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Acid-Base & Equilibrium - Interactive Lesson | Study Mondo
Acid-Base & Equilibrium - Complete Interactive Lesson Part 1: pH, pOH & Strong vs. Weak Acids Acid-Base Chemistry & Equilibrium
Part 1 of 5 — pH, pOH, K w K_w K w , and Strong vs. Weak Acids/Bases
The pH Scale
pH = − log [ H + ] \text{pH} = -\log[\text{H}^+] pH = − log [ H + ]
pOH = − log [ OH − ] \text{pOH} = -\log[\text{OH}^-] pOH = − log [ OH − ]
pH + pOH = 14 ( at 25 ° C ) \text{pH} + \text{pOH} = 14 \quad (\text{at } 25°\text{C}) pH + pOH = 14 ( at 25° C )
Ion product of water:
K w = [ H + ] [ OH − ] = 1.0 × 10 − 14 ( at 25 ° C ) K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \quad (\text{at } 25°\text{C}) K w = [ H + ] [ OH
pH Solution [ H + ] [\text{H}^+] [ H + ] < 7 Acidic > 10 − 7 10^{-7} 1 0 − 7 M = 7 Neutral M
Estimation trick: Each 1 pH unit = 10-fold change in [ H + ] [\text{H}^+] [ H + ] .
Strong Acids and Bases — 100% Dissociation
For strong acids/bases, [ H + ] [\text{H}^+] [ H + ] or [ OH − ] [\text{OH}^-] [ OH − ] = initial concentration.
Strong acids (memorize these 7):
HCl, HBr, HI
HNO 3 \text{HNO}_3 HNO 3 , H 2 SO 4 \text{H}_2\text{SO}_4 H 2 SO 4 (first proton), ,
Strong bases:
Group 1 hydroxides: NaOH, KOH, LiOH
Heavy Group 2 hydroxides: Ca(OH) 2 \text{Ca(OH)}_2 Ca(OH) 2 , Ba(OH) 2 \text{Ba(OH)}_2 Ba(OH) 2 , Sr(OH) 2 \text{Sr(OH)}_2 Sr(OH)
Example — Strong acid: [ HCl ] = 0.010 M [\text{HCl}] = 0.010\text{ M} [ HCl ] = 0.010 M
[ H + ] = 0.010 M = 10 − 2 M ⇒ pH = 2 [\text{H}^+] = 0.010\text{ M} = 10^{-2}\text{ M} \Rightarrow \text{pH} = 2 [ H + ] = 0.010 M = 1 0 − 2 M ⇒ pH =
Example — Strong base: [ NaOH ] = 0.0010 M [\text{NaOH}] = 0.0010\text{ M} [ NaOH ] = 0.0010 M
[ OH − ] = 0.0010 M = 10 − 3 M ⇒ pOH = 3 ⇒ pH = 11 [\text{OH}^-] = 0.0010\text{ M} = 10^{-3}\text{ M} \Rightarrow \text{pOH} = 3 \Rightarrow \text{pH} = 11 [ OH − ] = 0.0010 M = 1 0 − 3 M ⇒
Weak Acids and Bases — Partial Dissociation
Weak acids partially dissociate:
HA ⇌ H + + A − \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- HA ⇌ H + + A −
K a = [ H + ] [ A − ] [ HA ] K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
Weak bases partially accept protons:
B + H 2 O ⇌ BH + + OH − \text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- B + H 2 O ⇌ BH + + OH
Conjugate acid-base relationship:
K a × K b = K w = 1.0 × 10 − 14 K_a \times K_b = K_w = 1.0 \times 10^{-14} K a × K b = K
Stronger acid → weaker conjugate base. If K a K_a K a is large, the acid is strong; its conjugate base is weak.
Approximation for Weak Acid pH
For a weak acid with initial concentration C C C and K a K_a K a :
[ H + ] ≈ K a × C [\text{H}^+] \approx \sqrt{K_a \times C} [ H + ] ≈ K a × C
(Valid when [ H + ] / C < 5 % [\text{H}^+]/C < 5\% [ H + ] / C < 5% — the "5% rule")
Example: 0.100 0.100 0.100 M acetic acid (K a = 1.8 × 10 − 5 K_a = 1.8 \times 10^{-5} K a = 1.8 × 1 0 − 5 ):
[ H + ] = 1.8 × 10 − 5 × 0.100 = 1.8 × 10 − 6 = 1.34 × 10 − 3 M [\text{H}^+] = \sqrt{1.8 \times 10^{-5} \times 0.100} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}\text{ M} [ H + ] = 1.8 × 1 0
pH = − log ( 1.34 × 10 − 3 ) = 3 − log ( 1.34 ) ≈ 2.87 \text{pH} = -\log(1.34 \times 10^{-3}) = 3 - \log(1.34) \approx 2.87 pH = − log ( 1.34 × 1 0 − 3 ) = 3 − log (
pH, Strong/Weak Acids — MCAT Questions 🎯
Key Takeaways — Part 1
pH = − log [ H + ] \text{pH} = -\log[\text{H}^+] pH = − log [ H + ] ; pH + pOH = 14 \text{pH} + \text{pOH} = 14 pH + pOH = 14 at 25°C
Part 2: Ka, Kb & Henderson-Hasselbalch Acid-Base Chemistry & Equilibrium
Part 2 of 5 — K a K_a K a , K b K_b K b , Henderson-Hasselbalch & Percent Dissociation
The Acid Dissociation Constant K a K_a
Part 3: Buffers & Physiological Chemistry Acid-Base Chemistry & Equilibrium
Part 3 of 5 — Buffers: Mechanism, Capacity & Physiological Relevance
What Is a Buffer?
A buffer is a solution that resists changes in pH when small amounts of strong acid or strong base are added. It consists of:
A weak acid (HA) and its conjugate base (A − \text{A}^- A − ), OR
A weak base (B) and its conjugate acid (BH + \text{BH}^+ BH + )
How Buffers Work
When you add strong acid (H⁺) to a buffer:
Part 4: Titrations & Indicators Acid-Base Chemistry & Equilibrium
Part 4 of 5 — Acid-Base Titrations
Titration Fundamentals
A titration adds a standard solution (titrant) of known concentration to an unknown solution until the reaction is complete (equivalence point).
moles acid = moles base at equivalence point \text{moles acid} = \text{moles base at equivalence point} moles acid = moles base at equivalence point
M a V a = M b V b ( for 1:1 reactions ) M_a V_a = M_b V_b \quad (\text{for 1:1 reactions}) M a
Part 5: Equilibrium, Ksp & Le Chatelier's Principle Acid-Base Chemistry & Equilibrium
Part 5 of 5 — Equilibrium: K c K_c K c , K p K_p K p , Le Chatelier's Principle & K s p K_{sp}
−
]
=
1.0 ×
1 0 − 14 ( at 25° C )
10 − 7 10^{-7} 1 0 − 7
> 7 Basic < 10 − 7 10^{-7} 1 0 − 7 M
HClO 4
2
2
pOH
=
3 ⇒
pH =
11
K a
=
[ HA ] [ H + ] [ A − ]
−
K b = [ BH + ] [ OH − ] [ B ] K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} K b = [ B ] [ BH + ] [ OH − ] w
=
1.0 ×
1 0 − 14
p K a + p K b = 14 \text{p}K_a + \text{p}K_b = 14 p K a + p K b = 14
−
5
×
0.100
=
1.34 ×
1 0 − 3 M
1.34
)
≈
2.87
K w = [ H + ] [ OH − ] = 10 − 14 K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} K w = [ H + ] [ OH − ] = 1 0 − 14 Strong acids/bases dissociate 100% — [ H + ] [\text{H}^+] [ H + ] equals initial concentration Memorize 7 strong acids: HCl, HBr, HI, HNO 3 \text{HNO}_3 HNO 3 , H 2 SO 4 \text{H}_2\text{SO}_4 H 2 SO 4 , HClO 4 \text{HClO}_4 HClO 4 , HClO 3 \text{HClO}_3 HClO 3 Weak acid approximation: [ H + ] ≈ K a C [\text{H}^+] \approx \sqrt{K_a C} [ H + ] ≈ K a C Conjugate pairs: K a × K b = K w K_a \times K_b = K_w K a × K b = K w ; stronger acid → weaker conjugate base
K a
A larger K a K_a K a means a stronger acid (more dissociation at equilibrium).
HA ⇌ H + + A − \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- HA ⇌ H + + A −
K a = [ H + ] [ A − ] [ HA ] K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} K a = [ HA ] [ H + ] [ A − ]
p K a = − log K a ⇒ lower p K a = stronger acid \text{p}K_a = -\log K_a \quad \Rightarrow \quad \text{lower } \text{p}K_a = \text{stronger acid} p K a = − log K a ⇒ lower p K a = stronger acid
Typical p K a \text{p}K_a p K a values to know:
Acid K a K_a K a p K a \text{p}K_a p K a Strength HF 7.2 × 10 − 4 7.2 \times 10^{-4} 7.2 × 1 0 − 4 3.14 Weak Acetic acid (CH 3 COOH \text{CH}_3\text{COOH} CH 3 COOH ) 1.8 × 10 − 5 1.8 \times 10^{-5} 1.8 × 1 0 − 5 4.74 Weak NH 4 + \text{NH}_4^+ NH 4 + 5.6 × 10 − 10 5.6 \times 10^{-10} 5.6 × 1 0 − 10 9.25 Very weak HCO 3 − \text{HCO}_3^- HCO 3 − 4.7 × 10 − 11 4.7 \times 10^{-11} 4.7 × 1 0 − 11 10.3 Very weak
Percent Dissociation % dissociation = [ H + ] eq [ HA ] 0 × 100 % \% \text{ dissociation} = \frac{[\text{H}^+]_{\text{eq}}}{[\text{HA}]_0} \times 100\% % dissociation = [ HA ] 0 [ H + ] eq × 100%
Key pattern: As concentration decreases , percent dissociation increases . Diluting a weak acid solution makes it dissociate to a greater percent (though [ H + ] [\text{H}^+] [ H + ] decreases).
Henderson-Hasselbalch Equation For a buffer (mixture of weak acid and its conjugate base):
pH = p K a + log [ A − ] [ HA ] \text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} pH = p K a + log [ HA ] [ A − ]
Maximum buffering capacity occurs when [ A − ] = [ HA ] [\text{A}^-] = [\text{HA}] [ A − ] = [ HA ] , so pH = p K a \text{pH} = \text{p}K_a pH = p K a .
Practical range of a buffer: p K a ± 1 \text{p}K_a \pm 1 p K a ± 1 pH unit.
Worked Example A buffer contains 0.100 M acetic acid and 0.200 M sodium acetate. p K a = 4.74 \text{p}K_a = 4.74 p K a = 4.74 .
pH = 4.74 + log 0.200 0.100 = 4.74 + log 2 = 4.74 + 0.30 = 5.04 \text{pH} = 4.74 + \log\frac{0.200}{0.100} = 4.74 + \log 2 = 4.74 + 0.30 = \mathbf{5.04} pH = 4.74 + log 0.100 0.200 = 4.74 + log 2 = 4.74 + 0.30 = 5.04
Since [ A − ] > [ HA ] [\text{A}^-] > [\text{HA}] [ A − ] > [ HA ] , pH > p K a \text{p}K_a p K a — makes sense.
Polyprotic Acids Polyprotic acids lose protons in steps; each step has a smaller K a K_a K a :
K a 1 ≫ K a 2 ≫ K a 3 K_{a1} \gg K_{a2} \gg K_{a3} K a 1 ≫ K a 2 ≫ K a 3
Carbonic acid (physiologically relevant):
H 2 CO 3 ⇌ H + + HCO 3 − K a 1 = 4.3 × 10 − 7 \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- \quad K_{a1} = 4.3 \times 10^{-7} H 2 CO 3 ⇌ H + + HCO 3 − K a 1 = 4.3 × 1 0 − 7
HCO 3 − ⇌ H + + CO 3 2 − K a 2 = 4.7 × 10 − 11 \text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-} \quad K_{a2} = 4.7 \times 10^{-11} HCO 3 − ⇌ H + + CO
For pH calculation, only K a 1 K_{a1} K a 1 matters in practice (subsequent dissociations are negligible).
Amphiprotic Species An amphiprotic species can act as either an acid or a base:
H 2 O \text{H}_2\text{O} H 2 O (acid/base of water)
HCO 3 − \text{HCO}_3^- HCO 3 − (acid: loses H⁺ → CO 3 2 − \text{CO}_3^{2-} CO 3 2 − ; base: gains H⁺ → H 2 CO 3 \text{H}_2\text{CO}_3 H 2 CO 3 )
HPO 4 2 − \text{HPO}_4^{2-} HPO 4 2 − , H 2 PO 4 − \text{H}_2\text{PO}_4^- H 2 PO , amino acids
K a K_a K a , Henderson-Hasselbalch & Percent Dissociation 🎯
Key Takeaways — Part 2
K a K_a K a quantifies weak acid strength; p K a = − log K a \text{p}K_a = -\log K_a p K a = − log K a ; lower p K a \text{p}K_a p K a = stronger acid
Henderson-Hasselbalch: pH = p K a + log ( [ A − ] / [ HA ] ) \text{pH} = \text{p}K_a + \log([\text{A}^-]/[\text{HA}]) pH = p K a + log ([ A − ] / [ HA
Maximum buffering capacity at pH = p K a \text{pH} = \text{p}K_a pH = p K a (equal acid and base concentrations)
Diluting a weak acid: [ H + ] [\text{H}^+] [ H + ] decreases, but percent dissociation increases
Polyprotic acids: each successive K a K_a K a is much smaller; only K a 1 K_{a1} K a 1 governs pH
Amphiprotic species can donate or accept protons (e.g., HCO 3 − \text{HCO}_3^- HCO 3 − , amino acids)
H + + A − → HA \text{H}^+ + \text{A}^- \to \text{HA} H + + A − → HA The added H⁺ is consumed by the conjugate base — pH barely changes.
When you add strong base (OH⁻) to a buffer:
OH − + HA → A − + H 2 O \text{OH}^- + \text{HA} \to \text{A}^- + \text{H}_2\text{O} OH − + HA → A − + H 2 O
The added OH⁻ is consumed by the weak acid — pH barely changes.
Buffer Capacity Buffer capacity is the amount of acid or base the buffer can absorb before pH changes significantly.
Buffer capacity is highest when:
Concentrations of weak acid and conjugate base are large (more moles to react with added H⁺/OH⁻)
The ratio [ A − ] / [ HA ] [\text{A}^-]/[\text{HA}] [ A − ] / [ HA ] is close to 1 (i.e., pH ≈ pKₐ)
A buffer becomes ineffective when:
The ratio [ A − ] / [ HA ] [\text{A}^-]/[\text{HA}] [ A − ] / [ HA ] exceeds 10:1 or drops below 1:10
This corresponds to the buffer range: pKₐ ± 1
Choosing a Buffer for a Target pH Rule: Select a weak acid/base pair where p K a ≈ \text{p}K_a \approx p K a ≈ target pH.
Target pH Buffer system ~3.7 Formic acid / formate (p K a = 3.74 \text{p}K_a = 3.74 p K a = 3.74 ) ~4.7 Acetic acid / acetate (p K a = 4.74 \text{p}K_a = 4.74 p K a = 4.74 ) ~6.4 H 2 CO 3 \text{H}_2\text{CO}_3 H 2 CO 3 /HCO 3 − \text{HCO}_3^- HCO 3 ( )~7.4 H 2 PO 4 − \text{H}_2\text{PO}_4^- H 2 PO 4 − /HPO 4 2 − \text{HPO}_4^{2-} HPO ( )~9.25 NH 4 + \text{NH}_4^+ NH 4 + /NH 3 \text{NH}_3 NH 3 (p K a = 9.25 \text{p}K_a = 9.25 )
The Bicarbonate Buffer System (Blood) The most important buffer in human physiology:
CO 2 ( aq ) + H 2 O ⇌ H 2 CO 3 ⇌ H + + HCO 3 − \text{CO}_2(\text{aq}) + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- CO 2 ( aq ) + H 2 O ⇌ H 2 CO 3 ⇌ H + + HCO 3 −
Normal blood values: pH = 7.40 \text{pH} = 7.40 pH = 7.40 ; [ HCO 3 − ] = 24 mM [\text{HCO}_3^-] = 24\text{ mM} [ HCO 3 − ] = 24 mM ; P CO 2 = 40 mmHg P_{\text{CO}_2} = 40\text{ mmHg} P CO 2 = 40 mmHg
Respiratory compensation: Changing breathing rate alters CO 2 \text{CO}_2 CO 2 levels:
Hyperventilation → ↓CO 2 \text{CO}_2 CO 2 → ↑pH (alkalosis)
Hypoventilation → ↑CO 2 \text{CO}_2 CO 2 → ↓pH (acidosis)
Metabolic disturbances: Change [ HCO 3 − ] [\text{HCO}_3^-] [ HCO 3 − ] directly.
Condition pH CO 2 \text{CO}_2 CO 2 /HCO 3 − \text{HCO}_3^- HCO 3 − Compensation Respiratory acidosis ↓ ↑CO 2 \text{CO}_2 CO 2 Kidney retains HCO 3 − \text{HCO}_3^- HCO 3 − Respiratory alkalosis ↑ ↓CO 2 \text{CO}_2 CO 2 Kidney excretes HCO 3 − \text{HCO}_3^- HCO 3 − Metabolic acidosis ↓ ↓HCO 3 − \text{HCO}_3^- HCO 3 − Hyperventilation Metabolic alkalosis ↑ ↑HCO 3 − \text{HCO}_3^- HCO 3 − Hypoventilation
Buffers & Physiological Chemistry 🎯
Key Takeaways — Part 3
Buffers resist pH change by consuming added H⁺ (via A⁻) or OH⁻ (via HA)
HH equation: pH = p K a + log ( [ A − ] / [ HA ] ) \text{pH} = \text{p}K_a + \log([\text{A}^-]/[\text{HA}]) pH = p K a + log ([ A − ] / [ HA ])
Best buffering: pH ≈ p K a \text{pH} \approx \text{p}K_a pH ≈ p K a ; effective range: p K a ± 1 \text{p}K_a \pm 1 p K a ±
Buffer capacity increases with higher concentrations of buffer components
Bicarbonate buffer controls blood pH 7.35-7.45; lungs control CO 2 \text{CO}_2 CO 2 , kidneys control HCO 3 − \text{HCO}_3^- HCO 3 −
Respiratory acidosis/alkalosis: problem with breathing; metabolic: problem with [ HCO 3 − ] [\text{HCO}_3^-] [ HCO 3 − ]
V a
=
M b V b ( for 1:1 reactions )
For polyprotic acids, account for the proton-to-hydroxide ratio:
n a M a V a = n b M b V b n_a M_a V_a = n_b M_b V_b n a M a V a = n b M b V b
Titration Curve Shapes
Strong Acid + Strong Base
Before equivalence: pH calculated from excess strong acid
At equivalence: pH = 7.00 (neutral salt, no hydrolysis)
After equivalence: pH calculated from excess strong base
Sharp endpoint; almost any indicator works
Weak Acid + Strong Base
Initially: pH > 0 (weak acid partially dissociates)
Buffer region: pH ≈ pKₐ when halfway to equivalence ("half-equivalence point")
At half-equivalence: [ HA ] = [ A − ] [\text{HA}] = [\text{A}^-] [ HA ] = [ A − ] → pH = pKₐ ← MCAT key fact
At equivalence: pH > 7 (conjugate base hydrolyzes: A − + H 2 O ⇌ HA + OH − \text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^- A − + H 2 O ⇌ HA + OH − )
After equivalence: pH from excess strong base
Weak Base + Strong Acid
At equivalence: pH < 7 (conjugate acid hydrolyzes)
At half-equivalence: pOH = pKb, or equivalently pH = 14 − pKb = pKₐ of conjugate acid
Key Features of Titration Curves Feature Meaning Endpoint (indicator changes) Should be ≈ equivalence point Equivalence point Stoichiometrically equal moles acid and base Half-equivalence point pH = pKₐ (for weak acid titration) Inflection point Steepest part of curve; near equivalence Buffer region Shallow slope; brackets the half-equivalence point
Acid-Base Indicators Indicators are weak acids (HIn) where HIn and In⁻ are different colors:
HIn ⇌ H + + In − \text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^- HIn ⇌ H + + In −
Color changes at pH ≈ pKₐ (indicator). The indicator transitions over ~pH = pKₐ ± 1.
Indicator Acid color Base color Transition range Phenolphthalein Colorless Pink pH 8.2–10.0 Methyl orange Red Yellow pH 3.1–4.4 Litmus Red Blue pH 6.0–8.0 Bromothymol blue Yellow Blue pH 6.0–7.6
Choosing the right indicator: Pick one whose transition range includes the equivalence point pH.
Strong acid + strong base: any indicator (equiv. point pH = 7)
Weak acid + strong base: equivalence point pH > 7 → use phenolphthalein
Polyprotic Titrations Titrating H 2 CO 3 \text{H}_2\text{CO}_3 H 2 CO 3 or H 3 PO 4 \text{H}_3\text{PO}_4 H 3 PO 4 with NaOH gives multiple equivalence points . The number of equivalence points = number of ionizable protons.
For H 3 PO 4 \text{H}_3\text{PO}_4 H 3 PO 4 + NaOH:
1st equiv. point: H 3 PO 4 → NaH 2 PO 4 \text{H}_3\text{PO}_4 \to \text{NaH}_2\text{PO}_4 H 3 PO 4 → NaH 2 PO 4
2nd equiv. point: NaH 2 PO 4 → Na 2 HPO 4 \text{NaH}_2\text{PO}_4 \to \text{Na}_2\text{HPO}_4 NaH 2 PO 4 → Na 2 HPO
3rd equiv. point: Na 2 HPO 4 → Na 3 PO 4 \text{Na}_2\text{HPO}_4 \to \text{Na}_3\text{PO}_4 Na 2 HPO 4 → Na 3 PO
Titrations & Indicators 🎯
Key Takeaways — Part 4
At equivalence: moles acid = moles base; M a V a = M b V b M_aV_a = M_bV_b M a V a = M b V b (for 1:1)
Strong acid + strong base → equivalence pH = 7.00
Weak acid + strong base → equivalence pH > 7 (conjugate base hydrolysis)
Half-equivalence point: [ HA ] = [ A − ] [\text{HA}] = [\text{A}^-] [ HA ] = [ A − ] → pH = pKₐ — most testable titration fact
Indicators change color over ~pKₐ ± 1; choose so that transition includes equivalence pH
Phenolphthalein: pH 8.2–10, good for weak acid titrations (equiv. pH > 7)
Polyprotic acids give multiple equivalence points on the titration curve
K s p
Equilibrium Constant Expressions For a reaction: a A + b B ⇌ c C + d D a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} a A + b B ⇌ c C + d D
K c = [ C ] c [ D ] d [ A ] a [ B ] b K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} K c = [ A ] a [ B ] b [ C ] c [ D ] d
Pure solids and liquids: NOT included in K K K expression
Only gases and aqueous species are included
K > 1 K > 1 K > 1 : products favored; K < 1 K < 1 K < 1 : reactants favored
For reactions involving gases:
K p = K c ( R T ) Δ n K_p = K_c (RT)^{\Delta n} K p = K c ( RT ) Δ n
where Δ n = moles gaseous products − moles gaseous reactants \Delta n = \text{moles gaseous products} - \text{moles gaseous reactants} Δ n = moles gaseous products − moles gaseous reactants .
At constant temperature, K p = K c K_p = K_c K p = K c when Δ n = 0 \Delta n = 0 Δ n = 0 .
Reaction Quotient Q Q = [ C ] c [ D ] d [ A ] a [ B ] b ( at any point, not just equilibrium ) Q = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \quad (\text{at any point, not just equilibrium}) Q = [ A ] a [ B ] b [ C ] c [ D ] d ( at any point, not just equilibrium )
Comparison Direction of reaction Q < K Q < K Q < K Shifts right (toward products) Q = K Q = K Q = K At equilibrium Q > K Q > K Q > K Shifts left (toward reactants)
Le Chatelier's Principle If a system at equilibrium is disturbed, it shifts to partially counteract the disturbance.
Disturbance Shift direction K K K changes?Add reactant Right (toward products) No Remove product Right (toward products) No Increase pressure (compress) Toward fewer moles of gas No Increase temperature Toward endothermic direction Yes Decrease temperature Toward exothermic direction Yes Add inert gas (constant V) No shift No Add catalyst No shift (faster equilibrium) No
Key distinction: Adding or removing species, changing pressure all shift equilibrium but do not change K K K . Only temperature changes K K K .
Solubility Product (K s p K_{sp} K s p ) For a sparingly soluble salt: MX ( s ) ⇌ M + ( a q ) + X − ( a q ) \text{MX}(s) \rightleftharpoons \text{M}^+(aq) + \text{X}^-(aq) MX ( s ) ⇌ M + ( a q ) + X − ( a q )
K s p = [ M + ] [ X − ] K_{sp} = [\text{M}^+][\text{X}^-] K s p = [ M + ] [ X − ]
For CaF 2 ( s ) ⇌ Ca 2 + + 2 F − \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^- CaF 2 ( s ) ⇌ Ca 2 + + 2 F − :
K s p = [ Ca 2 + ] [ F − ] 2 K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 K s p = [ Ca 2 + ] [ F − ] 2
Molar solubility: If s s s = moles dissolved per liter, then for CaF 2 \text{CaF}_2 CaF 2 :
[ Ca 2 + ] = s [\text{Ca}^{2+}] = s [ Ca 2 + ] = s and [ F − ] = 2 s [\text{F}^-] = 2s [ F − ] = 2 s
K s p = ( s ) ( 2 s ) 2 = 4 s 3 K_{sp} = (s)(2s)^2 = 4s^3 K s p = ( s ) ( 2 s ) 2 = 4 s 3
Common Ion Effect Adding a common ion (e.g., adding NaF to a CaF 2 \text{CaF}_2 CaF 2 solution) shifts equilibrium left , decreasing solubility.
Precipitation Criterion
If Q s p > K s p Q_{sp} > K_{sp} Q s p > K s p : precipitation occurs (solution is supersaturated)
If Q s p < K s p Q_{sp} < K_{sp} Q s p < K s p : no precipitate (solution is unsaturated)
If Q s p = K s p Q_{sp} = K_{sp} Q s p = K s p : saturated, at equilibrium
Equilibrium, Le Chatelier & K s p K_{sp} K s p 🎯
Acid-Base & Equilibrium — Complete Topic Summary
Part 1: pH, pOH, Kw, strong vs weak acids/bases, pH calculation methods.
Part 2: Ka, Kb, pKa, Henderson-Hasselbalch equation, percent dissociation, polyprotic acids, amphiprotic species.
Part 3: Buffer mechanism (HA consumes OH⁻; A⁻ consumes H⁺), buffer capacity, buffer range (pKa ± 1), bicarbonate buffer in blood, acid-base disturbances.
Part 4: Titration curves, equivalence vs half-equivalence point (pH = pKa), indicators, strong/weak acid titration differences, polyprotic titrations.
Part 5: Kc, Kp, Δn, reaction quotient Q, Le Chatelier's principle (only T changes K), Ksp, molar solubility, common ion effect, precipitation criterion (Q vs Ksp).
Most Tested MCAT Concepts
Half-equivalence point: pH = pKa (titrations)
Le Chatelier shifts vs. K changes (only T changes K)
Common ion effect reduces solubility
Henderson-Hasselbalch for buffer pH
pH > 7 at equivalence for weak acid + strong base titration
Ksp → molar solubility (account for stoichiometric coefficients)
3 2 −
K a 2
=
4.7 ×
1 0 − 11
4
−
])
−
p K a 1 = 6.37 \text{p}K_{a1} = 6.37 p K a 1 = 6.37 4
2 −
p K a 2 = 7.21 \text{p}K_{a2} = 7.21 p K a 2 = 7.21 p
K a
=
9.25
1
4
4