Answer: 1.0×10⁻⁵ [H⁺] = 10⁻³ M = x. Ka ≈ x²/[HA]₀ = (10⁻³)²/0.10 = 10⁻⁶/10⁻¹ = 10⁻⁵.

Which best explains the pH of the NaF solution?

Answer: F- is the conjugate base of HF and hydrolyzes water NaF is the salt of a strong base (NaOH) and weak acid (HF). The fluoride ion (a weak base) abstracts a proton from water, giving slightly basic pH 7.

Percent ionization of a weak acid generally

Answer: Decreases as concentration increases Diluting shifts equilibrium toward more ionization (Le Chatelier on increased moles of solute particles).

AP Chemistry · Topic 8.3

Weak Acid and Base Equilibria Practice

Part of Acids and Bases.(SAP-9.C)

Practice questions

Want a predicted score for the whole AP CHEM exam? Take the 20-question diagnostic and Lumi will plan the rest.

Take diagnosticarrow_forward Practice all 8.3

Sample questions

5 of 27 — sign in to practice the rest with adaptive difficulty and mastery tracking.

Sample 1difficulty 2/5
$The pH of 0.10 M HA is 3.0.$
$What is Ka of HA?$
- A
  $1.0\times10⁻¹$
- B
  $1.0\times10⁻⁵$
  check_circle
- C
  $1.0\times10⁻³$
- D
  $1.0\times10⁻⁷$
Why
$[H⁺] = 10⁻³ M = x. Ka \approx x²/[HA]₀ = (10⁻³)²/0.10 = 10⁻⁶/10⁻¹ = 10⁻⁵.$
View full solution & statsarrow_forward
Sample 2difficulty 3/5
$Which best explains the pH of the NaF solution?$
- A
  $NaF is a strong base$
- B
  $F- is a strong acid$
- C
  $Na+ accepts H+ from water$
- D
  $F- is the conjugate base of HF and hydrolyzes water$
  check_circle
Why
$NaF is the salt of a strong base (NaOH) and weak acid (HF). The fluoride ion (a weak base) abstracts a proton from water, giving slightly basic pH > 7.$
View full solution & statsarrow_forward
Sample 3difficulty 3/5
$Percent ionization of a weak acid generally$
- A
  $Increases as concentration increases$
- B
  $Independent of concentration$
- C
  $Decreases as concentration increases$
  check_circle
- D
  $Always 100%$
Why
$Diluting shifts equilibrium toward more ionization (Le Chatelier on increased moles of solute particles).$
View full solution & statsarrow_forward
Sample 4difficulty 3/5
$For a diprotic acid H₂A, generally$
- A
  $K_a1 < K_a2$
- B
  $K_a1 ≫ K_a2$
  check_circle
- C
  $K_a1 = K_a2$
- D
  $Both are zero$
Why
$Removing the first proton is much easier than removing the second from the resulting anion.$
View full solution & statsarrow_forward
Sample 5difficulty 3/5
$For NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, K_b =$
- A
  $[NH₄⁺][OH⁻]/[NH₃]$
  check_circle
- B
  $[NH₃]/[NH₄⁺][OH⁻]$
- C
  $[OH⁻]/[NH₃]$
- D
  $[NH₄⁺]/[OH⁻]$
Why
$Standard expression for base dissociation; water omitted.$
View full solution & statsarrow_forward

Predict my AP scorearrow_forward