AP Chemistry - Chapter 17

Common Ion Effect

When the pH changes when a significant amount of conjugate base is present. Happens when a solution of acid is mixed with a solution of its conjugate base.

Common Ion Effect Example:

What is the effect on the pH of a 0.25M NH₃(aq) solution when NH₄Cl is added? NH₄⁺ is an ion that is COMMON to the equilibrium.

NH₃ + HCl <—> NH₄Cl [Cl is a spectator ion]

NH₃ + H⁺ <—> NH₄⁺ [the equation without a spectator]

NH₃ + H₂O <—> NH₄⁺ + OH^- [with water]

• equilibrium will shift to the left to reduce disturbance

• will result in the reduction of [OH^-], which will lower the pH

  • hint: NH₄⁺ is an acid

Step 1: Find the pH of 0.25M NH₃(aq) solution WITHOUT common ion [NH₄⁺]: K_b = 1.8×10^-5 (use ICE table)

pH = 11.33 for 0.25M NH₃

Step 2: Find the pH of 0.25M NH₃(aq) solution WITH common ion [0.10M NH₄Cl]

pH = 9.65 | pH drops from 11.33 due to the common ion

Common Ion Effect

When the pH changes when a significant amount of conjugate base is present. Happens when a solution of acid is mixed with a solution of its conjugate base.

Common Ion Effect Example:

What is the effect on the pH of a 0.25M NH₃(aq) solution when NH₄Cl is added? NH₄⁺ is an ion that is COMMON to the equilibrium.

NH₃ + HCl <—> NH₄Cl [Cl is a spectator ion]

NH₃ + H⁺ <—> NH₄⁺ [the equation without a spectator]

NH₃ + H₂O <—> NH₄⁺ + OH^- [with water]

• equilibrium will shift to the left to reduce disturbance

• will result in the reduction of [OH^-], which will lower the pH

  • hint: NH₄⁺ is an acid

Step 1: Find the pH of 0.25M NH₃(aq) solution WITHOUT common ion [NH₄⁺]: K_b = 1.8×10^-5 (use ICE table)

pH = 11.33 for 0.25M NH₃

Step 2: Find the pH of 0.25M NH₃(aq) solution WITH common ion [0.10M NH₄Cl]

pH = 9.65 | pH drops from 11.33 due to the common ion

Buffered Solution

a solution that resists a change in pH when either a hydroxide ions or protons (H⁺) are added is a buffered (or buffer) solution

buffered solutions contain

• weak acid and its salt

  • the salt will contain the anion of the weak acid, and the cation of a strong base

    • ex) HF → KF | HCN → NaCN

• weak base and its salt

  • the salt will contain the cation of the weak base, and the anion of a strong acid

    • ex) NH₃ → NH₄Cl | C₂H₅NH₂ → C₂H₅NH₃NO₃

Buffered Solution Example:

1) Calculate the pH of a solution that is 0.50M HAc and 0.25M NaAc (K_a = 1.8×10^-5) (use ICE table)

2) What is the pH of 1.0L of the previous solution when 0.010 mol of a solid NaOh is added? (make stoichiometry table, then ICE table)

HAc + Na⁺ <—> H⁺ + NaAc [Na is a spectator ion]

HAc <—> H⁺ + Ac^- [without spectator ion]

1) pH = 4.44

2) HAc <—> H⁺ + Ac^-

Before NaOH: 0.50 mol 0.25mol

After NaOH: 0.49 mol 0.26 mol

(adding 0.010mol OH^- will reduce the HAc and increase the Ac^- by 0.010 mole)

pH = 4.47

Graph of Titration of an Unbuffered Solution vs Buffered Solution

Henderson Hasselbalch Equation

Used to calculate the pH of a buffer solution (alternative to ICE table)

pH = pK_a + log([A^- or base]/[HA or acid])

pOH = pK_b + log([BH⁺ or acid]/[B or base])

Henderson-Hasselbalch Example

1) Calculate the pH of a solution of 0.75 M lactic acid (HC₃H₅O₃) and 0.25 M sodium lactate (K_a = 1.4×10^-4)

2) Calculate the pH of a solution of 0.25M NH₃ and 0.40M NH₄Cl (K_b = 1.8×10^-5)

1) pH = 3.38

2) pH = 9.05

Henderson Hasselbalch Equation and Buffer Capacity

the pH of a buffered solution is determined the ratio [A^-]/[HA]

pH = pK_a + log([base]/[acid])

1) What is the pH of a buffered solution determined by?

2) What is the optimal concentration of [A^-] and [HA] should be to make the “best“ buffer? (high or low) Why?

3) What do you think the optimal ratio of [A^-] and [HA] should be? Why?

1) The pH of a buffered solution is determined by the ratio [A^-]/[HA]. As long as the ratio doesn’t change, the pH won’t change.

2) The more concentrated the weak acid (HA) and its conjugate base (A^-), the more H⁺ and OH^- the solution will be able to absorb (more conc. = bigger buffer capacity)

3) 1:1 ratio - equal amounts, makes pH = pKa

What’s the relationship between pH and pK_a?

pH = pK_a + log([base]/[acid])

• pH < pK_a → [base]/[ACID]

• pH > pK_a → [BASE]/[acid]

• pH = pK_a → [base]/[acid] = 1

Two Ways to Create a Good Buffer

1) Find the RATIO of weak acid and salt (conj. base) and combine them directly. [more common method]

2) Start with one of the conjugate pairs and add a strong acid or strong base (at about ½ concentration) to create the other.

Buffer Solution Example: [Method 1]

What volumes of 0.50M HNO₂ and 0.50M NaNO₂ must be mixed to prepare 1.00L of a solution buffered at pH 3.55? (K_a = 4.5×10^-4)

pH = 3.55

pK_a = -log(4.5×10^-4) = 3.35

HNO₂ is a weak acid therefore the formula being used is:

pH = pK_a + log([base]/[acid])

3.55 = 3.35 + log(base/acid)

log(base/acid) = 0.20

base/acid = 10^0.20 = 1.58/1

[The salt (source of base) and the acid must be mixed in a 1.58/1 ratio and add up to 1.00 liters!]

1.50x + 1x = 1.00L

2.58x = 1000 mL

x = 388mL of acid

1.58x = 612 mL of base

Buffer Solution Example: [Method 2]

Describe how to prepare an optimal buffer solution using 1.0M NH₃ and 1.0M HCl (K_b = 1.8×10^-5). What will the pH of the buffer solution be?

NH₃ is a weak base therefore the formula being used is:

pOH = pK_b + log([acid]/[base])

NH₃ + H₂O <—> NH₄⁺ + OH^-

Before: 1.0L 0L 0L

After: 0.5L 0.5L 0L

[must be the same value for optimal ratio]

pOH = -log(1.8×10^-5) + log(0.5/0.5)

Equivalence Point

Precisely adding the amount of a solution of a known concentration (titrant) until the substance being tested (analyte) is completely consumed

Units of Molarity

mol/L or mmol/mL

Titration Curves - Weak Acid/Strong Base

1) Why is the equivalence point above pH 7?

2) What are [acid] and [base] at the halfway point?

3) What controls the pH BEFORE equivalence point?

4) What controls the pH HALFWAY to the equivalence point?

5) What controls the pH AT equivalence point?

6) What controls the pH AFTER the equivalence point?

1) The solution is basic. The weak acid produced a basic anion.

2) They are equal. also at halfway pH=pK_a (or pOH = pK_b)

3) The strong base is consumed therefore the pH is based on the remaining weak acid.

4) [HA] = [A^-] therefore the pH is based on pK_a

5) All HB and OH^- (from strong base) have been used up so the remaining B^- reacts with water, the only source of H⁺ left. This means the pH is based on H₂O and the conj. base.

B^- + H₂O <—> HB + OH^-

6) Excess OH^- from a strong base dominates any weak base therefore the pH is based on the excess strong base.

Titration Curves - Strong Acid/Strong Base

1) Why is the equivalence point at pH 7.

2) What controls the pH?

1) The solution is basic. The strong acid produces a neutral anion, the weak base produces a neutral cation.

2) The pH depends on the dominating remaining species (stoichiometry)

Titration Curves - Strong Acid/Weak Base

1) What is the equivalent point below pH 7?

2) What controls the pH BEFORE equivalence point?

3) What controls the pH HALFWAY to the equivalence point?

4) What controls the pH AT equivalence point?

5) What controls the pH AFTER the equivalence point?

1) The solution is acidic. The weak base produced a acidic cation.

2) The strong acid is consumed therefore the pH is based on the remaining weak base.

3) [B] = [BH⁺] so pOH = pK_b

• If you know pOH, you’ll know pH

• If you know pK_b you know pK_a

So the pH depends on pK_a

4) All B and H⁺ (from strong base) have been used up so the remaining HB⁺ reacts with water, the only significant base left. This means the pH is based on H₂O and the conj. acid.

HB⁺ + H₂O <—> B + H₃O⁺

5) Excess H₃O⁺ from a strong acid dominates any weak acid therefore the pH is based on the excess strong acid.

When do indicators change color?

Happens when the weak acids form their conjugate bases.

General Formula HIn <—> H⁺ + In^-

(red) (blue)

[endpoint is when the indicator changes color]

Is the color change of an indicator gradual or rapid? Why?

Since indicators involve an equilibrium, the color change is gradual.

Equilibrium position is controlled by pH.

When is color change noticeable for an indicator?

Color change is noticeable when ratio of [In^-]/[HI] or [HI]/[In^-] ≈ 1/10

Therefore, when picking an indicator for a desired equivalence point; pH = pK_a ± 1

How to pick an indicator for a desired equivalence point?

acid w/ base (pH goes up): pK_a(ind) = ^_desired pH + 1

base w/ acid (pH goes down): pK_a(ind) = ^_desired pH - 1

[makes noticing the color change (endpoint) happen near the equivalence point]

Calculating pH at the START of a Titration: WEAK Base with STRONG Acid

1) What is the substance that is controlling the pH?

2) 26.6mL sample of 0.308M triethylamine (C₂H₅)₃N, is titrated with 0.386M perchloric acid. What is the pH of the base BEFORE the titration? K_b = 5.2×10^-4

1) The weak base controls the pH because it is the only species present in the flask.

2) Since only triethylamine would be present since it is at the beginning of a reaction, you can use an ICE table.

The pH before the addition of any perchloric acid is 12.11

Calculating pH at the MIDPOINT (half-equivalence) of a Titration: WEAK Acid with STRONG Base

1) What is the substance in the flask that is controlling the pH?

2) When a 22.7 mL sample of a 0.366M aqueous acetic acid solution is titrated with a 0.442M aqueous barium hydroxide solution, what is the pH at the MIDPOINT in the titration? K_a = 1.8×10^-5

1) The weak acid controls the pH because it is the species remaining in the flask at this point in the reaction. The strong base is completely consumed.

2) Don’t ICE it. Use pH = pK_a

pH = pK_a = -log(1.8×10^-5) = 4.74

Calculating pH at the EQUIVALENCE point of a Titration: WEAK Acid with STRONG Base

1) What is the substance in the flask that is controlling the pH?

2) What is the pH at the EQUIVALENCE POINT in the titration of a 25.3mL sample of a 0.308 M aqueous hydrofluoric acid solution with a 0.399M aqueous barium hydroxide solution? K_a = 7.2×10^-4

1) The conjugate base of the weak acid controls the pH because it is the species remaining in the flask after the reaction.

2) Since only F^- is present, you can use an ICE table. pH = 8.32

Calculating pH at the of a Titration With a LIMITING REACTANT: WEAK Base with STRONG Acid

A 25.8 mL sample of 0.272M ammonia, NH₃, is titrated with a 41.9 mL sample of 0.266M of nitric acid. K_b= 1.8x10^-5

1) What substance in the flask is controlling the pH?

2) Determine the pH in this situation.

NH₃ + HNO₃→ NH₄⁺ + H₂O

mol NH₃: 0.0258L x 0.272 mol/L = 0.00702 mol NH₃

mol HNO₃: 0.0419L 0.266 mol/L = 0.0111 mol HNO₃

mol HNO₃ remaining: 0.0111 mol HNO₃ - 0.00702 mol NH₃ = 0.0041 mol HNO₃

1) 0.041 mol HNO₃, remaining strong acid, controls the pH.

0.0041 mol HNO₃/(0.0258 + 0.0419 L) = 0.061 M HNO₃

2) pH = -log(0.061) = 1.22

Calculating pH at the of a Titration With a LIMITING REACTANT: WEAK Acid with STRONG Base

A 40.3 mL sample of 0.399M acetic acid is titrated with a 13.1 mL sample of 0.414M of nitric acid. K_a= 1.8x10^-5

1) What substance in the flask is controlling the pH?

2) Determine the pH in this situation.

2 CH₃COOH + Ba(OH)₂ → 2 CH₃COO^- + 2H₂O

mol CH₃COOH: 0.0403L × 0.399mol/L = 0.0161 mol CH₃COOH

mol Ba(OH)₂: 0.0131L × 0.414mol/L = 0.00542 mol Ba(OH)₂

[Change Ba(OH)₂ to OH^- since there are 2 moles of OH^- per mole of compound]

mol OH^-: 0.00542 mol Ba(OH)₂ × 2 mol OH^-/1 mol Ba(OH)₂ = 0.0108 mol OH^-

mol remaining CH₃COOH: 0.0161 mol CH₃COOH - 0.0108 mol OH^- = 0.0053 mol CH₃COOH remaining

Before adding OH^-: CH₃COOH = 0.0161 mol | CH₃COO^- = 0 mol

Change (amount consumed by OH^-): CH₃COOH = 0.0161 mol - 0.0108 | CH₃COO^- = 0 mol + 0.0108

Amount remaining after adding OH^-: CH₃COOH = 0.0053 mol | CH₃COO^- = 0.0108 mol