Svante Arrhenius ◦ Acids produce H+ in

aqueous solutions

◦ Bases produce OH- in aqueous solutions

Johannes Bronsted and Thomas Lowry ◦ Acids are proton donors

◦ Bases are proton acceptors

HCl + H2O Cl- + H3O+

H2O + HA ↔ H3O+ + A-

The above equation represents a competition for the proton between two bases, H2O and A-

If H2O is much stronger (greater affinity for H+) the equilibrium position will be far to the right.

𝐾𝐴 = 𝐴𝑐𝑖𝑑 𝐷𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝐾𝐴 =[𝐻3𝑂][𝐴−]

[𝐻𝐴]

In a dilute solution we can assume that the concentration of liquid water remains

essentially constant when an acid is dissolved

The strength of an acid is defined by the equilibrium position of its dissociation (ionization) reaction

Strong acid – equilibrium position lies far to the right and yields a weak base

Property Strong Acid Weak Acid

Ka Value Ka is large Ka is small

Position of dissociation equilibrium

Far to the right Far to the left

Equilibrium concentration of [H+] compared with the

original [HA]

[H+] ≈ [HA]0 [H+] ≪ [HA]0

Strength of conjugate base compared with

water

A- weaker than H2O A- stronger than H2O

Common Strong Acids

H2SO4, HCl, HNO3, and HClO4

Common Weak Acids

H3PO4, HNO2, and HOCl

Kw = dissociation constant for water

pH = - log [H+]

pOH = - log [OH-]

pK = - log K

pH + pOH = 14

antilog (n) = 10n

Typically container labels indicate the substance(s) used to make up the solution but do not necessarily describe the solution components after dissolution.

major species – solution components present in relatively large amounts

Calculate the pH of 1.00 M HF (Ka = 7.2 x 10-4)

Step 1: Identify the Major Species

HF & H2O

Step 2: Identify which species furnishes the H+ ions

HF ↔ H+ + F- Ka = 7.2 x 10-4

H2O ↔ H+ + OH- Ka = 1.0 x 10-14

Step 3: Equilibrium Expression:

Ka = [𝐻+][𝐹−]

[𝐻𝐹]

Step 4: Initial Concentrations:

[HF] = 1.00 M [H+] = 0 [F-] = 0

Step 5: Changes!

[HF]0 – x = 1.00 – x

[F-]0 + x = 0 + x

[H+]0 + x = 0 + x

Ka = (𝑥)(𝑥)

1.00 −𝑥

Ka= 𝑥2

1.00

This is because we have a very weak acid that will

NOT dissociate completely.

Step 6: Math

x2 = (7.2 x 10-4)(1.00)

x = 2.7 x 10-2

Step 7: Check with the 5% Rule

𝑥

[𝐻𝐴]0 x 100% ≤ 5%

Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 x 10-10) and 5.00 M HNO2 (Ka = 4.0 x 10-4). Also calculate the

concentration of cyanide ions in this solution at equilibrium.

1. HNO2 has the highest Ka

Ka = 𝐻+ [𝑁𝑂2

−]

[𝐻𝑁𝑂2]

HNO2 ↔ H+ + NO2-

5.00 M 0 0

- x +x +x

5.00 – x x x

2. Build an equation:

Ka = 𝑥2

5.00 Remember! Weak acid = small x

3. Solve for x:

x = 4.5 x 10-2

4. Check with 5% Rule:

0.90 %

5. Find the concentration of cyanide

HCN ↔ H+ + CN-

Ka = [𝐻+][𝐶𝑁−]

[𝐻𝐶𝑁] =

(4.5 𝑥 10−2)(𝐶𝑁−)

1.00

[CN-] = 1.4 x 10-8 M

Percent Dissociation

% Dissociation = 𝑎𝑚𝑚𝑜𝑢𝑛𝑡 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑

𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑥 100%

Strong Bases

- dissociate completely

- group 1A and 2A are strong bases

- group 2A is not very soluble

Not all bases have hydroxide ions, but they do contribute to hydrogen ion concentration

Kb - reaction of a base with water to form conjugate acid and hydroxide ion

Kb = small values = weak bases

Polyprotic acids - Can furnish more than one proton and dissociate in a stepwise manner

Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentrations of the species

of H3PO4, H2PO4-, HPO4

2-, and PO43-

H3PO4 ↔ H+ + H2PO4- Ka = 7.5 x 10-3

R H3PO4 ↔ H+ + H2PO4- Ka = 7.5 x 10-3

I 5.0 0 0

C -x +x +x

E 5.0 – x x x Ka = 𝑥2

5.00

X = 0.19 = [H+]

pH = 0.72

Calculate the pH of Sulfuric Acid

1. Strong acid completely dissociates 1.0 M H2SO4 = 1.0 M H+ & 1.0 M HSO4

-

2. R HSO4- H+ + SO4

2-

I 1.0 1.0 0

C -x +x +x

E 1.0 – x 1.0 + x x

Ka = 1.2 x 10-2 = (1.0+𝑥)(𝑥)

(1.0 −𝑥) =

(1.0)(𝑥)

(1.0)

Calculate the pH of a 1.00 x 10-2 M H2SO4 solution

1. Strong Acid…Same Story

2. HSO4- ↔ H+ + SO4

2-

0.0100 0.0100 0

-x +x +x

0.0100 – x 0.0100 + x x

Ka = 1.2 x 10-2 = (0.0100+𝑥)(𝑥)

(0.0100 −𝑥) = 1.2 x 10-2 = 0.012

HOLD UP!

That don’t make no sense at all.

Check it.

Put that x value back into the rice chart and check it.

Because it does not work, we cannot negate x and need to use the quadratic formula to solve. Be careful that as you go along, you do not allow yourself to just go into auto pilot mode. Think about the problem and think about what you are trying to do. It is the only thing that separates us from the machines... …and sociopaths….

Salts that consist of the cations of strong bases and the anions of strong acids have no effect on [H+] when dissolved in water. (i.e. KCl, NaCl, NaNO3, and KNO3)

C2H3O2- + H2O ↔ HC2H3O2 + OH-

Conjugate base Acts as Conjugate acid

of acetic acid acid acid of acetate

Kb = 𝐻𝐶2𝐻3𝑂2 [𝑂𝐻−]

[𝐶2𝐻3𝑂2] Ka =

𝐶2𝐻3𝑂2 [𝐻+]

[𝐻𝐶2𝐻3𝑂2]

Kb x Ka

= 𝐻𝐶2𝐻3𝑂2 [𝑂𝐻−] 𝐶2𝐻3𝑂2 [𝐻+]

[𝐶2𝐻3𝑂2][𝐻𝐶2𝐻3𝑂2]

= [𝐻+][𝑂𝐻−] = Kw

pKa + pKb = pKw = 14.00

Buffered solution – one that resists a change in its pH

typically a weak acid/base and a salt

A buffered solution contains 0.50 M acetic acid (Ka = 1.8 x 10-5) and 0.5 M sodium acetate. Calculate the pH of the solution.

Initial concentrations Equil. Conc.

HC2H3O2 = 0.50 M 0.50 - x

C2H3O2 = 0.50 M 0.50 + x

H+ = 0 x

R HC2H3O2 ↔ H+ + C2H3O2 -

I 0.50 0 0.50

C -x +x +x

E 0.50 – x x 0.50 x

x ≈ 1.8 x 10-5

pH = 4.74

0.10 mole NaOH(s) added to 1.0 L of above solution. Compared to 0.010 mole NaOH(s) to 1.0 L H2O.

HC2H3O2 + OH- ↔ H2O + C2H3O2 –

0.50 mol 0.010mol 0.50 mol

0.49 mol 0 mol 0.51 mol

Weak Acid Strong Base

R HC2H3O2 ↔ H+ + C2H3O2 -

I 0.49 0 0.51

C - x + x +x

E 0.49 – x x 0.51 + x

Ka = 1.8 x 10-5 = 𝐶2𝐻3𝑂2 [𝐻+]

[𝐻𝐶2𝐻3𝑂2] =

(𝑥)(0.51)

0.49 ≈ 1.7 x 10-5

pH = 4.76

Now let’s check out the other solution combination…

In this case, we do not need to do an all out RICE chart.

Since we already know the concentration of OH- , we can use Kw to calculate the concentration of H+ ions.

[H+] = 𝐾𝑤

[𝑂𝐻−] =

1.0 𝑥 10−14

1.0 𝑥 10−2 = 1.0 x 10-12

Lastly, let’s compare. Originally the solution of acetic acid with the sodium acetate buffer gave us a pH of 4.74 Once we added the NaOH, it changes the pH to 4.76. This gives a change of 0.02 Adding NaOH to water (pH of 7.00) changes the pH to 12.00—a change of 5.00

The buffered solution yields a significantly lower change

So how does this work?

[HA] ↔ [H+] + [A-]

Well the common ion effect results in a large [A-]

This causes equilibrium to shift to the left; high [HA]

Large [HA] and [A-] results in large concentration compared to [OH-]

Similar reasoning applies when p+ are added to a buffered solution of a weak acid and a salt of its conjugate base

[H+] = (Ka)([𝐻𝐴]

[𝐴])

Henderson-Hasselbach Equation

pH = pKa + log[𝐴−]

[𝐻𝐴]= 𝑝𝐾𝑎 + log

[𝑏𝑎𝑠𝑒]

[𝑎𝑐𝑖𝑑]

The pH of a buffered solution is determined by the ratio [A-]/[HA]. The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-].

Large changes in ratio [A-]/[HA] produce large changes in pH

1 mmol = 1 𝑚𝑜𝑙

1000=

10−3𝑚𝑜𝑙

Molarity = 𝑚𝑚𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑒

𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Strong Acid-Strong Base Titrations

50.0 mL of 0.200 M HNO3 with 0.100 M NaOH

2 NH3 NH4+ + NH2

-

In liquid ammonia, the reaction represented above occurs. In the reaction, NH4

+ acts as

(A) a catalyst.

(B) both an acid and a base.

(C) the conjugate acid of NH3.

(D) the reducing agent.

(E) the oxidizing agent.

At 25 oC, aqueous solutions with a pH = 8 have a hydroxide concentration, [OH-], of

(A) 1 × 10-14 M

(B) 1 × 10-8 M

(C) 1 × 10-6 M

(D) 1 M

(E) 8 M

Which of the following oxides is a gas at 25 oC and 1 atm?

(A) Rb2O

(B) N2O

(C) Na2O2

(D) SiO2

(E) La2O3

The graph below shows the titration curve that results when 100. mL of 0.0250 M acetic acid is titrated with 0.100 M NaOH. Which of the following indicators is the best choice for this titration? pH Range of Indicator Color Change (A) Methyl Orange 3.2-4.4 (B) Methyl Red 4.8-6.0 (C) Bromothymol blue 6.1-7.6 (D) Phenolphthalein 8.2-10.0 (E) Alizarin 11.0-12.4

The graph below shows the titration curve that results when 100. mL of 0.0250 M acetic acid is titrated with 0.100 M NaOH. Which part of the curve corresponds to the optimum buffer action for the acetic acid / acetate ion pair? (A) Point V (B) Point X (C) Point Z (D) Along all of section WY (E) Along all of section YZ

How can 100. mL of sodium hydroxide solution with a pH of 13.00 be converted to a sodium hydroxide solution with a pH of 12.00?

(A) By diluting the solution with distilled water to a total volume of 108 mL

(B) By diluting the solution with distilled water to a total volume of 200 mL

(C) By diluting the solution with distilled water to a total volume of 1.0 L

(D) By adding 100. mL of 0.10 M HCl.

(E) By adding 100. mL of 0.10 M NaOH.

Mixtures that would be considered buffers include which of the following? I. 0.10 M HCl and 0.10 M NaCl II. 0.10 M HF and 0.10 M NaF III 0.10 M HBr and 0.10 M NaBr (A) I only (B) II only (C) III only (D) I and II (E) II and III

Ascorbic acid, H2C6H6O6(s), is a diprotic acid with K1 = 7.9 × 10-5 and K2 = 1.6 × 10-12. In a 0.005 M aqueous solution of ascorbic acid, which of the following species is present in the lowest concentration?

(A) H2O(l)

(B) H3O+(aq)

(C) H2C6H6O6(aq)

(D) HC6H6O6-(aq)

(E) C6H6O62-(aq)

Question 9-12 refer to aqueous solutions containing 1:1 mole ratios of the following pairs of substances. Assume all concentrations are 1 M.

The solution with the lowest pH

(A) NH3 and NH4Cl (B) H3PO4 and NaH2PO4 (C) HCl and NaCl (D) NaOH and NH3 (E) NH3 and HC2H3O2 (acetic acid)

Question 9-12 refer to aqueous solutions containing 1:1 mole ratios of the following pairs of substances. Assume all concentrations are 1 M.

The most nearly neutral solution

(A) NH3 and NH4Cl (B) H3PO4 and NaH2PO4 (C) HCl and NaCl (D) NaOH and NH3 (E) NH3 and HC2H3O2 (acetic acid)

Question 9-12 refer to aqueous solutions containing 1:1 mole ratios of the following pairs of substances. Assume all concentrations are 1 M.

A buffer at a pH > 8

(A) NH3 and NH4Cl (B) H3PO4 and NaH2PO4 (C) HCl and NaCl (D) NaOH and NH3 (E) NH3 and HC2H3O2 (acetic acid)

Question 9-12 refer to aqueous solutions containing 1:1 mole ratios of the following pairs of substances. Assume all concentrations are 1 M.

A buffer at a pH < 6

(A) NH3 and NH4Cl (B) H3PO4 and NaH2PO4 (C) HCl and NaCl (D) NaOH and NH3 (E) NH3 and HC2H3O2 (acetic acid)

Is a gas in its standard state at 298 K

(A) Lithium (B) Nickel (C) Bromine (D) Uranium (E) Fluorine

Reacts with water to form a strong base

(A) Lithium (B) Nickel (C) Bromine (D) Uranium (E) Fluorine

The safest and most effective emergency procedure to treat an acid splash on skin is to do which of the following immediately?

(A) Dry the affected area with paper towels (B) Sprinkle the affected area with powdered Na2SO4(s) (C) Flush the affected area with water and then with a dilute NaOH solution (D) Flush the affected area with water and then with a dilute NaHCO3 solution (E) Flush the affected area with water and then with a dilute vinegar solution

A molecule or an ion is classified as a Lewis acid if it

(A) accepts a proton from water (B) accepts a pair of electrons to form a bond (C) donates a pair of electrons to form a bond (D) donates a proton to water (E) has resonance Lewis electron-dot structures

A sample of 61.8 g of H3BO3, a weak acid is dissolved in 1,000 g of water to make a 1.0-molal solution. Which of the following would be the best procedure to determine to molarity of the solution? (Assume no additional information is available.) (A) Titration of the solution with standard acid (B) Measurement of the pH with a pH meter (C) Determination of the boiling point of the solution (D) Measurement of the total volume of the solution (E) Measurement of the specific heat of the solution

What is the H+(aq) concentration in 0.05 M HCN (aq) ? (The Ka for HCN is 5.0 x 10¯10)

(A) 2.5 x 10¯11 (B) 2.5 x 10¯10 (C) 5.0 x 10¯10 (D) 5.0 x 10¯6 (E) 5.0 x 10¯4

A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH¯(aq) in the resulting solution? (Assume that the volumes are additive) A) 0.10 M B) 0.19 M C) 0.28 M D) 0.40 M E) 0.55 M

HC2H3O2(aq) + CN¯(aq) <===> HCN(aq) + C2H3O2¯(aq)

The reaction represented above has an equilibrium constant equal to 3.7 x 104. Which of the following can be concluded from this information?

A) CN¯(aq) is a stronger base than C2H3O2¯(aq) B) HCN(aq) is a stronger acid than HC2H3O2(aq) C) The conjugate base of CN¯(aq) is C2H3O2¯(aq) D) The equilibrium constant will increase with an increase in temperature. E) The pH of a solution containing equimolar amounts of CN¯(aq) and HC2H3O2(aq) is 7.0.

The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution is approximately

A) 50.0 mL B) 60.0 mL C) 100. mL D) 110. mL E) 120. mL

HSO4- + H2O <===> H3O

+ + SO42-

In the equilibrium represented above, the species that act as bases include which of the following?

I. HSO4-

II. H2O III. SO4

2-

(A) II only (B) III only (C) I and II (D) I and III (E) II and III

H2C2O4 + 2 H2O 2 H3O+ + C2O4

2-

Oxalic acid, H2C2O4, is a diprotic acid with K1 = 5.36 x 10-2 and K2 = 5.3 x 10-5. For reaction above, what is the equilibrium constant?

(A) 5.36 x 10-2 (B) 5.3 x 10-5 (C) 2.8 x 10-6 (D) 1.9 x 10-10 (E) 1.9 x 10-13

Which of the following acids can be oxidized to form a stronger acid?

(A) H3PO4 (B) HNO3 (C) H2CO3 (D) H3BO3 (E) H2SO3

When dilute nitric acid was added to a solution of one of the following chemicals, a gas was evolved, This gas turned a drop of limewater, Ca(OH)2, cloudy, due to the formation of a white precipitate. The chemical was

(A) household ammonia, NH3 (B) baking soda, NaHCO3 (C) table salt, NaCl (D) epsom salts, MgSO4

. 7H2O (E) bleach, 5% NaOCl

What volume of 0.150-molar HCl is required to neutralize 25.0 millilters of 0.120-molar Ba(OH)2?

(A) 20.0 mL (B) 30 0 mL (C) 40.0 mL (D) 60.0 mL (E) 80.0 mL

A 1-molar solution of which of the following salts has the highest pH ?

(A) NaNO3 (B) Na2CO3 (C) NH4Cl (D) NaHSO4 (E) Na2SO4

What is the pH of a 1.0 x 10-2-molar solution of HCN? (For HCN, Ka = 4.0 x 10-10.)

(A) 10 (B) Between 7 and 10 (C) 7 (D) Between 4 and 7 (E) 4

Correct procedures for a titration include which of the following?

I. Draining a pipet by touching the tip to the side of the container used for the titration II. Rinsing the buret with distilled water just before filling it with the liquid to be titrated III. Swirling the solution frequently during the titration

(A) I only (B) II only (C) I and III only (D) II and III only (E) I, II, and III

To determine the molar mass of a solid monoprotic acid, a student titrated a weighed sample of the acid with standardized aqueous NaOH. Which of the following could explain why the student obtained a molar mass that was too large? I. Failure to rinse all acid from the weighing paper into the titration vessel II. Addition of more water than was needed to dissolve the acid III. Addition of some base beyond the equivalence point (A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III

A solution of calcium hypochlorite, a common additive to swimming-pool water, is (A) basic because of the hydrolysis of the OCl- ion (B) basic because Ca(OH)2 is a weak and insoluble base (C) neutral if the concentration is kept below 0.1 molar (D) acidic because of the hydrolysis of the Ca2+ ions (E) acidic because the acid HOCl is formed