The Mistake Bank
Chapter 7: Equilibrium
Balance is key. Both in reactions and in your logic.
The "pH = 8" Trap
Ionic EquilibriumScenario: Calculate the pH of a $10^{-8} \text{ M }$ $HCl$ solution.
Student applies formula blindly:
$$ pH = -\log[H^+] = -\log(10^{-8}) = 8 $$
(Impossible! An acid cannot have basic pH > 7.)
Don't Forget Water!
For very dilute acids ($< 10^{-6}M$), you must add the $H^+$ from water ($10^{-7}M$).
Total $[H^+] = 10^{-8} + 10^{-7}$
$$ [H^+] = 1.1 \times 10^{-7} $$
$$ pH = 7 - \log(1.1) \approx \mathbf{6.98} $$
Kp vs Kc Relationship
Chemical EquilibriumScenario: Write the relation between $K_p$ and $K_c$ for $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
Student calculates $\Delta n$ incorrectly or uses the wrong sign.
$$ \Delta n = (1+3) - 2 = 2 $$
$$ K_p = K_c(RT)^2 $$
(Wrong direction of subtraction!)
Products minus Reactants!
$$ \Delta n_g = n_p - n_r $$
$$ \Delta n_g = 2 - (1 + 3) = -2 $$
$$ K_p = K_c(RT)^{-2} $$
Note: Only count gaseous moles!
Solids in Equilibrium
Equilibrium ConstantScenario: Write the expression for $K_c$ for: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$
Student includes all species:
$$ K_c = \frac{[CaO][CO_2]}{[CaCO_3]} $$
(Pure solids and liquids have constant active mass.)
Ignore Pure Solids & Liquids!
Their concentration effectively remains constant.
$$ K_c = [CO_2] $$
$$ K_p = P_{CO_2} $$
Weak Acid Calculation
Ionic EquilibriumScenario: Find the pH of $0.1 \text{ M }$ Acetic Acid ($K_a = 1.8 \times 10^{-5}$).
Student treats it like a Strong Acid:
$$ pH = -\log(0.1) = 1 $$
(Weak acids do not dissociate completely!)
Use Ostwald's Dilution Law!
$$ [H^+] = \sqrt{K_a \cdot C} $$
$$ [H^+] = \sqrt{1.8 \times 10^{-5} \times 0.1} $$
$$ [H^+] \approx 1.34 \times 10^{-3} $$
$$ pH \approx 2.87 $$
Solubility Product (Ksp)
SolubilityScenario: Calculate Solubility ($S$) for $CaF_2$ given $K_{sp}$.
Student assumes simple stoichiometry:
$$ K_{sp} = S^2 $$
(This works for AgCl, but not for $MX_2$ type salts.)
Write the dissociation!
$$ CaF_2 \rightleftharpoons Ca^{2+} (S) + 2F^- (2S) $$
$$ K_{sp} = [Ca^{2+}][F^-]^2 $$
$$ K_{sp} = (S)(2S)^2 = 4S^3 $$
Le Chatelier & Inert Gas
Le Chatelier's PrincipleScenario: What happens if you add Inert Gas to a reaction at Constant Volume?
Student thinks: "Adding gas increases pressure, so reaction shifts to side with fewer moles."
No Effect at Constant Volume!
Adding inert gas at constant volume increases Total Pressure, but Partial Pressures of reactants/products remain unchanged.
Equilibrium does not shift.
(It only shifts if added at Constant Pressure.)
Identifying Buffers
Buffer SolutionsScenario: Which mixture forms a buffer?
A) $100mL$ $0.1M$ $HCl$ + $100mL$ $0.1M$ $NaOH$
B) $100mL$ $0.1M$ $CH_3COOH$ + $50mL$ $0.1M$ $NaOH$
Student selects A because it's a "standard neutralization".
Or rejects B because "strong base destroys the acid".
Buffer = Weak Component + Its Salt
A) Strong Acid + Strong Base = Neutral Salt ($NaCl$). Not a buffer.
B) The NaOH consumes half the Acetic Acid.
Result: Leftover $CH_3COOH$ + formed $CH_3COONa$.
This IS a Buffer!
Confess Your Sins!
"Chemical equilibrium is dynamic. Your understanding of it should be stable."
Did one of these catch you? Or do you have a different horror story from your last exam?
Scroll down to the comments section below and tell us:
No comments:
Post a Comment