Unit 8 Notes Topic 5 AP CHEMISTRY

⚗️ Strong & Weak Acid-Base Reactions

Unit 8 | Topic 5 – Acid Meets Base: What Happens?

💥 Strong Acid + Strong Base

These fully react (neutralization):

H⁺ + OH⁻ → H₂O

HCl + NaOH → NaCl + H₂O

To find pH, just check if one is in excess!

Example: What is the pH if 20 mL of 0.20 M NaOH is added to 15 mL of 0.20 M HCl?

• Moles OH⁻ = 0.020 × 0.20 = 0.004 mol
• Moles H⁺ = 0.015 × 0.20 = 0.003 mol
• Excess OH⁻ = 0.001 mol in 35 mL (0.035 L)
• [OH⁻] = 0.001 / 0.035 = 0.02857 M

pOH = –log(0.02857) = 1.54 → pH = 14 – 1.54 = 12.46

✅ Final pH: 12.46 (basic)

🌡️ Strong Acid + Weak Base

NH₃ + HCl → NH₄⁺ + Cl⁻

If base is in excess → BUFFER forms!

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

Example: What is the pH when 25 mL of 0.15 M NH₃ reacts with 15 mL of 0.15 M HCl?

• Moles NH₃ = 0.025 × 0.15 = 0.00375 mol
• Moles HCl = 0.015 × 0.15 = 0.00225 mol
• Excess NH₃ = 0.0015 mol, NH₄⁺ = 0.00225 mol
• pKb = 4.75 (NH₃), use formula below:

pOH = 4.75 + log(0.0015 / 0.00225) = 4.57 → pH = 14 – 4.57 = 9.43

✅ Final pH: 9.43 (slightly basic buffer)

🔥 Strong Base + Weak Acid

CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

If acid is in excess → BUFFER forms!

pH = pKa + log([A⁻]/[HA])

Example: What is the pH when 50 mL of 0.10 M NaOH is added to 75 mL of 0.20 M acetic acid?

• Moles NaOH = 0.050 × 0.10 = 0.005 mol
• Moles acid = 0.075 × 0.20 = 0.015 mol
• CH₃COOH left = 0.010 mol | CH₃COO⁻ formed = 0.005 mol

pH = 4.74 + log(0.005 / 0.010) = 4.74 + log(0.5) = 4.44

✅ Final pH: 4.44 (acidic buffer)

⚖️ Weak Acid + Weak Base

HA + B ⇌ A⁻ + BH⁺

Final pH depends on relative Ka and Kb!

• If Ka > Kb → solution is acidic
• If Ka < Kb → solution is basic

Example: HF + NH₃ reaction — is it acidic or basic?

• Ka (HF) = 6.6 × 10⁻⁴
• Kb (NH₃) = 1.8 × 10⁻⁵

Since Ka > Kb → solution is more acidic

💡 AP Tip: When acids and bases react, check who’s in excess. If none, then it's equilibrium time! Use ICE or buffer formula.