Acids and Bases — Practice Quiz

A Chemistry cheat sheet for Acids and Bases — every key formula with its symbols defined — plus a medium-level practice quiz to test recall.

Formulas & key concepts

Conjugate Acid-Base Pairs: Proton transfer between acid HX and water.

$$HX(aq) + H_2O(l) \rightleftharpoons X^-(aq) + H_3O^+(aq)$$

Proton Transfer in Water: If \(H_2O\) is a stronger base than \(X^-\), equilibrium favors products.

$$HX(aq) + H_2O(l) \rightleftharpoons X^-(aq) + H_3O^+(aq)$$

Acid Dissociation Constant: Equilibrium constant for weak acid ionization.

$$K_a = \frac{[H_3O^+][A^-]}{[HA]}$$ or $$K_a = \frac{[H^+][A^-]}{[HA]}$$

Where: \(K_a\) = acid dissociation constant

pH Definition: Negative logarithm of hydrogen ion concentration.

$$pH = -\log[H^+]$$

pOH Definition: Negative logarithm of hydroxide ion concentration.

$$pOH = -\log[OH^-]$$

pH-pOH Relationship: Sum of pH and pOH equals 14 at 25°C.

$$pH + pOH = 14.00$$

H+ from pH: Calculate hydrogen ion concentration from pH.

$$[H^+] = 10^{-pH}$$

OH- from pOH: Calculate hydroxide ion concentration from pOH.

$$[OH^-] = 10^{-pOH}$$

Percent Ionization: Fraction of acid molecules that donate protons.

$$\text{Percent ionization} = \frac{[H^+]_{\text{equilibrium}}}{[HA]_{\text{initial}}} \times 100\%$$

Weak Acid Ionization: General equation for weak acid dissociation.

$$HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)$$

Weak Acid Ionization Simplified: Simplified form without explicit water.

$$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$$

Weak Acid Ka Expression: Equilibrium constant expression for weak acid.

$$K_a = \frac{[H_3O^+][A^-]}{[HA]}$$ or $$K_a = \frac{[H^+][A^-]}{[HA]}$$

Weak Base Ionization: Base accepts proton from water.

$$B(aq) + H_2O(l) \rightleftharpoons HB^+(aq) + OH^-(aq)$$

Base Dissociation Constant: Equilibrium constant for weak base ionization.

$$K_b = \frac{[HB^+][OH^-]}{[B]}$$

Where: \(K_b\) = base dissociation constant

Ka-Kb Relationship: Product of acid and conjugate base constants equals \(K_w\) at 25°C.

$$K_a \times K_b = K_w = 1.0 \times 10^{-14}$$

pKa-pKb Relationship: Sum of pKa and pKb of conjugate pair equals 14 at 25°C.

$$pK_a + pK_b = pK_w = 14.00$$

Conjugate Base Reaction: Weak conjugate base reacts with water to produce weak acid and hydroxide.

$$X^-(aq) + H_2O(l) \rightleftharpoons HX(aq) + OH^-(aq)$$

<strong>Relative Strengths of Conjugate Acid-Base Pairs</strong><br> <table style="width:100%; border-collapse: collapse; font-size: 0.85em; margin-top: 5px;"> <tr style="border-bottom: 1px solid #ccc;"> <th style="text-align: left; padding: 3px;">Acid</th> <th style="text-align: left; padding: 3px;">Base</th> </tr> <tr><td>HCl (strong)</td><td>Cl⁻ (negligible)</td></tr> <tr><td>H₂SO₄ (strong)</td><td>HSO₄⁻ (weak)</td></tr> <tr><td>HNO₃ (strong)</td><td>NO₃⁻ (negligible)</td></tr> <tr><td>H₃O⁺</td><td>H₂O</td></tr> <tr><td>HF (weak)</td><td>F⁻ (weak)</td></tr> <tr><td>CH₃COOH (weak)</td><td>CH₃COO⁻ (weak)</td></tr> <tr><td>NH₄⁺ (weak)</td><td>NH₃ (weak)</td></tr> <tr><td>H₂O</td><td>OH⁻ (strong)</td></tr> </table>

<strong>Polyprotic Acids at 25°C</strong><br> <table style="width:100%; border-collapse: collapse; font-size: 0.85em; margin-top: 5px;"> <tr style="border-bottom: 1px solid #ccc;"> <th style="text-align: left; padding: 3px;">Acid</th> <th style="text-align: left; padding: 3px;">Ka₁</th> <th style="text-align: left; padding: 3px;">Ka₂</th> <th style="text-align: left; padding: 3px;">Ka₃</th> </tr> <tr><td>H₃PO₄ (Phosphoric)</td><td>7.5×10⁻³</td><td>6.2×10⁻⁸</td><td>4.2×10⁻¹³</td></tr> <tr><td>H₂SO₃ (Sulfurous)</td><td>1.7×10⁻²</td><td>6.4×10⁻⁸</td><td>—</td></tr> <tr><td>H₂CO₃ (Carbonic)</td><td>4.3×10⁻⁷</td><td>5.6×10⁻¹¹</td><td>—</td></tr> <tr><td>H₂C₂O₄ (Oxalic)</td><td>5.9×10⁻²</td><td>6.4×10⁻⁵</td><td>—</td></tr> </table>

<strong>Conjugate Acid-Base Pairs (Ka × Kb = Kw)</strong><br> <table style="width:100%; border-collapse: collapse; font-size: 0.85em; margin-top: 5px;"> <tr style="border-bottom: 1px solid #ccc;"> <th style="text-align: left; padding: 3px;">Acid</th> <th style="text-align: left; padding: 3px;">Ka</th> <th style="text-align: left; padding: 3px;">Base</th> <th style="text-align: left; padding: 3px;">Kb</th> </tr> <tr><td>HF</td><td>6.8×10⁻⁴</td><td>F⁻</td><td>1.5×10⁻¹¹</td></tr> <tr><td>HC₂H₃O₂</td><td>1.8×10⁻⁵</td><td>C₂H₃O₂⁻</td><td>5.6×10⁻¹⁰</td></tr> <tr><td>NH₄⁺</td><td>5.6×10⁻¹⁰</td><td>NH₃</td><td>1.8×10⁻⁵</td></tr> <tr><td>HCO₃⁻</td><td>5.6×10⁻¹¹</td><td>CO₃²⁻</td><td>1.8×10⁻⁴</td></tr> </table>

Practice quiz

  1. What is the pH of a solution with a hydrogen ion concentration of $2.5 \times 10^{-4} \text{ M}$?

    • $3.60$
    • $4.00$
    • $2.50$
    • $10.40$

    Answer: $3.60$

  2. In the reaction $CH_3COOH(aq) + H_2O(l) \rightleftharpoons CH_3COO^-(aq) + H_3O^+(aq)$, which species is the conjugate base of $CH_3COOH$?

    • $H_2O(l)$
    • $H_3O^+(aq)$
    • $CH_3COO^-(aq)$
    • $CH_3COOH(aq)$

    Answer: $CH_3COO^-(aq)$

  3. Which of the following is the correct expression for the acid dissociation constant ($K_a$) for a weak acid $HA$?

    • $K_a = \frac{[HA]}{[H_3O^+][A^-]}$
    • $K_a = \frac{[H_3O^+][A^-]}{[HA]}$
    • $K_a = [H_3O^+][A^-][HA]$
    • $K_a = \frac{[H_3O^+][HA]}{[A^-]}$

    Answer: $K_a = \frac{[H_3O^+][A^-]}{[HA]}$

  4. If the pOH of a solution is $5.25$ at $25^\circ C$, what is its pH?

    • $5.25$
    • $8.75$
    • $14.00$
    • $1.0 \times 10^{-5.25}$

    Answer: $8.75$

  5. A $0.10 \text{ M}$ solution of a weak acid $HA$ has an equilibrium $[H^+]$ concentration of $1.5 \times 10^{-3} \text{ M}$. What is the percent ionization of the acid?

    • $0.15\%$
    • $1.5\%$
    • $15\%$
    • $0.015\%$

    Answer: $1.5\%$

  6. A $0.20 \text{ M}$ solution of a weak acid $HX$ has a pH of $3.50$. Calculate the $K_a$ for $HX$.

    • $1.0 \times 10^{-7}$
    • $3.2 \times 10^{-4}$
    • $5.0 \times 10^{-7}$
    • $1.6 \times 10^{-6}$

    Answer: $5.0 \times 10^{-7}$

  7. Which of the following equations correctly represents the ionization of a weak base $B$ in water?

    • $B(aq) + H_3O^+(aq) \rightleftharpoons HB^+(aq) + H_2O(l)$
    • $B(aq) + H_2O(l) \rightleftharpoons HB^+(aq) + OH^-(aq)$
    • $B(aq) + OH^-(aq) \rightleftharpoons BO^-(aq) + H_2O(l)$
    • $B(aq) \rightleftharpoons B^+(aq) + e^-(aq)$

    Answer: $B(aq) + H_2O(l) \rightleftharpoons HB^+(aq) + OH^-(aq)$

  8. The $K_a$ for hydrofluoric acid ($HF$) is $6.8 \times 10^{-4}$. What is the $K_b$ for its conjugate base, $F^-$, at $25^\circ C$?

    • $6.8 \times 10^{-4}$
    • $1.0 \times 10^{-14}$
    • $1.5 \times 10^{-11}$
    • $1.5 \times 10^{-10}$

    Answer: $1.5 \times 10^{-11}$

  9. Consider the reaction: $HF(aq) + H_2O(l) \rightleftharpoons F^-(aq) + H_3O^+(aq)$. Given that $H_2O$ is a stronger base than $F^-$, which direction does the equilibrium favor?

    • The reactants
    • The products
    • Neither, it is at equilibrium
    • The reaction does not occur

    Answer: The products

  10. A solution has a pH of $9.30$ at $25^\circ C$. What is the hydroxide ion concentration, $[OH^-]$?

    • $9.30 \text{ M}$
    • $1.0 \times 10^{-14} \text{ M}$
    • $2.0 \times 10^{-5} \text{ M}$
    • $5.0 \times 10^{-10} \text{ M}$

    Answer: $2.0 \times 10^{-5} \text{ M}$

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