please recall: What is the concentration (M) of NaOH if .35 mole NaOH is dissolved in .120 L solution? What is the molarity of HCl if 12 g HCl is dissolved in .85 L of solution? What is [OH-] if .35 g Ba(OH)2 is dissolved in .250 L solution? What mass of H2SO4 is in 55 ml of .38 M H2SO4?

Please recall: What is the concentration (M) of NaOH if .35 mole NaOH is dissolved in .120 L solution? What is the molarity of HCl if 12 g HCl is dissolved in .85 L of solution? What is [OH-] if .35 g Ba(OH)2 is dissolved in .250 L solution? What mass of H2SO4 is in 55 ml of .38 M H2SO4? Did you notice?

Analyze these solutions
Contents pH [H+] [OH-] pOH Acidic or Basic 1 .023 mol HCl /L 2 1.5g NaOH /L 3 ?mol LiOH/50ml 8.5 4 ?mol KOH/25ml 2.5 5 ?gHClO4/150ml .02 6 ?molBa(OH)2/L .007

Right! Contents pH [H+] (M) [OH-] (M) pOH Acidic or Basic 1
.023 mol HCl /L 1.64 .023 4.3 x 10-13 12.36 Acidic 2 1.5g NaOH /L 12.57 2.7 x10-13 .0375 1.43 Basic 3 1.6 x10-7 mol LiOH/ 50ml 8.5 3.2 x10-9 3.2 x10-6 5.5 basic 4 7.9 x10 -5 mol KOH/25ml 11.5 3.2 x10-13 3.2 x10-3 2.5 5 .30 gHClO4 /150ml 1.70 .02 5.0 x10-13 12.30 acidic 6 .0035mol Ba(OH)2/L 11.85 1.4 x10-12 .007 2.15

Review question: 125 ml of a KOH solution is mixed so that the pH is 12.23 1) What is the pOH, [OH-] and [H+]? 2) What is the [KOH] ? 3) How many moles KOH was used? 4) What mass of KOH was used? (FMKOH= 56.1 g/mol)

Review question: 125 ml of a KOH solution is mixed so that the pH is 12.23 1)pOH=1.77;[OH-]=.0170M;[H+]=5.88x10-13M 2) [KOH]=[OH-]= .0170M (it’s a strong base!) 3)moles=MxV=.0170Mx.125L=.00213mol 4) massKOH =molesKOHx FMKOH = mol x 56.1 g/mol =.119 g

Strength of acids and bases.
HCl H2CO3 CH4

Strength is determined by amount of dissociation HCl -- strong acid H2CO3 -- weak acid CH4 --so weak it’s pathetic

HCl -- strong acid, it dissociates completely H2CO3 -- weak acid, it does not dissociate completely. CH4 --so weak it’s pathetic, it does not dissociate to any measurable extent. What about their conjugates?

Cl- HCO3- CH3-

Strength is determined by amount of association Cl- -- pathetic base HCO3- -- weak base CH3- --so strong a base, it associates completely with water, leaving hydroxide, a strong base.

Cl- -- pathetic base, it does not associate with water to any measurable extent. HCO3- -- weak base, it does not associate completely with water. CH3- --so strong a base, it associates completely with water, leaving hydroxide, a strong base. What about their conjugates?

The conjugate of a strong acid is a pathetic base The conjugate of a weak acid is a weak base The conjugate of a pathetic acid is a strong base

The conjugate of a strong acid is a pathetic base The conjugate of a weak acid is a weak base the stronger the acid, the weaker the base and vice versa The conjugate of a pathetic acid is a strong base

When comparing weak acids and bases…
For a weak acid, HA (aq)H+ (aq) +A- (aq) Ka=[H+][A-]/[HA] For a weak base, B- (aq) +H2O (l) HB (aq) +OH- (aq) Kb=[HB][OH-]/[B-] The position of the equilibrium is the strength of the acid or base.

Write the reaction and equilibrium constant expression for:
Ammonia associating with water Ammonium dissociating in water

Ammonia associating with water NH3(aq) + H2O (l)  NH4+ (aq) + OH- (aq) Ammonium dissociating in water NH4+ (aq)  H+ (aq) + NH3 (aq)

Ammonia associating with water NH3(aq) + H2O (l)  NH4+ (aq) + OH- (aq) Kb = [NH4+][OH-]/[NH3] Ammonium dissociating in water NH4+ (aq)  H+ (aq) + NH3 (aq) Ka = [H+ ][NH3]/[NH4+]

Kb Ka = ([NH4+][OH-][H+ ][NH3])/([NH4+][NH3])
And… Kb Ka = ([NH4+][OH-][H+ ][NH3])/([NH4+][NH3])

Kb Ka = ([NH4+][OH-][H+ ][NH3])/([NH4+][NH3])
And… Kb Ka = ([NH4+][OH-][H+ ][NH3])/([NH4+][NH3]) = [OH-][H+ ]=Kw This is true for any conjugate pair in a water solution

For example: Ka=1.8x10-5 for acetic acid Ka=6.5x10-5 for benzoic acid

For example: Ka=1.8x10-5 for acetic acid Ka=6.5x10-5 for benzoic acid
Benzoic acid is a stronger acid.

Benzoic acid is a stronger acid. .10M solutions of each would have a lower pH for benzoic acid.

Benzoic acid is a stronger acid. .10M solutions of each would have a lower pH for benzoic acid. An acetic acid solution could have a lower pH, but only at a higher concentration.

?pH What is the pH of a .25 M acetic acid solution?

?pH What is the pH of a .25 M acetic acid solution?
It’s a weak acid! [H+] is nowhere near .25M Use the ICE method. CH3COOH  CH3COO- + H+ I) .25 M 0M 0M

It’s a weak acid! [H+] is nowhere near .25M Use the ICE method. CH3COOH  CH3COO- + H+ I) .25 M 0M 0M C) -x +x +x

It’s a weak acid! [H+] is nowhere near .25M Use the ICE method. CH3COOH  CH3COO- + H+ I) .25 M 0M 0M C) -x +x +x E) .25-xM xM xM

Ka= [CH3COO-][H+] / [CH3COOH]

=(xM)(xM)/(.25-xM)=1.8 x 10-5

=(xM)(xM)/(.25-xM)=1.8 x 10-5 The two sides are equal where x=[H+]=.0021M, pH=2.68

For a weak acid solution…

x is very small. The .25 M doesn’t change very much. Try it.

x is very small. The .25 M doesn’t change very much. Try it. Ka=(x)(x)/(.25M)=1.8 x 10-5

x is very small. The .25 M doesn’t change very much. Try it. Ka=(x)(x)/(.25M)=1.8 x 10-5 The two sides are equal where x=[H+]=√[(Ka)(.25)]=.0021M, pH=2.67

What is the pH of .20 M carbonic acid? What is the pH of a 1.2 M NH3 solution? What is the pH of .80 M acetic acid which is also .35 M sodium acetate? What is the pH of a .30 M NH3 solution that is also 1.0 M ammonium chloride?

Titration

Titration An acid is neutralized by a base—at least one should be strong.

Titration An acid is neutralized by a base—at least one should be strong. The comparison of two volumes, with one known concentration, will give you the other concentration.

Titration An acid is neutralized by a base—at least one should be strong. The comparison of two volumes, with one known concentration, will give you the other concentration. That’s a titration.

One H+ neutralizes one OH-
H+ + OH-H2O Just remember– One H+ neutralizes one OH- It’s true for strong and weak acids and bases.

One H+ neutralizes one OH-
H+ + OH-H2O Just remember– One H+ neutralizes one OH- It’s true for strong and weak acids and bases. M1V1=M2V2 Moles of H+ Moles of OH-

For example: If ml of a .115M NaOH solution neutralizes ml of HCl solution, what is the concentration of the original acid?

For example: If ml of a .115M NaOH solution neutralizes ml of HCl solution, what is the concentration of the original acid? Step 1: Find moles of NaOH

For example: If ml of a .115M NaOH solution neutralizes ml of HCl solution, what is the concentration of the original acid? Step 1: Find moles of NaOH moles=MxV=.115Mx.02356L=

For example: If ml of a .115M NaOH solution neutralizes ml of HCl solution, what is the concentration of the original acid? Step 1: Find moles of NaOH moles=MxV=.115Mx.02356L= mol

For example: If ml of a .115M NaOH solution neutralizes ml of HCl solution, what is the concentration of the original acid? Step 1: Find moles of NaOH moles=MxV=.115Mx.02356L= mol Step 2: Find moles of HCl

For example: If ml of a .115M NaOH solution neutralizes ml of HCl solution, what is the concentration of the original acid? Step 1: Find moles of NaOH moles=MxV=.115Mx.02356L= mol Step 2: Find moles of HCl mol NaOHx1HCl/1NaOH=

For example: If ml of a .115M NaOH solution neutralizes ml of HCl solution, what is the concentration of the original acid? Step 1: Find moles of NaOH moles=MxV=.115Mx.02356L= mol Step 2: Find moles of HCl mol NaOHx1HCl/1NaOH=.00271mol

For example: If ml of a .115M NaOH solution neutralizes ml of HCl solution, what is the concentration of the original acid? Step 1: Find moles of NaOH moles=MxV=.115Mx.02356L= mol Step 2: Find moles of HCl mol NaOHx1HCl/1NaOH=.00271mol Step 3: Find molarity of HCl

For example: If ml of a .115M NaOH solution neutralizes ml of HCl solution, what is the concentration of the original acid? Step 1: Find moles of NaOH moles=MxV=.115Mx.02356L= mol Step 2: Find moles of HCl mol NaOHx1HCl/1NaOH=.00271mol Step 3: Find molarity of HCl M=Moles/L=.00271mol/ L

For example: If ml of a .115M NaOH solution neutralizes ml of HCl solution, what is the concentration of the original acid? Step 1: Find moles of NaOH moles=MxV=.115Mx.02356L= mol Step 2: Find moles of HCl mol NaOHx1HCl/1NaOH=.00271mol Step 3: Find molarity of HCl M=Moles/L=.00271mol/.01000L=.271 M HCl

For example: If ml of a .850M HNO3 solution neutralizes ml of a KOH solution, what is the concentration of the original base?

For example: If ml of a .850M HNO3 solution neutralizes ml of a KOH solution, what is the concentration of the original base? (.483M KOH)

For example: If 9.87 ml of a 1.32M H2SO4 solution neutralizes ml of a LiOH solution, what is the concentration of the original base?

For example: If 9.87 ml of a 1.32M H2SO4 solution neutralizes ml of a LiOH solution, what is the concentration of the original base? Did you notice?

For example: If 9.87 ml of a 1.32M H2SO4 solution neutralizes ml of a LiOH solution, what is the concentration of the original base? The ratio for the conversion from acid to base (Step 2) is : 2 LiOH 1 H2SO4 Did you notice?

What volume… What volume of a .365 M NaOH solution is needed to neutralize ml of a .211 M nitric acid solution?

What volume… What volume of a .365 M NaOH solution is needed to neutralize ml of a .211 M nitric acid solution? What happens if you add more NaOH?

Overtitration Once you’ve neutralized a weak acid or base, ignore it.
The pH is based on the excess of strong acid or base added afterwards. Subtract what was used in the neutralization, divide the excess by the total volume

Practice problem: 10.00 ml of .416 M HCl is titrated with .104M NaOH. What is the pH after: 0 ml? 10 ml? 20 ml? 30 ml? 40 ml? 50 ml?

Practice problem: 10.00 ml of .416 M HC2H3O2 is titrated with .104 M NaOH. What is the pH after: 0 ml? 10 ml? 20 ml? 30 ml? 40 ml? 50 ml?

Buffer solutions Resist changes in pH
Composed of significant amounts of a weak acid and its conjugate base Can be a partly neutralized weak acid, or mixed with the sodium salt as the base In the buffer range, the amount shifting is insignificant—ignore it. At the edges, use ICE

Salt hydrolysis Q: What happens to the pH of a solution when you add a salt?

Salt hydrolysis Q: What happens to the pH of a solution when you add a salt? A: It depends. Is it an acidic salt or a basic salt?

Salt hydrolysis Q: What happens to the pH of a solution when you add a salt? A: It depends. Is it an acidic salt or a basic salt? In the words of Glenda the Good: “I don’t know. Are you a good witch or a bad witch?”

Neutral salts If the cation is a group 1 metal, and the anion is the conjugate base of a strong acid, no effect.

Neutral salts If the cation is a group 1 metal, and the anion is the conjugate base of a strong acid, no effect. Examples: KCl, NaNO3, LiClO4

If not… The cation will show some tendency to associate itself with hydroxide—making more H+ in solution And/Or The anion will show some tendency to associate itself with H+, leaving more OH- in solution -

If not… The cation will show some tendency to associate itself with hydroxide—making more H+ in solution Examples: FeCl3, Al(NO3)3 And/Or The anion will show some tendency to associate itself with H+, leaving more OH- in solution Examples: NaF, Li2CO3

Acidic or basic solutions?
Aqueous: NaBr KNO3 ZnCl2 Al(NO3)3 CuCl2 Li3PO4 NaC2H3O2

Acidic or basic solutions?
Aqueous: NaBr KNO3 ZnCl2 Al(NO3)3 CuCl2 Li3PO4 NaC2H3O2 Acidic Neutral Basic

Solubility Equilibria
When a minimally soluble salt dissolves in water,

When a minimally soluble salt dissolves in water, Cu(OH)2 (s) Cu+2 (aq) + 2 OH - (aq) Ksp=[Cu+2] [OH-]2 No denominator--it’s a product of the ions! The solid is not part of the equilibrium. sp for “solubility product”

When a minimally soluble salt dissolves in water, In general: MxBy (s) X M+? (aq) + Y B -? (aq) Ksp=[M+?]X[B -?]Y No denominator--it’s a product of the ions! The solid is not part of the equilibrium. sp for “solubility product”

Write the reaction for dissolving a minimally soluble ionic compound
AgCl BaCO3 Fe(OH)2 Zinc (II) phosphate

Write the equilibrium expression
For: AgCl BaCO3 Fe(OH)2 Zinc (II) phosphate

Given a solubility—calculate Ksp
For: AgCl (solubility=1.34 x 10-5 mol/L) BaCO3 (solubility=1.40 x 10-2 g/L) Fe(OH)2 (solubility=5.23 x 10-5 g/100 ml) Zinc (II) phosphate

Given Ksp—calculate solubility
For: AgI Ksp=8.3 x 10-17 PbCO3 Ksp=3.3 x 10-14 Al(OH)3 Ksp=1.8 x 10-33 Zinc (II) phosphate (let’s not; and say we did)

Calculate solubility in a solution with one of the ions already.
Trying to dissolve lead chloride in a solution that already has chloride ions in it works even worse than dissolving in water Same thing for dissolving magnesium hydroxide in a solution that is already basic. This is the common ion effect.

Calculate solubility in a solution with one of the ions already.
Solve for the unknown ion. The common ion will not change much Try the previous two problems if [Cl-]= .85 M and pH= 13.63

Are you ready for a test?

Acids and Bases …and Airman 1st class HCl flew back over the front lines, confident that he had neutralized the enemy’s strongest base.

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Acids and Bases …and Airman 1st class HCl flew back over the front lines, confident that he had neutralized the enemy’s strongest base.

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