AP Chemistry Acids and Bases — Worked Answer Explanations

Unit 8 · 12 questions explained

1. Question 1 · Easy

   What is the pH of a $0.010\mathrm{M}$ solution of $\mathrm{HCl}$?

   • A
     $1.0$

     Why not A: That would be a 0.10 M solution of HCl.
   • B
     $2.0$Correct
   • C
     $3.0$

     Why not C: Off by a factor of 10; $0.010=10^{-2}$, so pH = 2.
   • D
     $12.0$

     Why not D: This is the pOH, not the pH, or the pH of a basic solution.
   Explanation

   HCl is a strong acid that ionizes completely: $[{\mathrm{H}}^{+}]=0.010\mathrm{M}=10^{-2}\mathrm{M}$. $\mathrm{pH}=-\log (10^{-2})=2.0$.

   Key takeaway

   For strong acids: $[\mathrm{H^+}]$ equals the initial acid concentration. $\mathrm{pH} = -\log[\mathrm{H^+}]$.

2. Question 2 · Easy

   Which of the following is the conjugate base of ${\mathrm{H}}_2\mathrm{P}{\mathrm{O}}_4^{-}$?

   • A
     ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$

     Why not A: ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$ is formed when ${\mathrm{H}}_2\mathrm{P}{\mathrm{O}}_4^{-}$ accepts a proton — that makes it the conjugate acid.
   • B
     $\mathrm{HP}{\mathrm{O}}_4^{2-}$Correct
   • C
     $\mathrm{P}{\mathrm{O}}_4^{3-}$

     Why not C: $\mathrm{P}{\mathrm{O}}_4^{3-}$ is two protons removed from ${\mathrm{H}}_2\mathrm{P}{\mathrm{O}}_4^{-}$, not one.
   • D
     ${\mathrm{H}}_2\mathrm{O}$

     Why not D: Water is not derived from ${\mathrm{H}}_2\mathrm{P}{\mathrm{O}}_4^{-}$ by proton loss.
   Explanation

   The conjugate base of an acid is formed by removing one proton (H${}^{+}$). ${\mathrm{H}}_2\mathrm{P}{\mathrm{O}}_4^{-}\stackrel{-H^{+}}{}\mathrm{HP}{\mathrm{O}}_4^{2-}$. The charge decreases by 1 and one H is removed.

   Key takeaway

   Conjugate base = acid $- \mathrm{H^+}$. Conjugate acid = base $+ \mathrm{H^+}$.

3. Question 3 · Easy

   A buffer solution is prepared containing $0.200\mathrm{M}$ acetic acid ($\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$, $p{K}_a=4.74$) and $0.200\mathrm{M}$ sodium acetate ($\mathrm{C}{\mathrm{H}}_3\mathrm{COONa}$). What is the pH of this buffer?

   • A
     $2.87$

     Why not A: This is the pH of pure 0.10 M acetic acid without the conjugate base.
   • B
     $4.74$Correct
   • C
     $7.00$

     Why not C: The buffer pH equals pKa only when acid and conjugate base concentrations are equal, which is here — giving 4.74, not 7.
   • D
     $9.26$

     Why not D: This is 14 $-$ pKa, the pKb of the conjugate base, not the buffer pH.
   Explanation

   Henderson-Hasselbalch: $\mathrm{pH}=p{K}_a+\log \frac{[{\mathrm{A}}^{-}]}{[\mathrm{HA}]}=4.74+\log \frac{0.200}{0.200}=4.74+\log (1)=4.74+0=4.74$.

   When $[{\mathrm{A}}^{-}]=[\mathrm{HA}]$, pH = pKa.

   Key takeaway

   Henderson-Hasselbalch equation: $\mathrm{pH} = pK_a + \log([\mathrm{A^-}]/[\mathrm{HA}])$. When acid = conjugate base, pH = pKa.

4. Question 4 · Easy

   A $25.0\mathrm{mL}$ sample of $0.100\mathrm{M}$ HCl is titrated with $0.100\mathrm{M}$ NaOH. What is the pH at the equivalence point?

   • A
     $1.0$

     Why not A: That would be the pH before any NaOH is added.
   • B
     $7.0$Correct
   • C
     $13.0$

     Why not C: This would be the pH after significant excess NaOH, not at the equivalence point.
   • D
     $4.74$

     Why not D: 4.74 is the pKa of acetic acid — relevant for weak acid titrations.
   Explanation

   Strong acid + strong base: $\mathrm{HCl}+\mathrm{NaOH}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}$. At the equivalence point, all acid and base are consumed, leaving only NaCl in water. NaCl is the salt of a strong acid and strong base — it does not hydrolyze, so the solution is neutral: pH = 7.0.

   Key takeaway

   Strong acid + strong base equivalence point: pH = 7.0 because the resulting salt does not hydrolyze.

5. Question 5 · Medium

   Acetic acid has $K_a=1.8\times 10^{-5}$. What is the pH of a $0.10\mathrm{M}$ $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ solution?

   • A
     $1.0$

     Why not A: Treated acetic acid as a strong acid; weak acids ionize only partially.
   • B
     $2.87$Correct
   • C
     $4.74$

     Why not C: This is the pKa, not the pH of this solution.
   • D
     $3.37$

     Why not D: Arithmetic error; likely used wrong formula or forgot to take the square root.
   Explanation

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}\rightleftharpoons \mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}+{\mathrm{H}}^{+}$

   $K_a=x^2/(0.10-x)\approx x^2/0.10=1.8\times 10^{-5}$

   $x=\sqrt{1.8\times 10^{-6}}=1.34\times 10^{-3}\mathrm{M}=[{\mathrm{H}}^{+}]$

   $\mathrm{pH}=-\log (1.34\times 10^{-3})\approx 2.87$

   Key takeaway

   Weak acid pH: set up ICE, $x \approx \sqrt{K_a \cdot C_a}$ when $K_a \ll C_a$. Check 5% rule.

6. Question 6 · Medium

   The $K_w$ for water at $25°\mathrm{C}$ is $1.0\times 10^{-14}$. If the pOH of a solution is $3.00$, what is the pH?

   • A
     $3.00$

     Why not A: pH = pOH only at neutrality; this solution is basic.
   • B
     $7.00$

     Why not B: 7.00 is neutral; pOH = 3 gives a basic solution.
   • C
     $11.00$Correct
   • D
     $14.0$

     Why not D: 14 is pKw, the sum pH + pOH, not either one alone.
   Explanation

   At $25°\mathrm{C}$: $\mathrm{pH}+\mathrm{pOH}=14.00$. If pOH $=3.00$, then $\mathrm{pH}=14.00-3.00=11.00$.

   Key takeaway

   pH + pOH = 14 at 25°C ($K_w = 1.0 \times 10^{-14}$).

7. Question 7 · Medium

   During the titration of a weak acid (HA) with a strong base (NaOH), the half-equivalence point is reached when half the acid has been neutralized. At the half-equivalence point:

   • A
     $\mathrm{pH}=7.0$ for all weak acids.

     Why not A: pH = 7 only at the equivalence point of a strong acid–strong base titration.
   • B
     $\mathrm{pH}=p{K}_a$ of the weak acid.Correct
   • C
     $[{\mathrm{A}}^{-}]=0$ because no conjugate base has formed yet.

     Why not C: At the half-equivalence point, half the acid has been converted to conjugate base: $[\mathrm{HA}]=[{\mathrm{A}}^{-}]$.
   • D
     $\mathrm{pH}$ is highest at this point during the titration.

     Why not D: pH continues to rise past the half-equivalence point and is highest after the equivalence point.
   Explanation

   At the half-equivalence point, $[\mathrm{HA}]=[{\mathrm{A}}^{-}]$. Henderson-Hasselbalch: $\mathrm{pH}=p{K}_a+\log ([{\mathrm{A}}^{-}]/[\mathrm{HA}])=p{K}_a+\log (1)=p{K}_a$. This is how $p{K}_a$ is determined experimentally from a titration curve.

   Key takeaway

   Half-equivalence point: $[\mathrm{HA}] = [\mathrm{A^-}]$, so pH = pKa. Used to determine Ka from titration curves.

8. Question 8 · Medium

   Identify the acid-base behavior of $\mathrm{A}{\mathrm{l}}^{3+}$ dissolved in water according to Lewis theory.

   • A
     $\mathrm{A}{\mathrm{l}}^{3+}$ is a Lewis base because it donates lone pairs to water.

     Why not A: Al${}^{3+}$ accepts lone pairs, it does not donate them.
   • B
     $\mathrm{A}{\mathrm{l}}^{3+}$ is a Lewis acid because it accepts lone pairs from water molecules.Correct
   • C
     $\mathrm{A}{\mathrm{l}}^{3+}$ is a Brønsted–Lowry acid because it donates protons to water.

     Why not C: Al${}^{3+}$ doesn't donate H${}^{+}$ directly; it is the [Al(H${}_2$O)${}_6$]${}^{3+}$ complex that indirectly releases protons.
   • D
     $\mathrm{A}{\mathrm{l}}^{3+}$ has no acid-base behavior in water because it is a metal cation.

     Why not D: Many metal cations act as Lewis acids and produce acidic solutions.
   Explanation

   Lewis acid = electron-pair acceptor. Al${}^{3+}$ has an empty valence orbital and accepts electron pairs from water ligands in $[\mathrm{Al}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$. This complex is then acidic (Brønsted sense) because the Al–O bond weakens the O–H bonds, facilitating proton release.

   Key takeaway

   Lewis acid = electron-pair acceptor. Metal cations like Al$^{3+}$ are Lewis acids; they accept lone pairs from water ligands.

9. Question 9 · Medium

   When $\mathrm{HF}$ ionizes in water, fluoride ion is formed. Why is ${\mathrm{F}}^{-}$ a stronger base than $\mathrm{C}{\mathrm{l}}^{-}$?

   • A
     Fluorine has a higher electronegativity, which makes the F–H bond stronger and F${}^{-}$ more reluctant to give it up.

     Why not A: This explains why HF is a weak acid, but the question asks about the basicity of F${}^{-}$ vs Cl${}^{-}$.
   • B
     ${\mathrm{F}}^{-}$ is smaller and holds its charge density over a smaller volume, making it a better proton acceptor (stronger base).Correct
   • C
     ${\mathrm{F}}^{-}$ has a higher negative charge than $\mathrm{C}{\mathrm{l}}^{-}$.

     Why not C: Both F${}^{-}$ and Cl${}^{-}$ have a $-1$ charge.
   • D
     F${}^{-}$ has more lone pairs available for proton donation.

     Why not D: Both have 4 lone pairs; basicity here is about proton acceptance, not donation.
   Explanation

   The conjugate base strength correlates inversely with the acid strength. HF is a weak acid ($K_a\approx 7\times 10^{-4}$) while HCl is strong ($K_a\gg 1$). Therefore F${}^{-}$ is a stronger base than Cl${}^{-}$. The reason: F${}^{-}$ is smaller, with higher charge density, and forms a stronger H–F bond — it holds onto protons more readily.

   Key takeaway

   Weaker acid → stronger conjugate base. Smaller anion with higher charge density (like F$^-$) is a stronger base than larger anion (like Cl$^-$).

10. Question 10 · Hard

    Which of the following aqueous solutions is the strongest base?

    • A
      $0.10\mathrm{M}$ $\mathrm{N}{\mathrm{H}}_3$ ($K_b=1.8\times 10^{-5}$)

      Why not A: Ammonia is a weak base; [OH${}^{-}$] ≈ 1.3 × 10${}^{-3}$ M.
    • B
      $0.10\mathrm{M}$ $\mathrm{NaOH}$Correct
    • C
      $0.10\mathrm{M}$ $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ ($K_b=4.4\times 10^{-4}$)

      Why not C: Methylamine is a weak base; stronger than ammonia but weaker than NaOH.
    • D
      $0.10\mathrm{M}$ $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$

      Why not D: Carbonate ion is a weak base through hydrolysis; NaOH ionizes completely.
    Explanation

    NaOH is a strong base that dissociates completely in water: $[\mathrm{O}{\mathrm{H}}^{-}]=0.10\mathrm{M}$, giving pOH = 1.0 and pH = 13.0. The other options are weak bases with much lower [OH${}^{-}$] at the same initial concentration.

    Key takeaway

    Strong bases ionize 100%; weak bases ionize partially. Same concentration → strong base has higher [OH$^-$] and higher pH.

11. Question 11 · Hard

    A buffer contains $0.300\mathrm{M}$ $\mathrm{N}{\mathrm{H}}_3$ and $0.200\mathrm{M}$ $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$. What is the pH? ($K_b(\mathrm{N}{\mathrm{H}}_3)=1.8\times 10^{-5}$; $p{K}_a(\mathrm{N}{\mathrm{H}}_4^{+})=9.26$)

    • A
      $9.44$Correct
    • B
      $9.08$

      Why not B: Inverted the ratio: used $\log ([\mathrm{HA}]/[{\mathrm{A}}^{-}])$ instead of $\log ([{\mathrm{A}}^{-}]/[\mathrm{HA}])$.
    • C
      $4.56$

      Why not C: Used pKa of acetic acid or the $K_b$ value directly; this buffer is basic.
    • D
      $9.26$

      Why not D: pH = pKa only when $[{\mathrm{A}}^{-}]=[\mathrm{HA}]$; here the ratio is 0.300:0.200.
    Explanation

    Treat as weak acid buffer with NH${}_4^{+}$ as the acid ($p{K}_a=9.26$) and NH${}_3$ as its conjugate base.

    $\mathrm{pH}=p{K}_a+\log \frac{[\mathrm{N}{\mathrm{H}}_3]}{[\mathrm{N}{\mathrm{H}}_4^{+}]}=9.26+\log \frac{0.300}{0.200}=9.26+\log (1.5)=9.26+0.176\approx 9.44$

    Key takeaway

    For an amine buffer, use $pK_a$ of the ammonium salt (conjugate acid) in Henderson-Hasselbalch: $\mathrm{pH} = pK_a + \log(\text{[base]/[acid]})$.

12. Question 12 · Hard

    A $100.0\mathrm{mL}$ sample of $0.100\mathrm{M}$ acetic acid ($p{K}_a=4.74$) is titrated with $0.100\mathrm{M}$ NaOH. After adding $50.0\mathrm{mL}$ of NaOH, what is the pH?

    • A
      $7.00$

      Why not A: After 50.0 mL NaOH, we are at the half-equivalence point, not the equivalence point.
    • B
      $4.74$Correct
    • C
      $2.87$

      Why not C: This is the pH of pure acetic acid before any NaOH is added.
    • D
      $8.72$

      Why not D: This would be the pH at the equivalence point, not the half-equivalence point.
    Explanation

    Initial moles HA = $0.100\mathrm{L}\times 0.100\mathrm{M}=0.0100\mathrm{mol}$.

    NaOH added: $0.0500\mathrm{L}\times 0.100\mathrm{M}=0.00500\mathrm{mol}$.

    After reaction: $0.00500\mathrm{mol}$ HA and $0.00500\mathrm{mol}$ A${}^{-}$ — equal amounts.

    Half-equivalence point: pH $=p{K}_a=4.74$.

    Key takeaway

    At the half-equivalence point of a weak acid titration (half the acid neutralized), pH = pKa.