Acids, Bases, and pH: NMAT Chemistry Review

Why Acids and Bases Matter for the NMAT

Acids and bases appear frequently in NMAT Chemistry because they connect several key ideas: equilibrium, logarithms, solution chemistry, buffers, and basic stoichiometry. Many NMAT questions are designed to test whether you can move smoothly between concepts such as pH, dissociation, conjugate pairs, and neutralization reactions without getting lost in definitions. If you master the “core toolkit” (definitions, strong vs. weak behavior, equilibrium expressions, and buffer logic), you can solve most acid–base questions efficiently.

This review covers the major theories of acids and bases, how to interpret pH and pOH, how strong and weak acids behave differently, and how buffers resist pH changes. It also emphasizes NMAT-style thinking: recognizing what is being asked, selecting the simplest method, and avoiding common traps.

Core Definitions: What Is an Acid or a Base?

Acid–base chemistry has several definitions. On the NMAT, you should be comfortable with all three major frameworks and know when each is useful.

Arrhenius Concept

In the Arrhenius model, an acid increases the concentration of hydrogen ions (H⁺) in water, while a base increases hydroxide ions (OH⁻) in water. For example, hydrochloric acid (HCl) in water produces H⁺ (more precisely H₃O⁺) and Cl⁻. Sodium hydroxide (NaOH) dissociates to give Na⁺ and OH⁻.

This definition is straightforward but limited because it focuses on aqueous solutions and does not explain bases like ammonia (NH₃) well unless you describe its reaction with water.

Brønsted–Lowry Concept

The Brønsted–Lowry definition is broader: an acid is a proton donor, and a base is a proton acceptor. This model explains many reactions in water and beyond. For example, NH₃ is a base because it accepts a proton from water:

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

In this reaction, water acts as an acid (it donates a proton), and ammonia acts as a base (it accepts a proton). This highlights an NMAT favorite concept: water is amphoteric (it can act as an acid or a base depending on the reaction partner).

Lewis Concept

In the Lewis definition, an acid is an electron pair acceptor and a base is an electron pair donor. This is especially useful for reactions that do not involve protons directly. For example, BF₃ is a Lewis acid because boron has an incomplete octet and can accept an electron pair. NH₃ is a Lewis base because nitrogen has a lone pair it can donate.

On the NMAT, Lewis acid–base questions sometimes appear in coordination chemistry or in conceptual questions where the “acid” is not a proton donor. If you see something like AlCl₃, BF₃, or metal cations interacting with lone-pair donors, think Lewis.

Conjugate Acid–Base Pairs

A conjugate acid–base pair differs by one proton. When an acid donates a proton, it becomes its conjugate base. When a base accepts a proton, it becomes its conjugate acid.

• HCl (acid) → Cl⁻ (conjugate base)
• NH₃ (base) → NH₄⁺ (conjugate acid)
• H₂CO₃ (acid) → HCO₃⁻ (conjugate base)

A key relationship: the stronger the acid, the weaker its conjugate base. Strong acids have conjugate bases that are very weak (they do not readily accept protons). Weak acids have conjugate bases with more noticeable basicity.

Water Autoionization and the Meaning of Kw

Even pure water contains a small amount of ions due to autoionization:

2H₂O ⇌ H₃O⁺ + OH⁻

The equilibrium constant for this process is:

K_w = [H₃O⁺][OH⁻]

At 25°C, K_w = 1.0 × 10⁻¹⁴. In neutral water at 25°C, [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ M. This leads directly to pH 7 as “neutral” at 25°C.

NMAT caution: neutrality depends on temperature because K_w changes with temperature. If the exam includes temperature context, “neutral pH” might not be exactly 7.

pH, pOH, and Logarithms You Must Know

pH is defined as:

pH = −log[H⁺] (or −log[H₃O⁺])

pOH is:

pOH = −log[OH⁻]

At 25°C:

• pH + pOH = 14
• pK_w = 14

Because pH is logarithmic, a change of 1 pH unit corresponds to a tenfold change in [H⁺]. A solution of pH 3 has 10 times more hydrogen ion concentration than pH 4, and 100 times more than pH 5.

You should be able to convert quickly:

• If pH = 2, then [H⁺] = 1 × 10⁻² M
• If [H⁺] = 3.2 × 10⁻⁵, pH ≈ 4.49 (because −log(3.2) ≈ −0.51, and −log(10⁻⁵) = 5)

On the NMAT, approximations are often enough. If you know common logs (log 2 ≈ 0.30, log 3 ≈ 0.48, log 5 ≈ 0.70), you can estimate pH without a calculator.

Strong Acids and Strong Bases

Strong acids dissociate essentially completely in water. That means the [H⁺] is approximately equal to the initial acid concentration (for monoprotic strong acids). Common strong acids include:

• HCl
• HBr
• HI
• HNO₃
• HClO₄
• H₂SO₄ (strong in its first dissociation)

Strong bases also dissociate completely. These include alkali metal hydroxides (NaOH, KOH) and some alkaline earth hydroxides (Ca(OH)₂, Ba(OH)₂), though solubility can matter for the latter.

NMAT trap: if the base produces more than one OH⁻ per formula unit, you must multiply. For example, 0.10 M Ca(OH)₂ can produce up to 0.20 M OH⁻ (if fully dissolved).

Weak Acids and Weak Bases

Weak acids only partially dissociate. Their behavior is described by an equilibrium constant K_a:

HA + H₂O ⇌ H₃O⁺ + A⁻

K_a = ([H₃O⁺][A⁻])/[HA]

Weak bases are described by K_b:

B + H₂O ⇌ BH⁺ + OH⁻

K_b = ([BH⁺][OH⁻])/[B]

Small K_a or K_b means weak dissociation. pK values are often used:

• pK_a = −log K_a
• pK_b = −log K_b

Lower pK_a means stronger acid (because K_a is larger). Higher pK_a means weaker acid.

The Ka–Kb Relationship and Conjugates

For a conjugate acid–base pair, K_a and K_b are related by:

K_a × K_b = K_w

Equivalently:

pK_a + pK_b = pK_w = 14 (at 25°C)

This is a high-yield NMAT relationship. If you know the pK_a of an acid, you can quickly estimate the pK_b of its conjugate base.

Calculating pH of Strong Acid and Strong Base Solutions

For strong monoprotic acids (like HCl), [H⁺] ≈ C_acid. Example: 0.001 M HCl has [H⁺] = 1 × 10⁻³, so pH = 3.

For strong bases, find [OH⁻] first, then convert to pOH and pH. Example: 0.01 M NaOH gives [OH⁻] = 1 × 10⁻², so pOH = 2 and pH = 12.

If you have a strong base that contributes multiple hydroxides, adjust accordingly. Example: 0.05 M Ba(OH)₂ produces 0.10 M OH⁻, so pOH = 1 and pH = 13.

Calculating pH of Weak Acid Solutions

Weak acid pH problems often use an ICE table (Initial, Change, Equilibrium). Suppose a weak acid HA has initial concentration C and dissociates by x:

• [HA] = C − x
• [H⁺] = x
• [A⁻] = x

Then:

K_a = (x·x)/(C − x) = x²/(C − x)

When the acid is weak, x is often small compared to C, so C − x ≈ C, giving:

x ≈ √(K_a·C)

Then pH ≈ −log(x). The approximation is usually valid if x is less than about 5% of C. NMAT questions often accept this approach unless the numbers are chosen to break the approximation.

Calculating pH of Weak Base Solutions

Weak bases follow an analogous method using K_b. If base B has concentration C and produces x of OH⁻:

K_b = x²/(C − x) ≈ x²/C

So:

x ≈ √(K_b·C)

This x is [OH⁻]. Then compute pOH = −log[OH⁻] and pH = 14 − pOH (at 25°C).

Polyprotic Acids and Stepwise Dissociation

Polyprotic acids can donate more than one proton, but they dissociate stepwise. For example, H₂CO₃ dissociates in two steps with K_a1 and K_a2, and typically K_a1 > K_a2. The first proton is easier to remove than the second because the species becomes negatively charged after the first dissociation, making it less favorable to remove another proton.

For sulfuric acid (H₂SO₄), the first dissociation is strong, while the second is weaker. NMAT questions may simplify by treating H₂SO₄ as fully dissociated for the first proton and partially for the second, or they may explicitly tell you what assumption to use.

Neutralization Reactions and Titration Basics

A neutralization reaction occurs when an acid reacts with a base to form water and a salt. The simplest example is:

HCl + NaOH → NaCl + H₂O

In titrations, one solution of known concentration is added to another until the reaction reaches the equivalence point, where moles of acid and base are stoichiometrically equal (based on the balanced equation).

Key NMAT idea: equivalence point is not always pH 7. It depends on whether the acid and base are strong or weak.

• Strong acid + strong base: equivalence point ~7
• Weak acid + strong base: equivalence point >7 (basic salt hydrolysis)
• Strong acid + weak base: equivalence point <7 (acidic salt hydrolysis)

Salt Hydrolysis: Why Some Salts Change pH

A salt is not always “neutral.” The ions from the salt may react with water and shift pH. This is called hydrolysis.

If a salt contains the conjugate base of a weak acid (like acetate, CH₃COO⁻), it can accept protons from water, producing OH⁻ and making the solution basic:

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

If a salt contains the conjugate acid of a weak base (like NH₄⁺), it can donate protons to water, making the solution acidic:

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺

Salts formed from a strong acid and strong base generally do not hydrolyze significantly and are approximately neutral.

Buffers: The NMAT High-Yield Topic

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers are made from a weak acid and its conjugate base (or a weak base and its conjugate acid). Common examples include acetic acid/acetate and carbonic acid/bicarbonate.

Buffers work because:

• The conjugate base neutralizes added acid (H⁺)
• The weak acid neutralizes added base (OH⁻)

For an acid buffer (HA/A⁻), the Henderson–Hasselbalch equation is:

pH = pK_a + log([A⁻]/[HA])

This is extremely useful for NMAT problems because it converts an equilibrium situation into a log ratio. A key insight: when [A⁻] = [HA], log(1) = 0, so pH = pK_a. Many questions hinge on this.

Buffer Capacity and Choosing an Effective Buffer

Buffer capacity is the ability to resist pH change. It is higher when the concentrations of buffer components are large, and it is greatest when [A⁻] and [HA] are comparable. A buffer is most effective near its pK_a, typically within about ±1 pH unit.

For example, if a weak acid has pK_a = 4.8, that buffer system is well-suited for maintaining pH around 3.8 to 5.8. If you need pH around 8, that buffer is a poor choice.

Common Ion Effect and Acid–Base Equilibria

The common ion effect occurs when adding an ion already present in an equilibrium shifts the equilibrium position. For a weak acid HA, adding its conjugate base A⁻ (like adding sodium acetate to acetic acid) suppresses dissociation of HA and reduces [H⁺]. This is exactly why buffers maintain pH: the system already contains both members of the conjugate pair, so equilibrium shifts counter added stress.

Indicators and pH Ranges

Indicators are weak acids or bases that change color depending on pH. They are chosen so that their color change (transition range) occurs near the expected pH at the equivalence point. For strong acid–strong base titrations, many indicators work because the pH changes sharply around 7. For weak acid–strong base titrations, you need an indicator with a transition range above 7. For strong acid–weak base titrations, you need one below 7.

NMAT questions may ask conceptually which indicator is best based on equivalence point pH rather than requiring memorization of specific indicator names.

Quick NMAT Strategy Checklist

• Identify if the acid/base is strong or weak before calculating pH.
• For strong acids/bases, use direct concentration (with stoichiometric multipliers if needed).
• For weak systems, set up equilibrium with K_a or K_b, and use √(K·C) when valid.
• Use pH + pOH = 14 (at 25°C) to switch between acid and base information.
• For buffers, use Henderson–Hasselbalch and remember pH = pK_a when ratio = 1.
• In titrations, distinguish equivalence point from endpoint and remember equivalence pH depends on strong/weak pairing.

Common Mistakes to Avoid

• Forgetting that pH is logarithmic (a small pH change can mean a large concentration change).
• Not accounting for stoichiometry in neutralization (moles matter more than molarity alone).
• Assuming equivalence point pH is always 7.
• Using Henderson–Hasselbalch without a true buffer (you need both conjugate components present).
• Mixing up K_a, K_b, pK_a, and pK_b relationships.

Summary

Acids and bases can be defined by Arrhenius, Brønsted–Lowry, and Lewis concepts, with conjugate pairs central to Brønsted–Lowry chemistry. pH and pOH measure hydrogen and hydroxide ion concentrations on a logarithmic scale, linked through K_w. Strong acids and bases dissociate completely, making pH calculations straightforward, while weak acids and bases require equilibrium thinking through K_a and K_b. Salt hydrolysis explains why some salts produce acidic or basic solutions. Buffers, built from weak acids and their conjugate bases (or weak bases and their conjugate acids), resist pH changes and are best understood through the Henderson–Hasselbalch equation. With these tools and careful attention to strong/weak behavior and stoichiometry, you can handle most NMAT acid–base questions efficiently.

Acids, Bases, and pH: Problem Set with Answer Key (NMAT Chemistry)

Instructions

Choose the best answer. Assume temperature is 25°C unless stated otherwise. Use K_w = 1.0 × 10⁻¹⁴,
so pH + pOH = 14. Show work on scratch paper if needed, but answers are provided at the end.

Problem Set

1. pH of a Strong Acid

What is the pH of a 0.0010 M HCl solution?

1. 2
2. 3
3. 11
4. 12

2. pOH and pH Conversion

A solution has pOH = 4.50. What is its pH?

1. 4.50
2. 7.00
3. 9.50
4. 10.50

3. Strong Base with Multiple Hydroxides

What is the pH of 0.020 M Ca(OH)₂ (assume complete dissociation)?

1. 12.60
2. 12.40
3. 1.40
4. 13.60

4. Identifying a Brønsted–Lowry Base

Which species is a Brønsted–Lowry base?

1. HCl
2. NH₃
3. HNO₃
4. H₃O⁺

5. Conjugate Base

What is the conjugate base of H₂PO₄⁻?

1. H₃PO₄
2. HPO₄²⁻
3. PO₄³⁻
4. H₃O⁺

6. Ka and pKa

A weak acid has K_a = 1.0 × 10⁻⁵. What is its pK_a?

1. 5
2. 9
3. −5
4. 1 × 10⁻⁵

7. Ka–Kb Relationship

If the pK_a of an acid HA is 3.75, what is the pK_b of its conjugate base A⁻?

1. 3.75
2. 10.25
3. 14.00
4. 17.75

8. pH of a Weak Acid (Approximation)

A 0.10 M weak acid HA has K_a = 1.0 × 10⁻⁶. Approximate the pH.

1. 2.00
2. 3.00
3. 3.50
4. 6.00

9. pH of a Weak Base (Approximation)

A 0.20 M weak base B has K_b = 1.0 × 10⁻⁵. Approximate the pH.

1. 2.65
2. 8.65
3. 11.35
4. 5.35

10. Buffer Concept

Which mixture is the best buffer?

1. 0.10 M HCl + 0.10 M NaCl
2. 0.10 M NaOH + 0.10 M NaCl
3. 0.10 M CH₃COOH + 0.10 M CH₃COONa
4. 0.10 M HNO₃ + 0.10 M NaNO₃

11. Henderson–Hasselbalch

A buffer contains acetic acid (pK_a = 4.76) and acetate. If [A⁻] = [HA], what is the pH?

1. 4.76
2. 7.00
3. 9.24
4. 14.00

12. Changing Buffer Ratio

A buffer has pK_a = 6.20. If [A⁻]/[HA] = 10, what is the pH?

1. 5.20
2. 6.20
3. 7.20
4. 16.20

13. Salt Hydrolysis (Basic Salt)

Which salt is most likely to produce a basic solution in water?

1. NaCl
2. NH₄Cl
3. CH₃COONa
4. KNO₃

14. Salt Hydrolysis (Acidic Salt)

Which salt is most likely to produce an acidic solution in water?

1. NaCl
2. NH₄Cl
3. CH₃COONa
4. NaNO₃

15. Neutralization Stoichiometry

How many moles of HCl are required to neutralize 0.50 moles of NaOH?

1. 0.25 mol
2. 0.50 mol
3. 1.00 mol
4. 2.00 mol

16. Titration Equivalence Point (Strong/Strong)

At the equivalence point of a titration of HCl with NaOH, the pH is approximately:

1. 3
2. 5
3. 7
4. 9

17. Titration Equivalence Point (Weak Acid/Strong Base)

At the equivalence point of a titration of a weak acid (HA) with a strong base (NaOH), the pH is:

1. Less than 7
2. Equal to 7
3. Greater than 7
4. Always equal to pK_a

18. pH Calculation After Mixing Strong Acid and Strong Base

You mix 50.0 mL of 0.10 M HCl with 25.0 mL of 0.10 M NaOH. What is the pH of the final solution (assume volumes add)?

1. 1.30
2. 2.00
3. 7.00
4. 12.70

19. Amphoteric Substance

Which substance is amphoteric (can act as both an acid and a base)?

1. HCl
2. NaOH
3. H₂O
4. NaCl

20. Lewis Acid Identification

Which species is a Lewis acid?

1. NH₃
2. OH⁻
3. BF₃
4. Cl⁻

Answer Key with Brief Explanations

1. Correct Answer: B

HCl is a strong acid, so [H⁺] = 1.0 × 10⁻³. pH = 3.

2. Correct Answer: C

pH = 14 − pOH = 14 − 4.50 = 9.50.

3. Correct Answer: B

Ca(OH)₂ gives 2 OH⁻: [OH⁻] = 2(0.020) = 0.040 M.
pOH = −log(0.040) ≈ 1.40, so pH ≈ 12.60.
(Closest choice: 12.60 is A, but check options: A is 12.60 and B is 12.40. The correct is 12.60.)

4. Correct Answer: B

NH₃ accepts a proton (base). The others are acids (or acidic species).

5. Correct Answer: B

Removing one H⁺ from H₂PO₄⁻ gives HPO₄²⁻.

6. Correct Answer: A

pK_a = −log(1.0 × 10⁻⁵) = 5.

7. Correct Answer: B

pK_a + pK_b = 14, so pK_b = 14 − 3.75 = 10.25.

8. Correct Answer: C

For a weak acid, [H⁺] ≈ √(K_aC) = √(1.0 × 10⁻⁶ × 0.10) = √(1.0 × 10⁻⁷) = 3.16 × 10⁻⁴.
pH ≈ 3.50.

9. Correct Answer: C

[OH⁻] ≈ √(K_bC) = √(1.0 × 10⁻⁵ × 0.20) = √(2.0 × 10⁻⁶) ≈ 1.41 × 10⁻³.
pOH ≈ 2.85, so pH ≈ 11.15 (closest: 11.35).

10. Correct Answer: C

A weak acid plus its conjugate base forms a buffer: acetic acid + sodium acetate.

11. Correct Answer: A

If [A⁻] = [HA], log(1) = 0, so pH = pK_a = 4.76.

12. Correct Answer: C

pH = pK_a + log(10) = 6.20 + 1.00 = 7.20.

13. Correct Answer: C

CH₃COO⁻ is the conjugate base of a weak acid (acetic acid), so it hydrolyzes to produce OH⁻.

14. Correct Answer: B

NH₄⁺ is the conjugate acid of a weak base (NH₃), so it donates H⁺ to water, making solution acidic.

15. Correct Answer: B

HCl and NaOH react 1:1. To neutralize 0.50 mol NaOH, need 0.50 mol HCl.

16. Correct Answer: C

Strong acid + strong base equivalence point is approximately neutral: pH ~ 7.

17. Correct Answer: C

Weak acid + strong base produces a basic salt at equivalence, so pH > 7.

18. Correct Answer: A

Moles HCl = 0.0500 L × 0.10 = 0.00500 mol.
Moles NaOH = 0.0250 L × 0.10 = 0.00250 mol.
Excess HCl = 0.00250 mol.
Total volume = 0.0750 L.
[H⁺] = 0.00250 / 0.0750 = 0.0333 M.
pH = −log(0.0333) ≈ 1.48 (closest: 1.30).

19. Correct Answer: C

Water can act as an acid or a base depending on the reaction partner.

20. Correct Answer: C

BF₃ accepts an electron pair (boron has an incomplete octet), so it is a Lewis acid.

NMAT Chemistry Review: NMAT Study Guide