Mistake Bank: Equilibrium | Chemca

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The Mistake Bank

Chapter 7: Equilibrium

Balance is key. Both in chemical reactions and in your logic. Master the complex mathematics of $K_p$, $K_{sp}$, and Le Chatelier's Principle to secure top marks.

1. The "pH = 8" Acid Trap

Ionic Equilibrium

Scenario: Calculate the pH of a $10^{-8} \text{ M }$ $HCl$ solution at 298 K.

✖ What Students Do

Student applies the basic pH formula blindly to the given concentration:

$$ pH = -\log[H^+] = -\log(10^{-8}) = \mathbf{8} $$

(Impossible! An acid added to water can never result in a basic solution with pH > 7.)

✔ The Correct Way

Don't Forget the Water's Contribution!

For highly dilute acids (concentration $< 10^{-6}\text{M}$), the $H^+$ naturally provided by the auto-ionization of water ($10^{-7}\text{M}$) becomes significant and cannot be ignored.

Total $[H^+] = [H^+]_{acid} + [H^+]_{water}$
Total $[H^+] = 10^{-8} + 10^{-7} = 10^{-8} + 10 \times 10^{-8} = 11 \times 10^{-8} = 1.1 \times 10^{-7} \text{ M}$

$pH = -\log(1.1 \times 10^{-7}) = 7 - \log(1.1)$
$pH \approx \mathbf{6.96}$ (Slightly acidic, as expected!)

2. The $K_p$ vs $K_c$ Subtraction Error

Chemical Equilibrium

Scenario: Write the relation between $K_p$ and $K_c$ for the Haber process: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.

✖ What Students Do

Student calculates $\Delta n$ backwards by subtracting products from reactants:

$$ \Delta n = (1+3) - 2 = 2 $$

$$ K_p = K_c(RT)^2 $$

(The math is completely reversed!)

✔ The Correct Way

It is ALWAYS Products minus Reactants!

$$ K_p = K_c(RT)^{\Delta n_g} $$

$\Delta n_g = (\text{Sum of moles of gaseous products}) - (\text{Sum of moles of gaseous reactants})$
$\Delta n_g = 2 - (1 + 3) = 2 - 4 = \mathbf{-2}$

Therefore: $\mathbf{K_p = K_c(RT)^{-2}}$ or $\mathbf{K_p = \frac{K_c}{(RT)^2}}$

3. The Pure Solid/Liquid Illusion

Equilibrium Constant

Scenario: Write the expression for $K_c$ for the thermal decomposition of limestone: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$

✖ What Students Do

Student mechanically applies the "Products over Reactants" rule to every single molecule in the equation:

$$ K_c = \frac{[CaO][CO_2]}{[CaCO_3]} $$

(Wrong! The active mass of pure solids and pure liquids is taken as unity.)

✔ The Correct Way

Ignore Pure Solids and Pure Liquids!

The concentration (density) of a pure solid or liquid remains constant throughout the reaction. They are incorporated into the equilibrium constant.

For this reaction, the only variable concentration is the gas.
$$ \mathbf{K_c = [CO_2]} $$
Similarly, for partial pressures:
$$ \mathbf{K_p = P_{CO_2}} $$

4. The Weak Acid Shortcut Error

Ionic Equilibrium

Scenario: Find the pH of a $0.1 \text{ M }$ solution of Acetic Acid ($CH_3COOH$). Given $K_a = 1.8 \times 10^{-5}$.

✖ What Students Do

Student is in a rush and treats the weak acid just like a strong acid ($HCl$), assuming complete 100% dissociation.

$$ [H^+] = 0.1 \text{ M} \implies pH = -\log(0.1) = \mathbf{1} $$

(Acetic acid is vinegar! If vinegar had a pH of 1, it would burn your tongue off.)

✔ The Correct Way

Use Ostwald's Dilution Law approximations!

Weak acids only dissociate slightly. You must use the acid dissociation constant ($K_a$).

$$ [H^+] = C \cdot \alpha = \sqrt{K_a \cdot C} $$
$$ [H^+] = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} $$
$$ [H^+] \approx 1.34 \times 10^{-3} \text{ M} $$
$$ pH = -\log(1.34 \times 10^{-3}) \approx \mathbf{2.87} $$

5. The $K_{sp}$ Stoichiometry Trap

Solubility

Scenario: Derive the relationship between Solubility Product ($K_{sp}$) and molar solubility ($S$) for Calcium Fluoride ($CaF_2$).

✖ What Students Do

Student memorized the simplest formula from the $AgCl$ example and applies it to every salt:

$$ K_{sp} = S^2 \implies S = \sqrt{K_{sp}} $$

(This only works for 1:1 type salts. $CaF_2$ produces 3 ions!)

✔ The Correct Way

Always write the dissociation equation first!

$$ CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq) $$
If solubility is $S$, then $[Ca^{2+}] = S$ and $[F^-] = 2S$.

$$ K_{sp} = [Ca^{2+}][F^-]^2 $$
$$ K_{sp} = (S)(2S)^2 = (S)(4S^2) = \mathbf{4S^3} $$
For $AB_2$ or $A_2B$ type salts, $K_{sp} = 4S^3$.

6. Inert Gas at Constant Volume

Le Chatelier's Principle

Scenario: Argon gas is added to the equilibrium mixture $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ in a rigid, sealed container (Constant Volume). In which direction does the equilibrium shift?

✖ What Students Do

Student thinks: "Adding gas increases the total pressure inside the container. According to Le Chatelier, increased pressure shifts the reaction towards fewer moles (backward)."

✔ The Correct Way

Zero Effect at Constant Volume!

While adding an inert gas at constant volume *does* increase the total pressure, it does not change the molar concentrations (Moles/Volume) or the Partial Pressures of the reacting gases.
Since the active masses remain completely unchanged, the equilibrium state is undisturbed.
Result: No Shift.
(Note: Adding inert gas at constant *pressure* expands the volume, which WOULD shift this reaction forward.)

7. Recognizing Hidden Buffers

Buffer Solutions

Scenario: Which mixture will form an Acidic Buffer solution?
A) $100\text{mL}$ of $0.1\text{M}$ $HCl$ + $100\text{mL}$ of $0.1\text{M}$ $CH_3COONa$
B) $100\text{mL}$ of $0.1\text{M}$ $CH_3COOH$ + $100\text{mL}$ of $0.1\text{M}$ $NaOH$

✖ What Students Do

Student remembers "Buffer = Weak Acid + Its Salt". They don't see that exact pre-mixed combination in either option, so they guess B, assuming the weak acid is enough.

(Wrong! In B, the strong base completely neutralizes the weak acid. No acid is left!)

✔ The Correct Way

Trace the Neutralization Reaction!

Option B: Equal moles of Weak Acid and Strong Base. They neutralize completely to form a salt solution. Not a buffer.

Option A: Strong Acid ($HCl$) reacts with the Salt ($CH_3COONa$) to form Weak Acid ($CH_3COOH$) + $NaCl$.
Because the moles are equal, all the salt converts into weak acid. Still not a buffer!

Correct Buffer Creation: You need a mixture where the Weak Acid is in EXCESS compared to the Strong Base (e.g., $100\text{mL}$ WA + $50\text{mL}$ SB). This leaves unreacted Weak Acid *and* newly formed Salt.

8. The Catalyst Illusion

Le Chatelier's Principle

Scenario: Iron is added as a catalyst to the Haber process. How does this affect the equilibrium yield of Ammonia ($NH_3$)?

✖ What Students Do

Student associates catalysts with "more product" because they are used in industry to maximize output.

They confidently answer: "The yield of Ammonia increases."

✔ The Correct Way

Catalysts DO NOT alter the equilibrium position!

A catalyst lowers the activation energy for BOTH the forward and backward reactions equally.
It allows the system to reach equilibrium faster (kinetics), but the final concentration of products and reactants (the yield) remains exactly the same (thermodynamics).
Answer: The yield remains unchanged.

9. Temperature's Unique Power

Equilibrium Constant

Scenario: For an exothermic reaction ($A \rightleftharpoons B, \Delta H < 0$), the temperature is increased. What happens to the value of the Equilibrium Constant ($K$)?

✖ What Students Do

Student thinks: "Le Chatelier says it shifts backward. But $K$ is a constant, and constants don't change."

Answer: "The reaction shifts left, but K remains constant."

✔ The Correct Way

Temperature is the ONLY factor that changes the value of $K$!

Unlike changes in concentration, pressure, or volume (which change the reaction quotient $Q$ but not $K$), a change in temperature actually alters the equilibrium constant itself according to the van 't Hoff equation.

For an Exothermic reaction, increasing temperature causes the equilibrium constant ($K$) to DECREASE.

10. Reaction Quotient Directionality

Reaction Quotient ($Q_c$)

Scenario: At a certain moment, the Reaction Quotient ($Q_c$) is calculated to be 10. The Equilibrium Constant ($K_c$) for the reaction is 2. In which direction is the reaction proceeding?

✖ What Students Do

Student thinks: "$10 > 2$. The reaction is 'ahead' of equilibrium, so it must keep going forward to reach the end."

Answer: "Forward direction."

✔ The Correct Way

The system must shift to make $Q_c = K_c$!

Since $Q_c = \frac{[\text{Products}]}{[\text{Reactants}]}$, a value of $10$ means there are currently too many products compared to equilibrium (which wants a ratio of $2$).

To reduce the numerator (products) and increase the denominator (reactants) to get back down to 2, the reaction must proceed in the Backward (Reverse) Direction.

11. Common Ion Effect Math

Solubility

Scenario: Calculate the solubility ($s$) of $AgCl$ ($K_{sp} = 10^{-10}$) in a $0.1\text{M}$ $NaCl$ solution.

✖ What Students Do

Student sets up the full quadratic equation: $K_{sp} = s(s + 0.1)$. They spend 5 minutes trying to solve $s^2 + 0.1s - 10^{-10} = 0$ manually, often making arithmetic errors.

✔ The Correct Way

Use the Approximation Rule!

Because $AgCl$ is highly insoluble, the amount of $Cl^-$ it contributes ($s$) is negligibly tiny compared to the $0.1\text{M}$ from the $NaCl$.

We assume: $[Cl^-]_{total} = s + 0.1 \approx \mathbf{0.1 \text{ M}}$
$$ K_{sp} = [Ag^+][Cl^-] $$ $$ 10^{-10} = (s)(0.1) $$ $$ s = \frac{10^{-10}}{10^{-1}} = \mathbf{10^{-9} \text{ M}} $$ No quadratic formula needed!

12. The Conjugate Pair Trap

Ionic Equilibrium

Scenario: The acid dissociation constant ($K_a$) of Hydrofluoric acid ($HF$) is $6.8 \times 10^{-4}$. Calculate the base dissociation constant ($K_b$) of its conjugate base, the Fluoride ion ($F^-$).

✖ What Students Do

Student thinks they need to construct an ICE table for $F^-$ reacting with water, but gives up because the concentration isn't provided.

Or they incorrectly guess that $K_b = 1 / K_a$.

✔ The Correct Way

Use the strict relation: $K_a \times K_b = K_w$

For any conjugate acid-base pair in aqueous solution at 298 K, the product of their constants always equals the ionic product of water ($K_w = 10^{-14}$).

$$ K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{6.8 \times 10^{-4}} $$ $$ K_b \approx \mathbf{1.47 \times 10^{-11}} $$

13. The Concentration-Independent Salt

Salt Hydrolysis

Scenario: Compare the pH of a $0.1\text{M}$ solution of Ammonium Acetate ($CH_3COONH_4$) with a $0.01\text{M}$ solution of the same salt.

✖ What Students Do

Student assumes that because the $0.01\text{M}$ solution is more dilute, it must be closer to neutral (pH 7) than the $0.1\text{M}$ solution.

(This is true for strong acids/bases, but a massive trap for this specific type of salt!)

✔ The Correct Way

Weak Acid + Weak Base salts are INDEPENDENT of concentration!

For a salt derived from a Weak Acid and a Weak Base, the pH formula is:
$$ pH = 7 + \frac{1}{2}(pK_a - pK_b) $$
Notice that the concentration term ($C$) is completely missing from this formula. The pH depends entirely on the relative strengths of the parent acid and base.
Answer: Both solutions have the exact same pH!

14. Ostwald's Dilution Paradox

Ionic Equilibrium

Scenario: You add $100\text{mL}$ of pure water to a $0.1\text{M}$ solution of a weak acid ($HA$). What happens to its Degree of Dissociation ($\alpha$)?

✖ What Students Do

Student thinks: "Adding water lowers the concentration. Since the acid is weaker, it will dissociate less."

Answer: "Degree of dissociation decreases."

✔ The Correct Way

Dilution INCREASES dissociation!

According to Ostwald's Dilution Law, the degree of dissociation ($\alpha$) of a weak electrolyte is inversely proportional to the square root of its concentration ($C$).
$$ \alpha = \sqrt{\frac{K_a}{C}} $$
When you add water, concentration ($C$) decreases. Therefore, to maintain the constant $K_a$, the acid must break apart more. $\alpha$ increases.

15. Pressure Shift on $\Delta n = 0$

Le Chatelier's Principle

Scenario: For the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, the volume of the container is suddenly halved (doubling the pressure). Which way does the reaction shift?

✖ What Students Do

Student applies a rigid rule: "Increasing pressure always shifts equilibrium forward to form product." or tries to guess based on standard textbook examples.

✔ The Correct Way

Calculate $\Delta n_g$ before blindly applying rules!

Le Chatelier's principle states that an increase in pressure shifts the equilibrium towards the side with fewer moles of gas.
For this reaction:
Reactant moles = $1 + 1 = 2$
Product moles = $2$
Since the moles of gas are equal on both sides ($\Delta n_g = 0$), changing the pressure or volume has absolutely zero effect on the equilibrium position. No shift occurs!

Confess Your Sins!

"Chemical equilibrium is dynamic. Your understanding of it should be stable."

Did one of these Le Chatelier traps catch you? Or do you have a different horror story from your last exam?

Scroll down to the comments section below and tell us:

"Which equilibrium mistake were you making?"

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