AP Chem Unit 8 Acids and Bases

Autoionization of Water

In pure water at 25'C, [H+] = 1.0 x 10^-7 and [OH-] = 1.0 x 10^-7

As [H3O+] = [OH-] the solution is normal

K(w) = [H3O+] [OH-]

K(w) = (1.0 x 10^-7) (1.0 x 10^-7)

K(w) = 1.0 x 10^-14 (at 25'C)

Constant and if know conc of [H3O+] u can calculate the conc of [OH-] or vise versa

H+ and H3O+ equations for acids

HA + H2O —> H3O+ + A-

H+ and H3O+ r used interchangeable for aqueous ion of hydrogen

pH

A solution with [H3O+] = 10^-4 had a pH of 4

pH and pOH Equations

pH=-log[H3O+]

[H+] = 10^(-pH)

pOH=-log[OH-]

[OH-] = 10^(-pOH)

Big 6 strong acids

HBr, HI, HNO3, HCl, HClO4, H2SO4

H+ from autoionization water is not important when finding pH

2 Reasons

1) 1 x 10^-7 M is a small number

2) Le Chatelier's Principle

• Increasing [H+] shifts the reaction fro the autoionization of water towards H2O, therby reducing the conc of H+ from this process

H2O <-> H+ + OH-

<————————- shifts left

Strong Bases

• Soluble compounds containing OH-

Possible cations:

• All group 1A Cations

• Ca 2+, Sr 2+, or Ba 2+

pK(a) = 14 at 25'C

K(w) = [H3O+] [OH-]

-log K(w) = (-log[H3O+]) + (-log[OH-])

pK(w) = pH + pOH = 14 at 25'C

In pure water, pH = pOH = 7.0 at 25'C

K(w) @ Temps other than 25'C

2H2O + heat <-> H3O+ + OH-

K(w) = [H3O+] [OH-]

• Auto-dissociation of water is an endothermic process

• When temp incr., equilibrium shifts to the right so [H3O+], [OH-], and K(w) increases

• When temp decr., equilibrium shifts to the left so [H3O+], [OH-], and K(w) decreases

• In any sample of pure water at any temp pH = pOH and [H3O+] = [OH-]

• Pure water at temp above 25'C will have a pH that is lower than 7.0 bc [H3O+] and [OH-] r larger than 1 x 10^-7M

• Pure water at temp below 25'C will have a pH that is higher than 7.0 bc [H3O+] and [OH-] r smaller than 1 x 10^-7M

Acid Ionization Constant K(a) and K(b)

the equilibrium constant for a reaction in which an acid donates a proton to water

K(a) = [H+][A-]/[HA]

• The stronger the acid, the larger the K(a)

and smaller the pK(a)

• The stronger the base, the larger the K(b) and the smaller the pK(b)

pK(a) and pK(b) equations

pK(a) = -log10 (K(a))

K(a) = 10^(-pK(a))

pK(b) = -log10 (K(b))

K(b) = 10^(-pK(b))

Weak bases

• weak bases react with water to produce OH-

• Equilibrium for weak base reactions normally lies to the left

Polyprotic Acids

they can donate more than one H+ in a solution (H2SO4 and H2CO3)

they have a different Ka value for each possible dissociation

(removing one H+ at a time) (Ka1 and Ka2)

(K(a1) = 4.3 x 10^-7) >> (K(a2) = 5.6 x 10^-11)

• Always use K(a1) to calculate [H+] and pH

• Most of H+ ions come from 1st ionization

• H+ from 1st ionization drive equilibrium for other ionization to the left

Acids - Bases reactions

• Reactions btwn acids and bases r called neutralization reactions

• If strong acid or base is involved, K(eq) > 1, reaction goes to completion and a 1 way arrow is used

• If weak acid-weak base reactions, K(eq) < 1, a state of equilibrium is established and 2 way arrows r used

• Strong Acids and strong bases experience 100% ionization

• H+ and OH- ions combine to form H2O

• K(eq) for reaction is 1 x 10^14 at 25'C

— goes to completion

— K(w) = 1 x 10^14 at 25'C

• Other parts of these acid and base act as spectator ions

• pH can be determined from the excess reactant

Weak acid - Strong base reactions

• Strong bases completely dissociate

• Each OH- ripe H+ ion off a weak acid molecule

• This produces water and CB of the weak acid

Henderson-Hasselbalch equation

pH = pKa + log [A-]/[HA]

pH = pKa + log [Base]/[Acid]

pK(a) = -log10 (K(a))

[A-] = molarity of CB

[HA] = molarity of weak acid (initial molarity)

log [Base] / [Acid] = Conjugate Acid-base pair and weak base and its conjugate acid

Strong acid - weak bade reactions

• Weak bases will accept protons from a strong acid

• Weak bases may be nitrogen containing compounds such as NH3, or the CB and weak acids

Weak acid - weak base reaction

• generally don't go to completion

• acids and bases r mostly undissociated

• We write these protons transfer reactions

Acid-Base Titrations

- often used to determine the conc. of an acid or base of an acid or base in a solution

• Strong acid of known conc. (titrant) is added to a base of unknown conc. (analyte), which is mixed with an indicator

• Strong base of known conc. (titrant) is added to an acid of unknown conc. (analyte), which is mixed with an indicator

— Indicator changes color to signal the arrival at the endpoint

Endpoint

indicator, which is mixed with the analyte, changes color to signal the arrival at the endpoint

• When the correct indicator is chosen, endpoint is very close to the equivalence point

equivalence point

At the equivalence point, the number of moles of titrant added is equal to the number of moles of analyte that were originally present

3 types of acid base titrants

1) Strong acid - Strong base titration

2) Titration of weak acid by strong base

3) Titration of weak base by strong acid

1) strong acid and strong base titration curve

2) Weak acid and strong base titration curve

Change is pH less than ~ 1.5 over region where most base required to reach equivalence pt is added. Change is pH is very large in the vicinity of equivalence pt

1) Initial pH for a strong acid is close to 1. Equivalence pt pH=7

2) Initial pH for a weak acid is greater than 1. Equivalence pt in basic range. pK(a) = pH at half equivalence pt.

Titration curves for polyprotic acids

- titration curves can be used to determine the following

• Number of acidic protons

• pK(a) value for each acidic proton of a weak polyprotic acid

• Major species present at any pt along curve

Acid strength

- Strong acids completely ionize in water

HA + H2O —> H3O+ + A-

• 1 way arrow as equilibrium lies right

- Weak acids partially ionize in water

HA + H2O <—> H3O+ + A-

• 2 way arrow as equilibrium lies left or middle

Weak bases containing Nitrogen

- Lone pair on nitrogen accepts protons in solution

• Any compound of nitrogen, that has 1 lone lair on nitrogen atom, can act as a weak base by accepting protons

Strength of acids and bases

• Strong acids always have very weak CB

• Very weak acids always have very strong CB

• Acids with mid-range strengths have CB with mid-range strengths

• Stronger base always accepts most protons in an acid/base reaction

= 2 bases

• Base on reactants side of equation

• CB on products side of equation

Strong acids have very weak CB

HBr + H2O —> H3O+ + Br-

• H2O and Br- compete for protons

• H2O is stronger base, so wins most of time, and reaction goes to completion

Weak acids have strong CB

NH4- + H2O <-> H3O+ + NH3

• H2O and NH3 compete for protons

• NH3 is stronger base, so wins most of the time, and equilibrium lies to the left

Strong bases have very weak CA

HF + OH- —> H2O + F-

• CA of strong base is H2O

• Strong base could be any soluble Group 1A or Group 2A hydroxide

Binary Acid Strengths

• acid strength increases when moving down a group

Acids strength increases ——>

HF << HCl < HBr < HI

Anion Radius Increases ———>

• According to Coulomb's law, the greater the distance between the nucleus of the anion and its outermost electrons, the smaller the attractive force on the H+ ion

CB strengths of Binary acids

- CB strength decreases when moving down a group

Base strength decreases ——->

F- > Cl- > Br- > I-

Anion Radius increases ——>

• Small radius has a greater ability to attract and accept H+ ions

induction, acid strength and stability of CB

Electronegative elements tend to stabilize CB

Inductive Effect

- Highly electronegative elements tend to "pull" or "induce" electrons toward themselves

- Inductive effect of electronegative elements can....

• Reduce size of electron clouds

• Reduce electron density in certain bonds

• Reduce electron density at certain locations on a molecule

Oxoacid Strengths

2 types of Oxoacids

1) An OH group bonded to an element that is not bound to other Oxygens (HOY acids)

H-O-Y

2) An OH group bonded to an element that is not bound to other Oxygens (HYO(n) acids)

O

H-O-Y-O

Inductive effect and HOY Oxoacids

- Acid strength increases as electronegativity of Y increases

- Electron density is pulled toward electronegative Y

electron density —->

H-O-Y

- Electron density is pulled out of the O-H bond bu the inductive effect of Y

• Weakens bond btwn H and O

• H+ falls off more easily, as force of attraction btwn hydrogen and oxygen is reduced

- Acid strength increases as the electronegativity of Y increases

HOI < HOBr < HOCl

Acid strength and K(a) increases —>

HOY Oxoacids and CB stability through Induction

EN of Y increases —>

HOI < HOBr < HOCl

Acid Strength of K(a) increases —>

OI- < OBr- < OCl-

CB stability increases

• When LP r pulled away from oxygen and toward electronegative Y, r less available to bond with an H+ ion

Inductive effect and HYO(n) Oxoacids

- Acid strength increases as more oxygens r added to the central Y

Acid strength and K(a) increase O added—->

• Increasing # of electronegative oxygen atoms increases electron density around Y

• Reduces electron density btwn hydrogen and oxygen in O-H bond

HYO(n) Oxoacids and CB stability through induction

• Acid strength and K(a) increase as oxygens r added

• CB stability increases

• As more oxygen atoms r added, electron density around oxygen atom that an H+ ion could form bond with its further reduced

• Stabilizes the base, as H+ ion have less ability to form a bond

CB stability through induction

• CB of 3 of 6 strong acids - H2SO4, HNO3, HClO4 - r stabilized by induction due to presence of multiple highly electronegative oxygen atoms

Carboxylic Acids Strength

• R weak organic acids

• They take this form

O

R - C - O - H

Where "R" can be just about anything

Denoted as RCOOH or RCO2H

Carboxylic acid strength and inductive effects

• More electronegative elements that make up "R", the stronger acid

• Electron density in O-H bond is reduced by electronegative fluorine atoms

• Decreases force of attraction on H+

CB of H2SO4 Stability through Resonance

• H+ ions r attracted to - charges

• Delocalized electrons cause -1 charge to migrate btwn 3 oxygen atoms, which makes it difficult for H+ to form a bond

• Stabilizes CB of H2SO4 - making it a weak base

CB stability through resonance

CB of 3 of 6 strong acids - H2SO4, HNO3, HClO4 - r stabilized by resonance due to presence of 1, 2, or 3 double bonds btwn oxygen and central atom

Common Ion Effect

HF + H2O <-> H3O+ + F-

• HF is a very weak acid. Equilibrium lies to left

(+ KF) KF —> K+ + F-

KF is soluble and experience 100% dissociation

• F- is common ion that causes the first reaction to shift to the left (Le Chatelier) and pH of system to rise

Common ion effect is used to create buffer solutions

- Buffers resist changes in pH

• When small quantities of strong acids or strong bases r added to a buffered solution, changes in pH r small

- Buffer

• Weak acid and its salt (extra CB)

• Weak base and its salt (extra CA)

Buffer solutions

• Equilibrium lies far to left for weak acid

CH3COOH <-> H+ + CH3COO-

• Salt completely dissociate

NaCH3CO2 —> Na+ + CH3COO-

• System has lots of acid to react with strong bases and lots of CB to react with strong acids

Buffering capacity

- Ability of a buffered solution to accept protons or hydroxide ions without a significant change in pH

- Higher conc. of Ha and a- in solution result in greater buffering capacities

• More weak acid to react with added OH-

• More CB to react with added H+

• Best buffers have equal conc. of HA and A-

Choosing Conjugate Acid-Base pairs

pH = pK(a) + log[A-]/[HA]

When [A-]/[HA]=1, log [A-]/[HA]=0 and pH=pK(a)

• Helpful when choosing a CAB pair

• If buffer needs to have a certain pH, 1 would choose a weak acid with a pK(a) value that is very close to the desired pH

Buffer Solutions and pH

• pH = pK(a) when [A-] = [HA], as log(1) = 0

• Buffers made from very weak acids and their salts have high pH values

— pH = pK(a) = -log(1x10^-10) = 10.0

— basic anion from salt increases [OH-]

• Buffers made from stronger weak acids and their salts have lower pH values

— pH = pK(a) = -log(1x10^-4) = 4.0

— Stronger acid increases [H+]

pH and [A-]:[HA] Ratio

pH = pK(a) + log [A-]/[HA]

When [A-]/[HA]=10, log [A-]/[HA]=1

When [A-]/[HA]=0.1, log [A-]/[HA]=-1

• Adding small amounts of acid or base to a buffered solution causes very small changes in pH

How acid-base indicators work

• Acid-base indicators r weak acids

HIn <-> H+ + In-

<——- Highly acidic sol. shift reaction left

• H+ from acid in sol. being titrated cuz shift

Highly basic sol. shift reaction right ——->

• Reduction of H+ in sol. being titrated causes shift, which caused by basic titrant (NaOH)

• In titration, large shift in pH as equivalence pt is passed

• Large shift in pH causes indicator's equilibrium to shift and its color to change

• Color starts to change when [HIn]~~[In-]

• Color change experienced by different indicators occurs over different pH ranges

— must select an indicator that changes color at a pH that is as close as possible to pH at equivalence pt

• When [HIn] = [In-]

pH = pK(a) + log[In-]/[HIn]

pH = pK(a)

Choose indicator that has:

pK(a)(indicator) ~~ pH(equivalence pt)

pH, pK(a), [HA], [A-]

• relative conc. of HA and A- in any solution, buffered or not, can be predicted by comparing the pH and pK(a) of acid

• When pH < pK(a), [A-]<[HA]

pH = pK(a) + log[A-]/[HA]

pH = pK(a) + log 1/2

pH = pK(a) + (-.3)

• When pH > pK(a), [A-] > [HA]

pH = pK(a) + log[A-]/[HA]

pH = pK(a) + log 2/1

pH = pK(a) + (0.3)

A 1.0 mole sample of HNO3 is added to water. Final volume of solution is 1.5 L and final temp. of solution is 25'C

C) Molar conc. of nitrate ion in final solution?

[NO3-]=0.67 M, as HNO3 is strong acid the experiences 100% dissociation

A 1.0 mole sample of HNO3 is added to water. Final volume of solution is 1.5 L and final temp. of solution is 25'C

D) Is solution acidic or basic? Explain

Solution is acidic as pH is less than 7

2.2 M solution of hydrocyanic acid, HCN, at 25'C. pK(a) = 9.21 at 25'C

c) Identify strongest base in this system

CN- is strongest base in this system. H2O and CN- compete for protons and CN- wins most of time. Bc of equilibrium lies far to left

0.50 M solution of HOCl at 25'C.

pK(a) = 7.46 at 25'C

Identify strongest base in this system

OCl- is strongest base in this system. H2O and OCl- compete for protons and OCl- wins most of time. Bc equilibrium lies to left

pH of distilled water at 25'C is 7.0. When temp. is increased to 37'C the pH drops to 6.8. At both temps. the water is considered to be neutral as [H3O+]=[OH-]. Explain why pH drops when temp. increases.

K(w) is temperature dependent and autodissociation of water is endothermic process.

H2O + heat <—> H+ + OH-

As heat is added, equilibrium shifts to the right to use up that heat. Causes [H3O+] and [OH-] to increase at the same rate. pH is lower bc [H3O+] is higher

A 0.45 M solution of propanoic acid, HC3H5O2, experiences 1.58% ionization. What conc. of HCl would produce a solution with the same pH as 0.45 M solution of propanoic acid, HC3H5O2? Justify

A 7.1x10^-3 M solution of HCl would have the same pH as a 0.45 M solution of propanoic acid. Bc HCl experiences 100% ionization

pH = -log[H+] = -log(7.1x10^-3 M) = 2.15

A 0.57 M solution of propanoic acid, HOC6H5, 0.0684% of acid has ionized. What conc. of HBr would produce a solution with the same pH as 0.57 M solution of propanoic acid, HOC6H5? Justify

A 3.9x10^-4 M solution of HBr would have the same pH as a 0.57 M solution of propanoic acid. Bc HBr experiences 100% ionization

pH = -log[H3O+] = -log(3.9x10^-4 M) = 3.41

K(a) for acetic acid is 1.8x10^-5, and K(a) for hypochlorous acid is 3.5x10^-8 at 25'C. If 500.0 mL of 1.0 M acetic acid was mixed with 500.0 mL of 1.0 M hypochlorous acid, which CB would have highest conc.? Justify

Acetic acid is stronger, as it has a larger K(a) value

• Acid strength increases as K(a) increases

• Larger K(a) value, further to the right the equilibrium position (more products)

• For this reason, [CH3COO-] > [ClO-]

pH of 0.25 M C5H5N solution at 25'C is 9.25.

C5H5N + H2O <-> C5H5NH+ + OH-

Is solution acidic or basic?

basic, as pH is greater than 7

In a 0.450 M HONH2 solution, [OH-] = 5.28x10^-6 M

HONH2 + H2O <-> HONH3+ + OH-

Find [HONH3+]

[OH-] = [HONH3+] = 5.28 x 10^6 M

In a 0.450 M HONH2 solution, [OH-] = 5.28x10^-6 M

HONH2 + H2O <-> HONH3+ + OH-

f) What conc. of NaOH required to make a solution with pH of 8.723

5.28 x 10^-6 M

In a 0.450 M HONH2 solution, [OH-] = 5.28x10^-6 M

HONH2 + H2O <-> HONH3+ + OH-

g) Find percent ionization of NaOH in above sol.

100% as NaOH is a strong base

In a 0.032 M NH3 solution, [OH-] = 1.27x10^-3M

NH3 + H2O <-> NH4+ + OH-

Find [NH4+]

[OH-] = [NH4+] = 1.27 x 10^-3 M

In a 0.032 M NH3 solution, [OH-] = 1.27x10^-3M

NH3 + H2O <-> NH4+ + OH-

What conc. of KOH would be required to make a solution with pH of 11.104

1.27 x 10^-3 M

In a 0.032 M NH3 solution, [OH-] = 1.27x10^-3M

NH3 + H2O <-> NH4+ + OH-

Find percent ionization of KOH

100% as KOH is a strong base

Citric Acid H3C6H5O7 is a polyprotic acid. K(a1) = 8.4x10^-4, and K(a2) = 1.8x10^-5 at 25'C

a) which species has lowest conc. in a 1.0 M H3C6H5O7 solution: H3C6H5O7, H2C6H5O7-, or HC6H5O7 2-. Justify

[H3C6H5O7 2-] is lowest, bc K2 is smaller than K1

Citric Acid H3C6H5O7 is a polyprotic acid. K(a1) = 8.4x10^-4, and K(a2) = 1.8x10^-5 at 25'C

b) which possesses highest conc. in a 1.0 M H3C6H5O7 solution: H3C6H5O7, H2C6H5O7-, or HC6H5O7 2-. Justify

[H3C6H5O7] is largest, bc it's weak acid. Equilibrium lies far to left (mostly reactants).

Citric Acid H3C6H5O7 is a polyprotic acid. K(a1) = 8.4x10^-4, and K(a2) = 1.8x10^-5 at 25'C

c) What us equilibrium constant for reaction below?

H3C6H5O7 + H2O <-> H3O+ + H2C6H5O7 -

K(a1) = 8.4 x 10^-4

Given 10 mL of a hydrochloric acid, HCl, solution with a pH of 1.0. Required to change the pH to 2.0 be adding water. How much water do u add?

• Add 90 mL of water

HCl is a string acid, thus # moles H+ will not change. To change pH we must change conc. H+

• Reducing conc. H+ by a factor of ten will cause pH to increase by 1. If vol is increased by a factor of ten, conc. is reduced by a factor of ten.

• Thus, adding 90 mL of water will raise pH from 1.0 to 2.0

Given 100 mL of a potassium hydroxide with a pH of 12.0. Required to change the pH to 11.0 be adding water. How much water do u add?

• Add 900 mL of water

KOH is a string base, thus # moles OH- will not change. To change pH we must change conc. OH-

• Reducing conc. OH- by a factor of ten will cause pOH to increase by 1 and pH to drop by 1. If vol is increased by a factor of ten, conc. is reduced by a factor of ten.

• Thus, adding 900 mL of water will reduce pH from 12.0 to 11.0

If pH of HBr solution is same as pH of a CH3COOH solution, is [HBr] less than, equal to, or greater than [CH3COOH]? Justify

[HBr] < [CH3COOH].

• HBr is a strong acid and experiences 100% ionization

• CH3COOH is a weak acid and experiences much less than 100% ionization

• Fewer moles of HBr r required to produce the same molar concentration of H3O+ in the solution

210.0 mL of 0.10 M HI is mixed with 100.0 mL of 0.1 M NaOH and 55.0 mL 0.30 M LiOH. Would pH of final solution at 25'C be less than 7, greater than 7, or equal to 7?

pH will be greater than 7, as [OH-] > [H+]

Equal vol. of 0.25 M CH3COOH and 0.25 M KOH r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

CH3COOH + OH- —> CH3COO- + H2O

• System contains equal # moles CH3COOH and KOH. React in 1:1 mole ratio and reaction goes to completion

• No an excess reactant and aqueous species that have the highest conc. at equilibrium r the acetate ion (CB of acetic acid) and K+ (spectator ion)

• Conc. of spectator ion will be slightly higher than that if the acetate ion

• As acetate is a base, it will accept some protons from water to form acetic acid

Equal vol. of 0.15 M H2SO3 and 0.30 M KOH r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

H2SO3 + 2 OH- —> SO3 2- + 2 H2O

• Mole ratio of H2SO3 to KOH is 1:2 and H2SO3 is a polyprotic acid with 2 labile protons

• React in a 1:2 mole ratio and reaction goes to completion

• Aqueous species that have highest conc. at equilibrium r SO3 2- (CB of H2SO3) and K+ (spectator ion)

• Conc. of K+ will be 0.15 M.

• Conc. slightly less than 0.075 M, as weak baseand will accept some protons from water

Equal vol. of 0.15 M H2SO3 and 0.15 M LiOH r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

H2SO3 + OH- —> HSO3- + H2O

• System contains equal # moles H2SO3 and LiOH. React in 1:1 mole ratio and reaction goes to completion

• aqueous species that have the highest conc. at equilibrium r the HSO3- (CB of H2SO3) and Li+ (spectator ion)

Odor is detected when solutions of Ammonium fluoride and KOH r combined. What is the odor?

NH4+ + OH- —> NH3 + H2O

Odor is ammonia

A 100.0 mL sample of 0.40 M HF is mixed with 100.0 mL of 0.40 M LiOH

a) write balanced net ionic equation

b) Will pH of final solution be less than 7, equal to 7, or greater than 7. Justify

HF + OH- —> H2O + F-

pH of final solution will be greater than 7. F- ion will act as base and remove H+ ions from water to form OH- ions. Increase in [OH-] will cause pH to rise

F- + H2O —> HF + OH-

Equal vol. of 1.0 M NaCH3CO2 and 1.0 M HCl r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

CH3CO2- + H3O+ —> CH3COOH + H2O

• System contains equal # moles NaCH3CO2 and HCl. React in 1:1 mole ratio and reaction goes to completion

• aqueous species that have the highest conc. at equilibrium r Na+ (spectator ion), Cl- (spectator ion), and acetic acid

• Conc. of spectator ion will be slightly higher than that of the acetate acid

Equal vol. of 0.2 M NH3 and 0.2 M HNO3 r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

NH3 + H3O+ —> NH4+ + H2O

• System contains equal # moles ammonia and nitric acid. React in 1:1 mole ratio and reaction goes to completion

• aqueous species that have the highest conc. at equilibrium r NH4+ and NO3- (spectator ion)

• Conc. of spectator ion will be slightly higher than that of the NH4+

Equal vol. of 0.2 M NaSO3 and 0.4 M HI r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

SO3 2- + 2H3O+ —> H2SO3 + 2H2O

• System contains twice as many moles HI as it does NaSO3

• Allows each SO3 2- to accept 2 H+ and reaction goes to completion with no limiting reagent

• aqueous species that have the highest conc. at equilibrium r H2SO4 and I- (spectator ion)

Equal vol. of 0.1 M Na2CO3 and 0.1 M H2SO4 r combined. Write net ionic equation for reaction and identify aqueous species that have highest conc. at equilibrium. Justify

CO3 2- + H3O+ —> HCO3- + H2O

• System contains equal # moles of Na2CO3 and H2SO4. React in a 1:1 mole ratio and reaction goes to completion

• aqueous species that have the highest conc. at equilibrium r Na+, HCO3 -, and HSO4 -

• Conc. of Na+ will be about twice that of HCO3 - or HSO4 -

Equal vol. of 0.2 M Na2CO3 and 0.4 M HCl r combined. Write net ionic equation for reaction

CO3 2- + 2H+ —> H2CO3 —> H2O + CO2

CO3 2- + 2H+ —> H2O + CO2

Solutions Ammonium nitrate and NaCN r poured into a beaker. Write net ionic equation for reaction

NH4+ + CN- <-> NH3 + HCN

Solutions of acetic acid and sodium fluoride r poured into a beaker

CH3COOH + F- <-> CH3COO- + HF

Acid dissociation constants for HOI and HC3H5O3 at 298 K r 2x10^-11 and 1.38x10^-4 respectively. Which solution is more basic: 1.0 M NaOI or 1.0 M NaC3H5O3? Justify

HOI is weaker acid, as 2x10^-11 < 1.38x10^-4

• Its CB, OI-, is stronger

• Weaker acids have stronger CB

• Equilibrium for OI- + H2O <-> HOI + OH- lies further to right than it does in...

C3H5O3- + H2O <-> HC3H5O3 + OH-

• Thus, 1.0 M NaOI produces a higher conc. of OH-, making a more basic solution

Equal volumes of 0.2 M CH3COOH (K(a) = 1.8x10^-5) and 0.2 M C6H5NH2 (K(b) = 3.8x10^-10) r mixed at 25'C

a) Which aqueous compound will have the highest conc. when equilibrium is established in final solution? Justify

Equal # moles CH3COOH and C6H5NH2 r present in solution

• K values for both reactions r less than 1, so both equilibrium lie to the left

• [C6H5NH2] > [C6H5NH3+] and [CH3COOH] > [CH3COO-]

• Bc K(a) value for CH3COOH is larger than K(b) value for C6H5NH2, the equilibrium for following reaction lies further to the left: C6H5NH2 + H+ <-> C6H5NH3 +

• [C6H5NH2] > [CH3COOH], and thus, C6H5NH2 is aqueous compound that has highest conc.

Equal volumes of 0.2 M CH3COOH (K(a) = 1.8x10^-5) and 0.2 M C6H5NH2 (K(b) = 3.8x10^-10) r mixed at 25'C

b) Is final solution acidic or basic? Justify

Equal #s moles of CH3COOH and C6H5NH2 r present in solution

• K(a) for CH3COOH is 1.8x10^-5 and K(b) for C6H5NH2, CH3COOH is a stronger acid than C6H5NH2, is a base

• Equilibrium for reaction lies further to right: CH3COOH <-> CH3COO- + H+

• [H+] > [OH-] so final solution would be acidic

Conc. of C5H5NH+ as 1.8x10^-5 M in 0.25 M C5H5N solution at 25'C and conc. of C6H5NH+ is 1.2x10^-4 M in a 0.25 M C6H5NH2 solution at 25'C. Identify stronger base. Justify

C6H5NH2 is stronger base, as accepts a greater # of protons from water

A 45 mL sample of 0.175 M KOH is titrated with 0.200 M HI

e) what is the pH at the equivalence pt

pH is 7.0 at equivalence pt. All of the added H+ and OH- have reacted to form water. Solution only contains K+ and I- ions, neither of which affect the pH of solution

5.0 M solution of HNO3 is titrated with 0.30 M NaOH. Identify species that have highest conc. in the solution being titrated halfway to the equivalence pt

NO3-, Na+, and H3O+

28.0 mL sample of 0.500 M NaHSO4 is titrated with 0.250 M NaOH at 25'C

d) identify species that has highest conc. in this solution at half equivalence pt

Na+, HSO4-, SO4 2-, and H3O+

65 mL sample of 0.35 M NH3 is titrated with 0.25 M HCl at 25'C

e) Identify species that have highest conc. in this solution at half equivalence pt

NH3, NH4 +, Cl-, and OH-

65 mL sample of 0.35 M NH3 is titrated with 0.25 M HCl at 25'C

f) Identify species that have highest conc. in this solution at equivalence pt

NH4 +, Cl-, and H3O +

65 mL sample of 0.35 M NH3 is titrated with 0.25 M HCl at 25'C

g) Explain why the solution is acidic at the equivalence pt of the titration. Use a chemical equation

When NH3 is titrated with a strong acid the reaction produces NH4+

• At the equivalence pt, NH4 +, which has developed high conc., reacts with water to produce H3O+

• Increased conc. of H3O+ makes solution acidic

NH4+ + H2O <-> NH3 + H3O+

A beaker containing 125 mL of 0.120 M HOCl is titrated using 0.250 M NaOH K(a) for HOCl is 3.5x10^-8

b) Is solution acidic or basic at equivalence pt? Jusitfy

HOCl is a weak acid, so its CB, OCl- is a weak base.

• At equivalence pt, equal #s moles HOCl and OH- have reacted

• Only species remaining in solution that can affect pH is OCl-, which is basic

• OCl- accepts protons from water molecules, thereby increasing conc. OH- in solution according to the reaction below

OCl- + H2O —> HOCl + OH-

pH rises into basic range as [OH-] increases

A 55 mL solution of 0.15 M HF is titrated with 0.30 M NaOH

c) Is pH at equivalence pt for situation (28mL) greater than, equal to, or less than that for the situation (14 mL)?? Justify

At equivalence pt, only species that will have an effect on pH is basic F-

• F- will react with water to produce OH- according to equation

F- + H2O —> HF + OH-

# moles OH- produced will be about the same in both situations

• Conc. OH- ions in 2 situations will be different, as overall vol. of final solutions r different

• Situation (a) has a larger vol. [55 mL + 28 mL = 83 mL] than situation (b) [55 mL + 14 mL = 69 mL]

• (b) has higher pH as it's a higher conc. of OH- ions

A solution of the amino acid alanine, NC3O2H8+, was created and titrated with NaOH. The data from this experiment was used to plot following titration curve

d) Identify species that have highest conc. in this solution at first half equivalence pt

e) Identify species that have highest conc. in this solution at first equivalence pt

d) NC3O2H8+, NC3O2H7, Na+, and H3O+

e) NC3O2H7, Na+, and H3O+