Exam 3

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A nitric acid solution has a pH of 1.75. What is the molar concentration of this nitric acid solution?

0.0178

Determine the pOH pf a 0.169 M Ca(OH)₂ solution.

0.471

What volume of 0.885 M KOH solution is required to make 3.73 L of a solution with pH of 12.1?

0.531 L

Monoprotic

1 proton

What is Kw at room temperature (25°C)?

1 x 10⁻¹⁴

What is the magnitude of Kw?

1 x 10⁻¹⁴

The effective pH range of a buffer is pKa _____.

± 1

A buffer will be more effective when 0.1 (>/=/<) [base]:[acid] (>/=/<) 10. It is most effective if [base]:[acid] (>/=/<) 1.

<; <; =

Calculate the pH of a mixture formed by adding 20.0 mL of 0.015 M Ba(OH)₂ with 40.0 mL of 8.2 x 10⁻³ M KOH.

Find moles in each solution 20 mL of 0.15 M Ba(OH)₂: (0.020 L) x (0.015 moles/1 L) x (2 moles OH/1 mole) = 0.0006 moles OH 40 mL of 8.2 x 10⁻³ M KOH: (0.04 L) x (8.2 x 10⁻³ mol/1 L) x (1 mol OH/1 mol KOH) = 0.000328 moles OH Total moles: 0.0006 + 0.000398 = 0.000928 moles OH [OH] = (0.000928 moles/0.06 L) = 0.015 M pOH = -log[0.015] = 1.82 pH = 14 - 1.82 = 12.18

A 0.200 M solution of a weak acid, HX, is 9.4 percent ionized. Using this information, calculate Ka for HX. a) 1.8 × 10⁻³ b) 2.0 × 10⁻³ c) 4.4 × 10⁻² d) 8.3 × 10⁻² e) 1.9 × 10⁻²

b) 2.0 × 10⁻³ % ionization = [x]/[initial HX] x 100 x/0.2 x 100 = 9.4 x = [H+] = 0.0188 Ka = [0.0188]²/[0.200 - 0.0188] = 2.0 x 10⁻³

An amphoteric molecule paired with an acid can act as: a. An acid b. A base c. Both d. Neither

b. A base

A Bronstead-Lowry acid or base is also a: a. Arrhenius b. Lewis c. Both d. Neither

b. Lewis

Which one of the following cannot act as a Lewis base? a) CN⁻ b) NH₃ c) Cr³⁺ d) Cl⁻ e) F⁻

c) Cr³⁺ - is a cation so it is a Lewis acid that can accept electrons To act as a Lewis base, a compound must be able to donate electrons: Lewis bases usually have a negative charge

Which of the following has the strongest conjugate base? a) acetic acid, Ka = 1.8 x 10⁻⁵ b) HNO₂, Ka = 4.5 x 10⁻⁴ c) HPO₄²⁻, Ka = 4.8 x 10⁻¹³ d) Vitamin C, Ka = 8.0 x 10⁻⁵ e) HClO (Ka = 3.5 x 10⁻⁸)

c) HPO₄²⁻, Ka = 4.8 x 10⁻¹³ -If an acid has a strong conjugate base, it is a weak acid. So this is just another way of asking for the weakest acid. ***Remember that the stronger the acid, the larger the value of Ka. The weaker the acid, the smaller the value of Ka

An Arrhenius acid or base is also a: a. Bronstead-Lowry b. Lewis c. Both d. Neither

c. Both

An amphoteric molecule paired with another amphoteric molecule can act as: a. An acid b. A base c. Both d. Neither

c. Both

Which of the following aqueous solutions will form a basic solution? a) AlCl⁻ b) NaBr c) CH₃COONa d) K₂SO₄ e) All of the above

c. Na has no effect. But CH₃COO⁻ is the conjugate base of CH₃COOH. In H₂O, CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ -so it forms a basic solution. a. Cl⁻ is the anion of a strong acid so it has no effect on the pH. Al³⁺ is a Lewis acid and in H₂O it reacts: Al³⁺ + 3H₂O ⇌ Al(OH)₃ + 3H⁺ -so it forms H⁺ ions ∴ acidic solution. b. Na⁺ is the cation of a strong base and Br⁻ is the anion of a strong acid. Neither will have an effect on the pH. d. SO₄²⁻ is the conjugate base of a strong acid and K⁺ is the cation of a strong base so once again, neither will have an effect on the pH.

The hydrogen sulfate or bisulfate ion HSO₄⁻ can act as either an acid or a base in water solution. In which of the following equations does HSO₄⁻ act as an acid? a) HSO₄⁻ + H₂O ⇌ H₂SO₄ + OH⁻ b) HSO₄⁻ + H₃O⁺ ⇌ SO₃²⁻ + 2H₂O c) HSO₄⁻ + OH⁻ ⇌ H₂SO₄ + O⁻ d) HSO₄⁻ + H₂O ⇌ SO₄²⁻ + H₃O⁺

d) a) HSO₄⁻ acts as a base (accepts a proton) c) HSO₄⁻ acts as a base (accepts a proton) d) HSO₄⁻ acts as an acid (donates a proton)

A Lewis acid or base is also a: a. Arrhenius b. Bronstead-Lowry c. Both d. Neither

d. Neither

What change will be caused by the addition of a small amount of HCL to a HF/NaF buffer? H+ + F- → HF a. [H3O+] will increase significantly b. [F-] and [H3O+] will both increase c. [HF] will decrease and [F-] will increase d. [F-] will decrease and [HF] will increase e. The fluoride ions will precipitate out of solution as its acid salt

d. [F-] will decrease and [HF] will increase

How is pH related to pOH and pKw?

pH + pOH = pKw = 14

Calculate the pH of 0.023 M HCL.

pH = -log(0.023) = 1.64

Whose definition? A base is a proton acceptor

Bronsted-Lowry

What is the relationship between pH and pOH?

-pH + pOH = 14 at 25° -If you know one, you can determine the other

Bronsted-Lowry Definition of a Base

-A base is a proton acceptor -Must have a pair of non-bonding electrons

Arrhenius Definition of a Base

-A substance that increases the OH⁻ concentration when dissolved in water -Strong bases: complete dissociation -Weak bases: equilibrium/partial dissociation

Arrhenius Definition of an Acid

-A substance that increases the concentration of H⁺ (H₃O⁺) when dissolved in water -Strong acids: complete dissociation -Weak acids: equilibrium/partial dissociation

pOH

-A way of expressing the acidity/basicity of a solution -pOH = -log[OH⁻] OR [OH⁻} = 10^(-pOH) -Basic < 7 < acidic -pOH 7 = neutral

pK

-A way of expressing the strength of an acid or base through its pK -pKa = -log(Ka); 10^(-pKa) -pKb = -log(Kb); 10^(-pKb)

Percent Ionization

-A way to measure the strength of an acid by determining the percentage of acid molecules that ionize when dissolved in water Molarity of ionized acid ----------------------- x 100 initial molarity of weak acid

Why doesn't the increase in H₃O⁺ keep up with the increase in HA?

-According to Le Châtelier's principle, if we reduce the concentrations of all the (aq) components, the equilibrium should shift to the right to increase the total number of dissolved particles. The (aq) concentrations can be reduced by using a more dilute initial acid concentration. -The result will be a larger [H3O+] in the dilute solution compared to the initial acid concentration. This will result in a larger percent ionization.

Binary Acids

-Acids that have acidic hydrogens attached to a nonmetal atom -Examples: HCl and HF

What is an amphoteric substance?

-Amphoteric (or amphiprotic) substances are those that can act as an acid or a base. -Has an H⁺ and a ⁻ HSO₄⁻: it can act as an acid by donating a proton to become SO₄²⁻ or as a base by accepting a proton to become H₂SO₄ H₂O: t can act as an acid by donating a proton to become OH⁻ or as a base by accepting a proton to become H₃O⁺

Bronsted-Lowry Definition of an Acid

-An acid is a proton donor -Must have a removable proton

Whose definition? An acid is a proton donor

Bronsted-Lowry

Lewis Base

-An electron donor -Electron rich ∴ nucleophile -Must have a lone pair of electrons -Anions are better than neutral atoms -More electronegative → less willing to be a base

Lewis Acid

-An electron acceptor -Electron deficient from being bonded to electronegative atoms or not having an octet ∴ electrophile -Must have an empty orbital

What does the p in pH and pOH stand for?

-log[x]

Weak Acids

-Anything with a proton it can give up -Donate a small fraction of their hydrogen atoms -[weak acid] does NOT = [H⁺] in solution -Partial ionization (<1%) -Establishes equilibrium -[H₃O⁺] << [OH⁻] -Ka < 1

Lewis Acid-Base Reactions

-Base donates a pair of electrons to the acid → convalent bond -Forms an adduct

What are prefixes/roots/suffixes of common weak bases?

-Carbonates -ines -Ammonia

Henderson-Hasselbach Equation for Basic Buffers

-Chemical equation must be looked at like an acid reaction -React conjugate acid with water to turn it into a Ka expression pOH = pKb + log([conjugate acid]/[base]) 14 - pOH = pH

What are conjugate acids and bases?

-Conjugate acid: the molecule you get after adding 1 proton (H⁺) to a base -Conjugate base: the molecule you get after taking 1 proton (H⁺) away from an acid -The difference between acids and bases and their conjugate acids and conjugate bases is always 1 proton (H⁺)

Ionization in Polyprotic Acids

-Each H has a separate Ka value -Generally only need to use the first ionization to determine pH -[A²⁻] = Ka₂ as long as the second ionization is negligible

How do you find the pH of a strong mono hydroxyl ionic basic solution?

-Find the pOH = -log [OH⁻] -pH + pOH = 14

Lewis Acid-Base Theory

-Focuses on the transferring of electron pairs -Does not require H atoms to be classified as an acid

Weak Bases

-Has a negative sign and can accept a proton -No OH⁻ -An NH₃ bonded to 3 things -[OH⁻] << [weak base] -Kb < 1

Acidic Solutions

-Have a larger [H₃O⁺] than [OH⁻] -[H₃O⁺] > 1.00 x 10⁻⁷ -[OH⁻] < 1.00 x 10⁻⁷

Basic Solutions

-Have a smaller [H₃O⁺] than [OH⁻] -[H₃O⁺] < 1.00 x 10⁻⁷ -[OH⁻] > 1.00 x 10⁻⁷

Oxyacids

-Have acidic hydrogens bonded to an oxygen atom -H-O-Y -More electronnegative Y = stronger oxyacid

Neutral Solutions

-Have equal [H₃O⁺] and [OH⁻] -[H₃O⁺] = [OH⁻] = 1.00 x 10⁻⁷

What does the value of Ka tell us about the acid?

-How strong it is/how dissociated it is -High Ka = strong = high dissociation = more [H⁺] -Low Ka = weak = low dissociation = fewer [H⁺]

Base Ionization Constant

-Kb -Strength measured by the size of the equilibrium constant when it reacts with H₂O

Strengths of Acids and Bases

-Measured by determining the equilibrium constant of a substance's reaction with water -The farther the equilibrium position lies toward the products, the stronger the acid or base. -The position of equilibrium depends on the strength of attraction between the base form and the H⁺

Strong Bases

-OH⁻ + a group 1/2 element -LiOH, NaOH, KOH, Sr(OH)₂, Ca(OH)₂, Ba(OH)₂ -Kb > 1 -[OH-] = [strong base]

Hydronium Ion

-Produced when acids react with water because H⁺ ions (protons) that are produced by an acid cannot exist in water

Acid Ionization Constant

-Strength is measured by the size of the equilibrium constant when it reacts with H₂O -Increases with strength

Strong Acids

-Strong electrolyte -Completely ionize: donate all of their hydrogen atoms -HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄ -Ka > 1 -[H₃O⁺] = [strong acid]

Reaction Quotient (Q)

-The concentration ratio of the products to the reactants, each raised to the power of their coefficients -Used to determine which direction the reaction will proceed when it is not in equilibrium -Compares the current concentration ratios to the equilibrium constant

Ion Product of Water

-The product of the H₃O⁺ and OH⁻ concentrations is always the same number at room temperature (25°C) = 1 × 10⁻¹⁴ -Kw -A.K.A. the dissociation constant of water -[H₃O⁺] and [OH⁻] are inversely proportional

pH

-The term to express the acidity or basicity of a solution -One pH unit corresponds to a factor of 10 difference in acidity -pH = -log[H₃O⁺] OR [H₃O⁺] = 10^pH -Acidic < 7 < basic -7 = neutral

Autoionization of Water

-Water can be an acid or a base ∴ a few ions are present -All aqueous solutions contain both H₃O⁺ and OH⁻ with equal concentrations -[H₃O⁺] = [OH⁻] = 10⁻⁷ M at 25°C

5% Rule

-When the equilibrium constant is very small, the position of equilibrium favors the reactants. -For relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium. -If the approximate value of x is less than 5% of the initial concentration, the approximation is valid. (approximate x) ---------------- x 100 < 5% initial concetration

What does pH stand for?

-log[H⁺]

How do you calculate the pH of a weak acid or base?

1. Ice table 2. Ka expression 3. pH equation

How can you tell which direction the equilibrium lies in an acid-base reaction?

1. Look at the bases 2. Use a provided chart to determine which is stronger 3. Equilibrium shifts away from the stronger base

What two factors does a buffer's effectiveness depend on?

1. The relative amounts of acid and base 2. The absolute concentrations of acid and base

How can you get an acid buffer?

1. Weak acid and its conjugate base 2. Weak acid and a strong base in a lesser amount

What are the two forms that a buffer comes in?

1. Weak acid and its conjugate base 2. Weak base and its conjugate acid

How can you get a base buffer?

1. Weak base and its conjugate acid 2. Weak base and a strong acid in a lesser amount

Phenol is a weak acid with one acidic hydrogen. The pH of a 0.005 M solution of phenol is 6.09. Calculate Ka.

1.32 x 10⁻⁷

Lactic acid had 1 acidic hydrogen. A 0.10 M solution of lactic acid has a pH of 2.44. Calculate Ka.

1.36 x 10⁻⁴

The Ka of nicotinic acid is 1.4 x 10⁻⁵. What is the percent ionizaiton of nicotinic acid in a 0.10 M solution?

1.8%

A 0.150 M solution of a weak base has a pH of 11.22. Determine Kb for the base.

1.9 x 10⁻⁵

How can you find [H⁺] if you have pH?

10^(-pH)

In an ICE table, you can assume no change in your initial acid concentration if Kb/Ka is _____ smaller than the initial concentration.

10³

What mass of HI should be present in 0.200 L of solution to obtain a solution with each of the following pH's? pH= 1.20

10⁻¹²⁰ = 0.063 M 0.063 M x 0.200 L = 0.0126 mol 0.0126 mol x 127 g/mol = 1.6 g

How can you calculate pOH if you have pH?

14 - pH = pOH

How can you calculate pH if you have pOH?

14 - pOH = pH

If 15.0 mL of glacial acetic acid (pure HC₂H₃O₂) is diluted to 1.70 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL.

15 mL * 1.05 g/mL = 15.75 g = 0.2623 moles concentration = 0.2623 moles/1.7 L = 0.154 M Ka = 1.8 x 10^-5 [H+]² / 0.154 = 1.8 x 10⁻⁵ [H+] = 0.00167 M pH = 2.78

Calculate pH of 2.250 g of HClO₄ dissolved in 250 mL of solution.

2.250 g HClO₄ x (1 mole/100.51 g) = 0.0224 moles [H+] = (0.0224 moles/0.250 L) = 0.089 M pH = -log[0.089] = 1.05

What is the pH of the resulting solution when 0.500 moles of acetic acid are dissolved in water and diluted to 1.00 L? Ka = 1.8 x 10⁻⁵ at 25°C

2.52

What is the pH of a 0.100 M solution of acetic acid (Ka = 1.8 x 10⁻⁵)?

2.87

A 8.0×10⁻² M solution of a monoprotic acid has a percent dissociation of 0.60%. Determine the acid ionization constant (Ka) for the acid.

2.9 x 10⁻⁶

Calculate the pH of 0.023 acetic acid. Ka = 1.8 x 10⁻⁵

3.19

What is the pH of the resulting solution when 100 mL of 0.500 M hypochlorous acid is mixed with 100 mL water? Ka = 3.0 x 10⁻⁸ at 25°C

4.05

Lewis Definition of a Base

A base is an electron pair donor

Stoichiometry Calculation

A stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other: a. Added acid reacts with the A- to make more HA B\b. Added base reacts with the HA to make more A-

A Bronsted-Lowry base is defined as a substance that _____.

Accepts a proton

A base produces a conjugate (acid/base).

Acid

In an equilibrium reaction, the original base has an extra H⁺ after the reaction, so it will act as a/n (acid/base) in the reverse process. H-A + :B ↔ _____ + _____

Acid

What are the two sources of H₃O⁺ in an aqueous solution of a strong acid?

Acid and water

Acidic, Neutral, or Basic? [H₃O⁺] > 1.00 x 10⁻⁷

Acidic

Acidic, Neutral, or Basic? [OH⁻] < 1.00 x 10⁻⁷

Acidic

If the salt cation is a highly charged metal ion and the anion is the conjugate base of a strong acid, a/n (acidic/neutral/basic) solution will be formed.

Acidic

If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, a/n (acidic/neutral/basic) solution will be formed.

Acidic

Salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are (acidic/basic).

Acidic

Give the conjugate acid for the following Bronsted-Lowry bases: a. O²⁻ b. H₂PO₄⁻ c. SO₄²⁻ d. CH₃NH₂

Add a proton a. OH⁻ b. H₃PO₄ c. HSO₄⁻ d. CH₃NH₃⁺

A substance that is capable of acting as both an acid and as a base is called _____.

Amphoteric or amphiprotic

Lewis Definition of an Acid

An acid is an electron pair acceptor

Equilibrium Calculation

An equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]

Arrhenius and Brønsted-Lowry acid-base reactions (are not/are) Lewis acid-base reactions.

Are

Whose definition? A base is a substance that increases the OH⁻ concentration when dissolved in water

Arrhenius

Whose definition? An acid is a substance that increases the concentration of H⁺ (H₃O⁺) when dissolved in water

Arrhenius

What is the pH of a 0.030 M Ba(OH)₂ solution? a) 1.52 b) 12.78 c) 12.48 d) 1.22 e) 0.03

Ba(OH)₂ is a strong base so in H₂O, Ba(OH)₂ completely dissociates to form: Ba(OH)₂ → Ba²⁺ + 2 OH⁻ For every one mole of Ba(OH)₂, there are 2 moles of OH⁻. [OH⁻]= 2 [Ba(OH)₂] = 2(0.030 M) = 0.060 M pOH= -log[OH⁻] = -log[0.060] =1.22 pH=14-1.22=12.78

Write a chemical equation for Ba(OH)₂ (aq) showing how it is an acid or a base according to the Arrhenius definition.

Ba(OH)₂ → Ba²⁺ + 2OH⁻

An acid produces a conjugate (acid/base).

Base

In an equilibrium reaction, the original acid has a lone pair of electrons after the reaction, so it will act as a/n (acid/base) in the reverse process. H-A + :B ↔ _____ + _____

Base

What are the two sources of OH⁻ in an aqueous solution of a strong base?

Base and water

Acidic, Neutral, or Basic? [H₃O⁺] < 1.00 x 10⁻⁷

Basic

Acidic, Neutral, or Basic? [OH⁻] > 1.00 x 10⁻⁷

Basic

An anion that is the conjugate base of a weak acid is (acidic/neutral/basic).

Basic

If the salt cation is the counterion of a strong base and the anion is the conjugate base of a weak acid, a/n (acidic/neutral/basic) solution will be formed.

Basic

Salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid create (acidic/basic) solutions.

Basic

What is something that must be considered when determining the pH of a diprotic strong acid (H₂SO₄)?

Both H⁺ ions must be accounted for

What theory (definition) does the following demonstrate? H-A + :B ↔ :A⁻ + H-B⁺

Bronsted-Lowry acid-base reactions H-A = acid :B = base :A⁻ = conjugate base H-B⁺ = conjugate acid

Find the pH of a 0.250 M NaC₂H₃O₂ solution. (Ka = 1.8 x 10⁻⁵)

CH₃COO⁻ + H₂O ↔ CH₃COOH + OH⁻ 0.25 0 0 -x +x +x ~0.25 x x Kb = (1.0 x 10⁻¹⁴)/(1.8 x 10⁻⁵) = 5.55 x 10⁻¹⁰ 5.55 x 10⁻¹⁰ = [CH₃COOH][OH⁻]/[CH₃COO⁻] = x²/0.25 = 1.17 x 10⁻⁵ = [OH⁻] pOH = 4.93 pH = 14 - 4.93 = 9.07 (basic)

Cations as Weak Acids

Can be conjugate acid of a weak base, while some can be counterions of a strong base

How should you go about choosing an acid to make a buffer?

Choose one that has a pKa closest to the pH of the buffer

In a Bronsted-Lowry acid-base reaction, the original base becomes a _____ in the reverse reaction.

Conjugate acid

In a Bronsted-Lowry acid-base reaction, the original acid becomes a _____ in the reverse reaction.

Conjugate base

True or false: The stronger the H-X bond, the more acidic the compound

False The stronger the H-X bond, the weaker the acid.

True or false: pH is only 1 - 14

False pH can be negative (very acidic) or larger than 14 (very alkaline)

Caproic acid (C₅H₁₁COOH) is found in small amounts in coconut and palm oils and is used in making artificial flavors. A saturated solution of the acid contains 11g/L and has a pH of 2.94. Calculate Ka for the acid.

Convert g/L to mol/L (11g/L) x (1 mol/116 g) = 0.095 M [H⁺] = 10^-pH = 10^-2.94 = 1.148 x 10⁻³ Ka = [1.148 x 10⁻³]²/[0.095 M - (1.148 x 10⁻³)] = 1.40 x 10⁻⁵

Calculate [OH-] in a solution of 6.33x10-3 g/mL solution of KOH.

Convert g/mL to mol/L (6.33 x 10⁻³ g/mL) x (1 mol KOH/56.01 g KOH) x (1000 mL/1 L) = 0.113 M

Acidity of oxyacids (decrease/increase) down a group on the periodic table.

Decrease

As [H₃O⁺] increases, [OH⁻] must (decrease/increase).

Decrease

Increasing the initial concentration of acid results in (decreased/increased) percent ionization.

Decreased

Henderson-Hasselbach Equation for Acidic Buffers

Derived from the Ka expression that allows us to calculate the pH of a buffer solution as long as the 5% rule approximation is valid

Binary acid Electronegativity and acidity are (inversely/directly) related.

Directly

Which of the following ions will act as a weak base in water? A) OH⁻ B) Cl⁻ C) NO₃⁻ D) Ca²⁺ E) ClO⁻

E) Conjugate base of HClO. It will act as a base in H₂O. A) Even though OH⁻ is a base, it is a strong base not a weak base. B) Cl⁻ is the anion of a strong acid. It will not act as a base. C) NO₃⁻ is also the anion of a strong acid. D) Cation of a strong base. Does not act as acid or base.

Why does electronegativity weaken oxyacids?

Electronegativity means stronger hydrogen bonds

For a monoprotic strong acid, the acid concentration is (less than/equal to/more than) the hydronium concentration.

Equal to

True or false: strong acids can be written in equilibrium form

False

True or false: the stronger the acid, the stronger the conjugate base.

False

True or false: An anion that is the conjugate base of a weak acid will decrease the pH.

False A base increases the pH.

True or False: Every Lewis acid is also a Bronsted-Lowry acid

False Bronsted-Lowry acid is a more specific category of acids. All Lewis acids are not Bronsted-Lowry acids.

True or false: The chloride ion is a stronger base than the fluoride ion.

False The chloride ion does not act as a base in water because it is the anion of a strong acid

Which is less acidic? H-F or H-I

H-F

Which is more acidic? H-C or H-F

H-F

The pH of a 0.10 M solution of formic acid, HCOOH, at 25 °C is 2.38. Calculate Ka for formic acid at this temperature.

HCOOH ↔ H⁺ + HCOO⁻ 0.1 ~0 0 -x +x +x 0.1-x x x [H⁺] = 10⁻²³⁸ = 4.2 x 10⁻¹ = x Ka = ((4.2 x 10³)²)/(0.1-(4.2 x 10⁻³)) = 1.81 x 10⁻⁴

Calculate the percent ionization of formic acid (HCO₂H) in a solution that is 0.219 M in formic acid. The Ka of formic acid is 1.77 × 10⁻⁴. A) 2.87 B) 3.94 × 10⁻⁵ C)12.2 D) 0.280 E) 2.20

HCOOH ↔ H⁺ + HCOO⁻ 0.219 0 0 -x -x -x ~0.219 x x Ka = (x²)/0.219 = 1.77 x 10⁻⁴ x² = 3.88 x 10⁻⁵ x = 6.23 x 10⁻³ M % ionization = ((6.23 x 10⁻³)/0.219( x 100 = 2.87

Write a chemical equation for HCl (aq) showing how it is an acid or a base according to the Arrhenius definition.

HCl → Cl⁻ + H⁺

Which is stronger? HClO or HIO

HClO

Which is stronger? HClO₂ or HClO₃

HClO₃

Write a chemical equation for HC₆H₅O(aq) showing how it is an acid or a base according to the Arrhenius definition.

HC₆H₅O → C₆H₅O⁻ + H⁺

Find the pH of a mixture that is 0.150 M in HF (Ka = 3.5 x 10⁻⁴) and 0.100 M in HClO. (Ka = 2.9 x 10⁻⁸)

HF ↔ H⁺ + F⁻ 0.15 ~0 0 -x +x +x 0.15-x x x x²/0.15 = 3.5 x 10⁻⁴ x = 7.25 x 10⁻³ % ionization = (7.5 x 1⁻³)/(0.15) x 100 = 4.83%

Find the H₃O⁺ concentration and the pH of a 0.250 M hydrofluoric acid solution. Ka = 3.5 x 10⁻⁴

HF ↔ H⁺ + F⁻ 0.25 ~0 0 -x +x +x 0.25-x x x Ka = (3.5 x 10⁻⁴) = [H⁺][F⁻]/[HF] = ((x)(x))/0.25 = x²/0.25 = 9.35 x 10⁻³ 9.35 x 10⁻³ = [H⁺] = -log(9.35 x 10⁻³) = 2.02 ((9.35 x 10⁻³)/0.25) x 100 = 3.74% → 5%<

Designate the acid and the base and the conjugate acid and conjugate base. Also, predict whether equilibrium lies predominantly to the left or to the right. HF + H₂O ↔ F⁻ + H₃O⁺

HF: acid H₂O: base F⁻: conjugate base H₃O⁺: conjugate acid Equilibrium lies on the left

How do you know when to use an ICE table vs the Henderson-Hasselbach equation?

HH is good when the approximation 5% rule is valid: (a) initial concentrations of acid and salt are not very dilute (b) the Ka is fairly small (initial concentrations are10² to 10³ times larger)

How can you find Kw if you have [H⁺] and [OH⁻]?

Kw = [H][OH]

Designate the acid and the base and the conjugate acid and conjugate base. Also, predict whether equilibrium lies predominantly to the left or to the right. HNO₃⁺ + H₂O ↔ H₃O⁺ + NO₃⁻

HNO₃⁺: acid H₂O: base H₃O⁺: conjugate acid NO₃⁻: conjugate base Equilibrium lies on the right

Designate the acid and the base and the conjugate acid and conjugate base. Also, predict whether equilibrium lies predominantly to the left or to the right. HSO₄⁻(aq) + HCO₃⁻(aq) ↔ SO₄²⁻(aq) + H₂CO₃(aq)

HSO₄⁻: Acid HCO₃⁻: Base SO₄²⁻: Conjugate base H₂CO₃: Conjugate acid Equilibrium lies on the right

Designate the acid and the base and the conjugate acid and conjugate base. Also, predict whether equilibrium lies predominantly to the left or to the right. HSO₄⁻ + NH₃ ↔ SO₄²⁻ + NH₄⁺

HSO₄⁻: acid NH₃: base SO₄²⁻: conjugate base NH₄⁺: conjugate acid Equilibrium lies on the right

Which acid would you use to make a buffer with pH = 7.35? If you have 500.0 mL of a 0.10 M solution of the acid, what mass of the corresponding sodium salt of the conjugate base do you need to make the buffer? Chlorous acid: pKa = 1.95 Formic acid: pKa = 3.74 Nitrous acid: pKa = 3.34 Hypochlorous acid: pKa = 7.54

Hypochlorous acid: pKa = 7.54 7.35 = 7.54 + log([base]/[0.10 M]) -0.19 = log([base]/[0.10 M] 10^(-0.19) = [base]/[0.10 M] 0.646 = [base]/[0.1 M] 0.0646 = [base] 0.646 mol/L x 0.500 L = 0.0323 mols NaClO 0.0323 mols x (74.44 g/1 mol) = 2.40 g NaClO

Which is stronger? H₃BO₃ or H₂CO₃

H₂CO₃

How can you find [H⁺] if you have [OH⁻] and Kw?

Kw = [H][OH]

How can you find [OH⁻] if you have [H⁺] and Kw?

Kw = [H][OH]

Increasing the initial concentration of acid results in (decreased/increased) [H₃O⁺] at equilibrium.

Increased

Acidity of oxyacids (decreases/increases) to the right across a period on the periodic table.

Increases

As you move down the periodic table, binary acidity (decreases/increases).

Increases

Binary acid strength (decreases/increases) as you move right across a period on the periodic table.

Increases

Binary acid strength (decreases/increases) down a column of the periodic table.

Increases

According to the Arrhenius concept, an acid is a substance that _____.

Increases H⁺

Calculate the pH of a buffer solution containing 0.100 mol HC₂H₃O₂ and 0.100 mol NaC₂H₃O₂ per liter of solution (Ka = 1.8 x 10⁻⁵) after the addition of 0.015 mol NaOH.

Initially: pH = -log(1.8 x 10⁻⁵) + log([0.1]/[0.1]) = -log(1.8 x 10⁻⁵) = 4.74 With addition: pH = -log(1.8 x 10⁻⁵) + log([0.115]/[0.085]) = 4.88 HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O I 0.100 0.015 0.100 - C -0.015 -0.015 +0.015 - E 0.085 0 0.115 -

Bond strength and binary acidity are (inversely/directly) related.

Inversely

[H₃O⁺] and [OH⁻] are (directly/inversely) proportional

Inversely

If Q = K, the reaction _____.

Is at equilibrium

Why isn't HF more acidic than it is?

It has hydrogen bonding, so it is more difficult for it to actually release its hydrogen

How can you tell if you have a good buffer?

It is able to neutralize moderate amounts of added acid or base *** there is still a limit to how much can be added before the pH changes significantly

What does a buffer do?

It neutralizes the acid or base that is added to the buffered solution and applying Le Chatelier's principle to weak acid equilibrium

How do you calculate Ka or Kb?

K = [products]/[reactants]

Determine the percent ionization of a 0.130 M HCN solution.

Ka = 4.9 x 10⁻¹⁰ = x²/ 0.13-x x = 7.98 x 10⁻⁶ M = [H+] = [CN-] % ionization = 7.98 x 10⁻⁶ x 100/0.13 = 0.0061

What is the relationship between Ka of an acid and Kb of its conjugate base?

Ka x Kb = Kw = [H₃O⁺][OH⁻]

How can you get from Kb to pKa?

KaKb = Kw = (1 x 10⁻¹⁴) ∴ Kw/Kb = Ka ∴ (1 x 10⁻¹⁴)Kb = Ka -logKa = pKa OR Kb -logKb = pKb pKa + pKb = 14 ∴ 14 - pKb = pKa

How are pKa and pKb related to one another?

KaKb = Kw ∴ pKa + pKb = pKw

At 1000 K the value of Kp for the reaction 2 SO₃(g) ↔ 2 SO₂(g) + O₂(g) is 0.338. Calculate the value of Qp, and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures are PSO₃ = 0.16 atm; PSO₂ = 0.41 atm; PO₂ = 2.5 atm.

Kp = [P][P]/[P] = 0.338 Qp = (0.41)²(2.5)/(0.16)² = 16.4 16.4 > 0.338 Reaction will shift to the left

The molar concentration of hydronium ion in pure water at 25°C is _____.

Kw = 1 x 10⁻⁴ = [H⁺][OH⁻] 1 x 10⁻⁷

At normal body temperature, 37°C, Kw = 2.4 × 10⁻¹⁴. Calculate the pH if [OH−] = 1.3 × 10⁻⁹ M at this temperature.

Kw = [H⁺][OH⁻] [H⁺] = (2.4 x 10⁻¹⁴)/(1.3 x 10⁻⁹) = 1.85 x 10⁻⁵ M pH = -log(1.85 x 10⁻⁵) = 4.73 *Note that Kw is different at different temperatures. So you can't take the -log[OH-] to get the pOH and subtract it from 14 to get the pH (That only applies when Kw= 1 x 10⁻¹⁴).

The (smaller/larger) the oxidation number of the central atom, the stronger the oxyacid.

Larger

The higher the concentration of products, the (smaller/larger) the Ka.

Larger

Whose definition? A base is an electron pair donor

Lewis

Whose definition? An acid is an electron pair acceptor

Lewis

When lithium oxide (Li₂O) is dissolved in water, the solution turns basic from the reaction of the oxide ion (O²⁻) with water. Write the reaction that occurs, and identify the conjugate acid-base pairs.

Li₂O + H₂O O²⁻ + H₂O → OH⁻ + OH⁻ base + acid → conjugated acid + conjugated base

How can you estimate pH off without a calculator via the -log equation?

Look at the magnitude of the concentration 1 x 10⁻² ↑ ↑ This tells you that your pH should be 2 If your multiplier is not 1: 1.3 x 10⁻² ↑ This tells you that your pH is around two, but will be slightly smaller than exactly 2

Find [OH⁻] and pH of a 0.33 M methyl amine solution. (Kb = 4.4 x 10⁻⁴)

Methyl + H₂O ↔ Hmethyl⁺ + OH⁻ 0.33 0 0 -x +x +x 0.33-x x x x²/0.33 = 4.4 x 1⁻⁴ x² = 1.45 x 10⁻⁴ x = 1.2 x 10⁻² = [OH⁻] pOH = 1.92; pH = 12.08

The more polarized the bond, the (less/more) acidic the bond.

More

Write a chemical equation for NH₄⁺ (aq) showing how it is an acid or a base according to the Arrhenius definition.

NH₄⁺ → NH₃ + H⁺

For a strong acid or base, the contribution of the water to the total [H₃O⁺] or [OH⁻] is (negligible/significant).

Negligible

A cation that is the counterion of a strong base is (acidic/neutral/basic)

Neutral

Acidic, Neutral, or Basic? [H₃O⁺] = 1.00 x 10⁻⁷

Neutral

Acidic, Neutral, or Basic? [OH⁻] = 1.00 x 10⁻⁷

Neutral

An anion that is the conjugate base of a strong acid is (acidic/neutral/basic).

Neutral

If the salt cation is the counterion of a strong base and the anion is the conjugate base of a strong acid, a/n (acidic/neutral/basic) solution will be formed.

Neutral

Find the [OH⁻] and pH of a 0.010 M Ba(OH)₂ solution

No ice table → strong base Ba(OH)²⁺ → Ba²⁺ + 2OH⁻ [OH⁻] = 2(0.01) = 0.20 M pOH = -log(0.20) = 1.70 pH = 14 - 1.70 = 12.30

The hydride ion, H-, is a strogner base than the hydroxide ion, OH-. The product(s) of the reaction of hydride with water is/are _____.

OH- and H₂

The farther the equilibrium position lies toward the (reactants/products), the stronger the acid or base.

Products

If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the _____ of the acid and base.

Relative strengths

Give the conjugate base of the following Bronsted-Lowry acids: a. HIO₃ b. NH₄⁺ c. HCOOH d. HPO₄⁻

Remove a proton a. IO₃⁻ b. NH₃ c. HCOO⁻ d. PO₄⁻

As you move (left/right) across the periodic table, binary acidity increases.

Right

Reaction of acid + base → _____ + _____.

Salt; water

If Q > K, the reaction _____.

Shifts to the left; increase products, decrease reactants; reverse direction

If Q < K, the reaction _____.

Shifts to the right; decrease products, increase reactants; forward direction

The higher the concentration, the (smaller/higher) the pKa.

Smaller

The larger the Ka, the (smaller/larger) the pKa.

Smaller

The larger the Kb, the (smaller/larger) the pKb.

Smaller

The stronger the acid, the (smaller/larger) the pKa.

Smaller

The stronger the base, the (smaller/larger) the pKb.

Smaller

How do you find the pH of a weak acid?

Solve an equilibrium problem using an ice format for the reaction that defines the acidity of the acid

How do you calculate the pH of a buffer change when an acid or base is added?

Stoichiometry calculation followed by equilibrium calculation

(Weak/Strong) acids donate all their hydrogen atoms, or ionize completely.

Strong

A strong acid produces a (weak/strong) conjugate base.

Strong

A weak acid produces a (weak/strong) conjugate base.

Strong

The higher the percent ionization, the (weaker/stronger) the acid.

Stronger

The larger the Kb, the (weaker/stronger) the base.

Stronger

The stronger the attraction between the base form and the H⁺, the (weaker/stronger) the base and (weaker/stronger) the acid.

Stronger; weaker

The weakest acids have the (weakest/strongest) conjugate bases.

Strongest

The weakest bases have the (weakest/strongest) conjugate acids.

Strongest

What happens if a strong acid is added to a basic buffer?

The added acid is neutralized by the conjugate base component of the buffer

What happens if a strong base is added to an acidic buffer?

The added base is neutralized by the weak acid component of the buffer

Buffering Capacity

The amount of acid or base a buffer can neutralize

If a strong acid is added to water, what determines the pH of the resulting solution?

The hydronium ion (H⁺ or H₃O⁺)

Buffering Range

The pH range at which the buffer can be effective

Conjugate Pair

The reactant and its corresponding product in Bronsted-Lowry acid-base reactions

True or false: Every Bronsted-Lowry acid is also a Lewis acid

True

True or false: Le Chatelier's principle states that if a system at equilibrium is disturbed, the equilibrium will shift to minimize the disturbance.

True

True or false: The percent ionization of a weak acid in water increases as the concentration of acid decreases

True

True or false: a Lewis base is an electron-pair acceptor, and a Lewis acid is an electron-pair donor

True

True or false: weak acids and bases can be written in equilibrium form

True

True or false: every anion can potentially be a base.

True Every anion can be thought of as the conjugate base of an acid.

Diprotic

Two protons

Calculate the pH of a buffer solution that is 0.200 M HC₂H₃O₂ and 0.100 M NaC₂H₃O₂. Ka = 1.8 x 10⁻⁵

WA CB HC₂H₃O₂ ↔ H⁺ + C₂H₃O₂⁻ I 0.200M 0 0.1 C -x +x +x E 0.2 - x +x 0.1 - x 0.200 0.100 Ka =[H⁺][C₂H₃O₂⁻]/[HC₂H₃O₂] 1.8 x 10⁻⁵ = [x][0.100]/[0.200] x = 3.6 x 10⁻⁵ M= [H⁺] pH = -log(3.6 x 10⁻⁵) = 4.44 OR pH = -log(1.8 x 10⁻⁵) + log([0.1 M]/[0.2 M] = 4.44

(Weak/Strong) acids donate a fraction of their hydrogen atoms.

Weak

The less electronegative the Y atom in an oxyacid, the (weaker/stronger) the oxyacid.

Weaker

The stronger the H-X bond, the (weaker/stronger) the binary acid.

Weaker

The fewer oxygens attached to Y, the (weaker/stronger) the oxyacid. Why?

Weaker More oxygens weaken and polarize the H-O bond

The strongest acids have the (weakest/strongest) conjugate bases.

Weakest

The strongest bases have the (weakest/strongest) conjugate acids.

Weakest

Cations of small, highly charged metals are (weakly/never/strongly) acidic.

Weakly

What does percent ionization tell you?

What percent of the reactant is ionized into the products

A 0.120 M solution of a weak acid (HA) has a pH of 3.31. Calculate the acid ionization constant (Ka) for the acid.

[H+] = 10^-3.31 = 0.0004898 M HA <-----> H+ + A- [H+] = [A-] [HA] = 0.120 - 0.0004898 = 0.120 K = ( 0.0004898)²/0.120 = 1.999 x 10⁻⁶

Calculate the number of moles of HCl required to give a pH of 2.10 in a 5.0 L solution.

[H⁺] = 10 ^-ph = 10^-2.10 = 7.94 x 10⁻³ M (7.94 x 10⁻³ mol/L) = (moles/5 L) = 0.0397 moles H⁺

If a weak acid is added to water, what determines the pH of the resulting solution?

[H₃O⁺] = Ka[X⁻]/[HA]

How do you find the pH of a strong poly hydroxyl ionic basic solution?

[OH⁻] is equal to the number of OH⁻ ions in the base

Calculate the concentration of [OH-] in a solution in which a. [H₃O⁺] = 2.0 x 10⁻⁶ b. [H₃O⁺] = 100 x [OH⁻] c. [H₃O⁺] = [OH⁻]

a. b. 1 x 10⁻⁸ c.

Determine the pH of each of the following solutions. a. 9.62 × 10⁻² M HClO₄ b. a solution that is 4.7 × 10⁻² M in HClO₄ and 4.8 × 10⁻² M in HCl c. a solution that is 1.10% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

a. -log(9.62 x 10⁻²): 1.017 b. (4.7 x 10⁻²) + (4.8 x 10⁻²) = 0.095 -log(0.095) = 1.02 c. 1 liter = 1010 g HCl = 1.10% of 1010 g = 11.11 g HCl 11.11 g HCl = 0.309 mole pH = -log[0.309] = 0.516

a. Determine the [OH−] of a 0.21 M ammonia solution. b. Determine pH of a 0.21 M ammonia solution. c. Determine pOH of a 0.21 M ammonia solution.

a. 0.0019 b. 11.28 c. 2.72

Like all equilibrium constants, the value of Kw depends on temperature. At body temperature (37°C), Kw=2.4 x 10⁻¹⁴. a. What is the [H₃O⁺] in pure water at body temperature?

a. 1.5 x 10⁻⁷ b. 6.81

Determine the pH of an HF solution of each of the following concentrations. a. 0.300 M b. 4.70 × 10⁻² M c. 2.60 × 10⁻² M d. In which cases can you not make the simplifying assumption that x is small?

a. 1.99 b. 2.41 c. 2.55 d. b & c

a. Determine the [OH⁻] of a solution that is 0.140 M in F⁻. b. Determine the pH

a. 2.0 x 10⁻⁶ b. 8.3

Complete the following table. [H₃O⁺] [OH−] pH Acidic or Basic 3.5×10⁻³,_________,______,_____ _________,3.8×10⁻⁷,______,_____ 1.8×10⁻⁹,_________,______,_____ _________,_________,7.15__,_____

a. 2.9x10⁻¹², 2.46, acidic b. 2.6x10⁻⁸, 7.58, basic c. 5.6x10⁻⁶, 8.74, basic d. 7.1x10⁻⁸, 1.4x10⁻⁷, basic

Calculate [H₃O⁺] at 25 °C for each solution and determine if the solution is acidic, basic, or neutral. a. [OH⁻] = 1.5 x 10⁻² b. [OH⁻] = 1.0 x 10⁻⁷ c. [OH⁻] = 8.2 x 10⁻¹⁰

a. 6.7 x 10⁻¹³; basic b. 1 x 10⁻⁷; neutral c. 1.2 x 10⁻⁵; acidic

Determine the pH of each of the following solutions. a. 0.23 M KCHO₂ b. 0.23 M CH₃NH₃I c. 0.19 M KI

a. 8.55 b. 5.64 c. 7

An amphoteric molecule paired with a base can act as: a. An acid b. A base c. Both d. Neither

a. An acid

Classify each anion as a weak base or pH-neutral. (a) CHO₂⁻ (b) ClO₄⁻

a. Basic b. Neutral

Write the formula for the conjugate base for each of the following: a. HClO₄ b. H₂PO₄⁻ c. PH₄⁺ d. H₂S

a. ClO₄⁻ b. HPO₄²⁻ c. PH₃ d. HS⁻

Calculate the pH of a buffer solution that is 0.250 M in HCN and 0.170 M in KCN. Ka = 4.9 x 10⁻¹⁰; pKa = 9.31) Use both the equilibrium approach and Henderson-Hasselbach.

a. HCN → CN- + H+ I 0.250 0.170 0 C -x +x +x E 0.25 - x 0.17 - x x ~0.25 ~0.17 Ka = 4.9 x 10⁻¹⁰ = [x][0.17]/[0.25] x = 7.21 x 10⁻¹⁰ M pH = -log(7.21 x 10⁻¹⁰) = 9.14 b. pH = pKa + log([base]/[acid]) = 9.31 + log([0.170]/[0.250]) = 9.14

Both HC₂O₄⁻ and HS⁻ are amphoteric. a. Write an equation to show how HC₂O₄⁻ can act as a base with HS⁻ acting as an acid. b. Write an equation to show how HC₂O₄⁻ can act as a acid with HS⁻ acting as an base.

a. HC₂O₄⁻ + HS⁻ → H₂C₂O₄ + S²⁻ b. HC₂O₄⁻ + HS⁻ → C₂O₄²⁻ + H₂S

Write the formula for the conjugate acid of each of the following: a. HSO₃⁻ b. F⁻ c. PO₄³⁻ d. CO

a. H₂SO₃ b. HF c. HPO₄²⁻ d. HCO⁺

If a strong acid is added to water, identify what happens to a. the hydronium ion concentration b. the hydroxide ion concentration c. the pH d. the pOH

a. Increases b. Decreases c. Decreases d. Increases

Write the formula for the conjugate acid of each of the following bases. a. NH₃ b. ClO₄⁻ c. HPO₄²⁻ d. CO₃²⁻

a. NH₄⁺ b. HClO₄ c. H₂PO₄⁻ d. HCO₃⁻

a. Write a balanced net ionic equation for the reaction between NH₄⁺ and NaOH. b. What are the conjugate acid/base pairs in this reaction? c. Predict whether equilibrium will lie to the right or to the left.

a. NaOH(aq) + NH₄⁺(aq) ↔ Na⁺(aq) + H₂O(l) + NH₃(g) Break up the molecules into ions: Na⁺(aq) + OH⁻(aq) + NH₄⁺(aq) ↔ Na⁺ (aq) + H₂O(l) + NH₃(g) Cross out spectator ions: OH⁻(aq) + NH₄⁺(aq) ↔ H₂O(l) + NH₃(g) b. OH⁻(aq) + NH₄⁺aq) ↔ H₂O(l) + NH₃(g) base + acid ↔ conj acid + conj base c. OH⁻(aq) + NH₄⁺(aq) ↔ H₂O(l) + NH₃(g) base + acid ↔ conj acid + conj base -Comparing base strength, equilibrium lies away from strong base. OH⁻ is the strong base so equilibrium lies to the right (products' side).

Classify each cation as weak acid or pH-neutral: (a) Li⁺ (b) CH³NH³ (c) Fe³⁺

a. Neutral b. Acidic c. Acidic

Determine if each of the following combinations result in a buffer or not: a. 0.1 M HCl/0/1 M NaCl b. 0.1 M HCl/0.1 M NaOH c. 0.1 M HCN/0.03 M NaOH d. 0.1 M HCN/0.03 M HCl e. 0.1 M HCN/0.1 M NaCN

a. SA/CB = not a buffer b. SA/SB = neutralization = not a buffer c. WA/Sb = buffer d. A/A = no reaction = not a buffer e. WB/CA = buffer

For each of the following, classify as acid or base, strong or weak, or amphoteric, and then write a balanced equation for its ionization in water: a. F⁻ b. HSO₃⁻ c. HCl d. HCO₃⁻ e. CH₃COOH

a. Weak base b. Amphoteric c. Strong acid d. Amphoteric e. Weak acid

What is the [H₃O⁺] of a solution with a. pH = 8.37 b. pH = 1.55

a. [H₃O⁺] = 10^(-8.37) = 4.27 x 10⁻⁹ b. [H₃O⁺] = 10^(-1.55) = 2.82 x 10⁻²

15. a) Write or complete the acid-base reaction that occurs for each of the following and label the conjugate acids and bases. i. Methylamine (CH₃NH₂), an organic base, reacts with water. CH₃NH₂ (aq) + H₂O (l) ⇌ CH₃NH₃⁺ (aq) + OH⁻ (aq) ii. NH₄⁺ (aq) + CN⁻ (aq) ⇌ NH₄⁺ (aq) + CN⁻ (aq) ⇌ HCN (aq) + NH₃ (aq) iii. The hydrogen oxalate ion (HC₂O₄⁻) acts as an acid in water. HC₂O₄⁻ (aq) + H₂O (aq) ⇌ C₂O₄²⁻ + H₃O⁺ (aq) b) for the first two reactions above, predict in which direction the equilibrium lies.

a. i. Base + acid ↔ conjugate acid + conjugate base ii. Acid + base ↔ conjugate acid + conjugate base iii. Acid + base ↔ conjugate base + conjugate acid b. i. Lie on the left ii. Lie on the right

Calculate the pH of each solution and indicate whether the solution is acidic or basic. (a) [H₃O⁺] = 9.5 x 10⁻⁹ (b) A sample of lemon juice has [H₃O⁺] = 3.8 x 10⁻⁴ (c) A window-cleaning solution has [OH-] = 1.9 x 10⁻⁶

a. pH = -log(9.5 x 10⁻⁹) = 8.02 b. pH = -log(3.8 x 10⁻⁴) = 3.4 c. pH = -log(1.9 x 10⁻⁶) = 5.7 14 - 5.7 = 8.3

A buffer contains 0.13 mol of CH₃COONa and 0.10 mol of CH₃COOH in 1.0 L of solution. a) Determine the pH of the buffer. b) Write the complete ionic equation for the reaction that occurs when a few drops of KOH are added to the buffer. c) What is the pH of the buffer after the addition of 0.02 mol of KOH? d) Write the complete net ionic equation for the reaction that occurs when a few drops of HNO₃ are added to the buffer. e) What is the pH of the buffer if 0.02 mol of HNO₃ are added to the buffer?

a. pH = pKa + log[base]/[acid] pH = 4.74 + log[0.13]/[0.10] pH = 4.85 b. If a base is added to the buffer, it will react with the acid of the buffer: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O c. CH₃COOH + OH⁻ → CH₃COO⁻ before: 0.10 0 0.13 add: -0.02 +0.02 after: 0.08 0 0.15 pH = 4.74 + log[0.15]/[0.08] = 5.01 d. If an acid is added to a buffer, it will react with the base of the buffer: CH₃COO⁻ + H⁺ → CH₃COOH e. CH₃COO⁻ + HNO₃ → CH₃COOH B 0.13 0 0.10 A -0.02 +0.02 A 0.11 0 0.12 pH 4.74 + log[0.11]/[0.12] = 4.70

If the concentration of a weak acid equals the concentration of a conjugate base, how does the Henderson-Hasselbach equation change?

pH = pKa

How can you use the Henderson-Hasselbach equation to estimate the lowest pH in a buffer range?

pH = pKa + log(0.10) pH = pKa - 1

How can you use the Henderson-Hasselbach equation to estimate the highest pH in a buffer range?

pH = pKa + log(10) pH = pKa + 1

What happens to the pH of a buffer change when an acid is added? Why?

pH decreases because the conjugate base decreases and the conjugate acid increases

What happens to the pH of a buffer change when a base is added? Why?

pH increases because the conjugate base increases and the conjugate acid decreases

How do you calculate Ka from pH?

pH → [H⁺] → ice → [other] → Ka expression


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