Acid Base Investigation

The Power of 10

0

2

4

7

9

11

14

pH Logarithm

pH = −log[H+]

Ion Relationship

14 = pH + pOH

Math Mastery: [H+] → pH

Example: Find pH if [H+] = 2.5 × 10−4 M.

1

Formula: pH = −log[H+]

2

Setup: pH = −log(2.5 × 10−4)

3

Final: pH = 3.60

Math Mastery: pH → [H+]

Example: Find [H+] if pH = 8.42.

1

Formula: [H+] = 10−pH

2

Setup: [H+] = 10−8.42

3

Final: [H+] = 3.8 × 10−9 M

Neutralization

Neutralization is a specific reaction resulting in:

Acid + Base
Water + Salt

Example Reaction

HCl + NaOH H2O + NaCl

Net Result: H+ + OH− → H2O

Anchoring Phenomenon

A Sand Dollar Dissolves

Observing the real-time effects of decreased seawater pH on marine life and calcification.

youtu.be/GL7qJ5-BWGQ

Challenge: How does increased H+ impact CaCO3 stability?

Visual pH Evidence Map

Color the bar below and write the substance names in the columns above their pH values.

01234567891011121314

Case File Analysis

1. Which substances were clearly BASIC? How did you know?

2. Which item had the LOWEST concentration of hydrogen ions?

3. How much more acidic is pH 3 than pH 5? (10x rule)

4. Why are cleaning supplies often basic?

Cations (+)

Protocol Alpha

8.26

2.33

12.7

6.00

13.7

12.6

9.48

4.84

6.49

10.1

3.99

11.8

3.00

10.8

1.98

5.13

14.0

9.91

9.83

8.16

0.88

0.51

8.41

7.00

5.92

6.31

1.12

0.17

0.99

5.98

1.47

6.31

4.91

10.9

13.6

2.81

7.51

4.25

1.77

12.3

3.81

1.85

2.69

6.84

8.94

13.0

1.20

10.2

13.9

5.66

2.61

11.6

0.01

7.80

12.9

Unit 12: Aqueous Systems - pH Puzzle Pursuit Solver

© 2026 Chemistry Investigations Lab

6.31

4.91

10.9

13.6

2.81

7.51

4.25

1.77

12.3

3.81

1.85

2.69

6.84

8.94

13.0

1.20

10.2

13.9

5.66

2.61

11.6

0.01

7.80

12.9

Unit 12: Aqueous Systems - pH Puzzle Pursuit Key

© 2026 Chemistry Investigations Lab - Internal Use Only

Scaffolded

A chemistry student titrates \(15.0\text{ mL}\) of an unknown concentration of hydrochloric acid (\(\text{HCl}\)) with \(22.5\text{ mL}\) of \(0.150\text{ M Ca(OH)}_2\) to reach the phenolphthalein end point. What is the molarity of the acid?

A) Reactants & Ion Count

Acid: \(\text{HCl} \implies \boldsymbol{n_A = }\) ____

Base: \(\text{Ca(OH)}_2 \implies \boldsymbol{n_B = }\) ____

Check the number of \(\text{H}^+\) and \(\text{OH}^-\) ions!

B) Setup Variables

\(M_A =\) _________

\(V_A =\) _________

\(M_B =\) _________

\(V_B =\) _________

C) Calculate Answer

\(M_A =\) ____________

2 Calculating Volume of Acid Required

Independent

How many milliliters of \(0.350\text{ M HNO}_3\) are needed to completely neutralize \(45.0\text{ mL}\) of \(0.120\text{ M KOH}\)?

Show All Calculation Work

Final Answer

________ mL

4. Advanced Stoichiometry Challenge

Sometimes neutralizations occur outside aqueous solution-to-solution titrations. For example, solid bases are often used to clean up liquid acid spills. Solve this stoichiometric scenario step-by-step:

"A lab technician spills \(50.0\text{ mL}\) of \(3.00\text{ M H}_2\text{SO}_4\) on the counter. They decide to neutralize it completely using solid sodium bicarbonate (\(\text{NaHCO}_3\), Molar Mass = \(84.01\text{ g/mol}\))."

\(\text{H}_2\text{SO}_4(aq) + 2\,\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\,\text{H}_2\text{O}(l) + 2\,\text{CO}_2(g)\)

Step A

Find moles of spill acid:

Use \(\text{Moles} = M \times V_{\text{L}}\)

___________ moles

Step B

Use the Mole Ratio:

Convert moles acid → moles base

___________ moles

Step C

Convert to solid mass:

Use MM = \(84.01\text{ g/mol}\)

___________ grams

Unit: Acid-Base Investigation Page 2 of 2

3. Guided Math Practice - Solutions

1 Determining Acid Molarity via Titration

Scaffolded

A chemistry student titrates \(15.0\text{ mL}\) of an unknown concentration of hydrochloric acid (\(\text{HCl}\)) with \(22.5\text{ mL}\) of \(0.150\text{ M Ca(OH)}_2\) to reach the phenolphthalein end point. What is the molarity of the acid?

A) Reactants & Ion Count

Acid: \(\text{HCl} \implies \boldsymbol{n_A = 1}\)

Base: \(\text{Ca(OH)}_2 \implies \boldsymbol{n_B = 2}\)

Hydrochloric acid produces 1 H⁺ ion, and calcium hydroxide produces 2 OH⁻ ions.

B) Setup Variables

\(M_A = \text{ ? (Unknown)}\)

\(V_A = 15.0\text{ mL}\)

\(M_B = 0.150\text{ M}\)

\(V_B = 22.5\text{ mL}\)

C) Calculate Answer

\(M_A \times 15.0 \times 1 = 0.150 \times 22.5 \times 2\)

\(15.0 \times M_A = 6.75\)

\(M_A = 6.75 / 15.0\)

\(M_A = \mathbf{0.450\text{ M}}\)

2 Calculating Volume of Acid Required

Independent

How many milliliters of \(0.350\text{ M HNO}_3\) are needed to completely neutralize \(45.0\text{ mL}\) of \(0.120\text{ M KOH}\)?

Calculation Work

Equation: \(\text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O}\) (\(\boldsymbol{n_A = 1}\), \(\boldsymbol{n_B = 1}\))

\(M_A \times V_A \times n_A = M_B \times V_B \times n_B\)

\(0.350 \times V_A \times 1 = 0.120 \times 45.0 \times 1\)

\(0.350 \times V_A = 5.40 \implies V_A = 5.40 / 0.350 = 15.428\text{ mL}\)

Final Answer

15.4 mL

4. Advanced Stoichiometry Challenge - Solutions

Sometimes neutralizations occur outside aqueous solution-to-solution titrations. For example, solid bases are often used to clean up liquid acid spills. Solve this stoichiometric scenario step-by-step:

"A lab technician spills \(50.0\text{ mL}\) of \(3.00\text{ M H}_2\text{SO}_4\) on the counter. They decide to neutralize it completely using solid sodium bicarbonate (\(\text{NaHCO}_3\), Molar Mass = \(84.01\text{ g/mol}\))."

\(\text{H}_2\text{SO}_4(aq) + 2\,\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\,\text{H}_2\text{O}(l) + 2\,\text{CO}_2(g)\)

Step A

Find moles of spill acid:

\(\text{Moles} = M \times V_{\text{L}}\)
\(= 3.00 \times 0.0500\)
\(= 0.150\text{ mol}\)

0.150 moles

Step B

Use the Mole Ratio:

Ratio: \(1\text{ mol H}_2\text{SO}_4 : 2\text{ mol NaHCO}_3\)
\(0.150\text{ mol acid} \times 2\)
\(= 0.300\text{ mol base}\)

0.300 moles

Step C

Convert to solid mass:

\(\text{Mass} = \text{mol} \times \text{Molar Mass}\)
\(= 0.300 \times 84.01\)
\(= 25.2\text{ g}\)

25.2 grams

Unit: Acid-Base Investigation • Teacher Answer Key Page 2 of 2

4. Neutralization Mastery

Acid + Base → Water + Salt

The Particle Battle

Neutralization occurs when moles of H+ equals moles of OH−.

nAMAVA = nBMBVB

• n: Number of H's or OH's in formula

• M: Molarity (mol/L)

• V: Volume (any unit, must match)

• A/B: Acid and Base

Example 1: Equal Strength

Find mL of 0.2 M NaOH to neutralize 50 mL of 0.2 M HCl.

Both have 1 active ion. Molarities are equal. Result: 50 mL.

Example 2: Step-by-Step

Calculate volume of 1.0 M KOH to neutralize 40 mL of 2.0 M H2SO4.

1. Ratio: 1 mole H2SO4 reacts with 2 moles KOH.

2. Acid Moles: 0.040 L × 2.0 M = 0.08 moles.

3. Base Moles: 0.08 acid × 2 = 0.16 moles base needed.

4. Volume: 0.16 moles / 1.0 M = 0.16 L = 160 mL.

5. Logarithmic logic (pH)

pH = −log[H+]

[H+] = 10−pH

Shortcut: If [H+] = 1.0 × 10−7, pH is 7.

Unit Practice Arena

Part A: Conjugate Twins

Conjugate Acid of CN−

Conjugate Base of H2SO4

Part B: Stoichiometry Showdown

1. How many moles of LiOH are required to neutralize 0.20 L of 3.0 M HCl?

2. If 10 mL of 0.5 M Ba(OH)2 neutralizes 20 mL of HNO3, what is the acid molarity?

Review Self-Test

1. Which property is characteristic of a Brønsted-Lowry base?

A. Donates H+

B. Accepts H+

C. pH < 7

D. Taste sour

2. An [H+] concentration of 1.0 × 10−11 M results in:

A. pH = 11

B. pH = −11

C. pH = 3

D. pH = 7

Answer Key: 1. B (Accepts protons), 2. A (pH 11). Rationales: 1. B-L focuses only on proton transfer. 2. pH is negative log of concentration.