Solutions, Gases & Electrolysis

Solutions, Gases & Electrolysis

Part 1 of 3 — Solutions: Molarity, Dilution & Solution Stoichiometry

Where this fits. The Stoichiometry: Moles & Reactions lesson built the mole conversions and the balanced-equation mole ratio (moles of one substance → moles of another, using the equation's coefficients). This lesson carries that same bookkeeping into solutions, gases and electrolysis.

Concentration Units

MCAT emphasis: Molarity is by far the most frequently tested. Molality appears in colligative property problems.

Preparing a Solution

To make a solution of known molarity, find the moles it must hold from $n = M \times V$ (volume in litres), then weigh out that many moles of solute.

Example — Preparing a Solution

How many grams of $\text{NaOH}$ ($M = 40.0$ g/mol) make 250 mL of a 0.500 M solution? Convert the volume to litres first, find the moles, then the mass:

$n = M \times V_{\text{L}} = 0.500\text{ M} \times 0.250\text{ L} = 0.125\text{ mol}$

$m = 0.125\text{ mol} \times 40.0\text{ g/mol} = \mathbf{5.00\text{ g}}$

Dilution

When a solution is diluted (solvent added), the amount of solute doesn't change:

$M_1 V_1 = M_2 V_2$

Problem: 25.0 mL of 12.0 M HCl is diluted to 300 mL. What is the final concentration?

$M_2 = \frac{M_1 V_1}{V_2} = \frac{12.0 \times 25.0}{300} = \mathbf{1.00\text{ M}}$

Lab safety note: Always add acid to water, never water to acid (heat released causes dangerous spattering).

Solution Stoichiometry

Combine molarity calculations with balanced-equation ratios: volume × molarity gives moles, the mole ratio carries them across the equation, and moles ÷ molarity gives back a volume.

Normality counts reacting equivalents per litre: $N = M \times$ (H⁺ or OH⁻ per formula unit), so 0.200 M $\text{H}_2\text{SO}_4$ is 0.400 N, and at neutralization $N_aV_a = N_bV_b$.

Worked Example — Neutralizing a Base with Sulfuric Acid

Problem. 50.0 mL of 0.300 M $\text{NaOH}$ is neutralized with 0.200 M $\text{H}_2\text{SO}_4$:

$\text{H}_2\text{SO}_4 + 2\text{NaOH} \to \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

• (a) How many moles of NaOH must be neutralized?
• (b) How many moles of $\text{H}_2\text{SO}_4$ does that take?
• (c) What volume of the acid holds that many moles?
• (d) Does $N_aV_a = N_bV_b$ give the same volume?

(a) Moles of base. Moles NaOH $= 0.300 \times 0.0500 = 0.0150$ mol

(b) Mole ratio. The equation uses 2 NaOH per $\text{H}_2\text{SO}_4$, so the 0.0150 mol from (a) needs: $0.0150/2 = 0.00750\text{ mol}$

(c) Volume of acid. With 0.00750 mol from (b): $V = n/M = 0.00750/0.200 = 0.0375\text{ L} = \mathbf{37.5\text{ mL}}$

(d) Normality check. The acid gives 2 H⁺ each, so it is 0.400 N; NaOH is 0.300 N. $V_a = (0.300)(50.0)/0.400 = 37.5$ mL, matching (c).

Colligative Properties (Molality-Based)

Properties that depend on the number (not identity) of solute particles:

$i$ = van 't Hoff factor (number of particles per formula unit after dissociation):

• Nonelectrolytes (glucose, urea): $i = 1$
• NaCl: $i = 2$; $\text{CaCl}_2$: $i = 3$; $\text{Al}_2(\text{SO}_4)_3$: $i = 5$

Example — Osmotic pressure:
$1.0\text{ M}$ NaCl solution at 37°C ($T = 310$ K):

$\Pi = iMRT = 2 \times 1.0 \times 0.0821 \times 310 \approx \mathbf{50.9\text{ atm}}$

Solution Concentration & Dilution 🎯

Net Ionic Equations

In solution, strong electrolytes dissociate completely. A net ionic equation shows only the species that actually participate in the reaction (spectator ions are omitted).

Worked Example — Precipitating Silver Chloride

Problem. Aqueous $\text{AgNO}_3$ and aqueous NaCl are mixed, and a white solid, AgCl, forms.

• (a) What is the molecular equation?
• (b) What is the complete ionic equation?
• (c) Which ions are spectators, and what is the net ionic equation?

(a) Molecular equation. $\text{AgNO}_3(aq) + \text{NaCl}(aq) \to \text{AgCl}(s) + \text{NaNO}_3(aq)$

(b) Complete ionic. Split every soluble ionic compound into its ions; the solid AgCl stays whole: $\text{Ag}^+(aq) + \text{NO}_3^-(aq) + \text{Na}^+(aq) + \text{Cl}^-(aq) \to \text{AgCl}(s) + \text{Na}^+(aq) + \text{NO}_3^-(aq)$

(c) Cancel spectators. $\text{Na}^+$ and $\text{NO}_3^-$ appear unchanged on both sides, so they cancel: $\boxed{\text{Ag}^+(aq) + \text{Cl}^-(aq) \to \text{AgCl}(s)}$

Solubility Rules (MCAT-Relevant)

Always soluble: nitrates ($\text{NO}_3^-$), alkali metals ($\text{Na}^+$, $\text{K}^+$), ammonium ($\text{NH}_4^+$)

Usually insoluble: $\text{Ag}^+$ halides (except $\text{AgF}$), $\text{Ba}^{2+}$/$\text{Pb}^{2+}$/$\text{Ca}^{2+}$ sulfates, most carbonates and phosphates (except Group IA and $\text{NH}_4^+$), most hydroxides (except Group IA and $\text{Ba}^{2+}$; $\text{Ca}^{2+}$ and $\text{Sr}^{2+}$ hydroxides are only slightly soluble)

Predicting a Precipitate ($Q$ vs. $K_{sp}$)

The solubility rules are tendencies. When concentrations and $K_{sp}$ are given, compute $Q$ (each ion's concentration raised to its coefficient, multiplied) and compare: a precipitate forms only if $Q > K_{sp}$, even for an "insoluble" salt.

Net Ionic Equations & Precipitation 🎯

Key Takeaways — Part 1

• Molarity and dilution. $M = n/V$ with $V$ in litres; dilution keeps the moles fixed, so $M_1V_1 = M_2V_2$.
• Solution stoichiometry. Get moles from $n = MV$, carry them across the balanced equation's ratio, then convert back.
• Colligative properties. $\Delta T_b$, $\Delta T_f$ and $\Pi$ scale with particle count, $i$ times concentration, not identity.
• Net ionic and precipitation. Cancel the spectator ions; a precipitate forms when $Q > K_{sp}$.

Common traps

• Millilitres in $n = MV$. Convert mL to L first: 250 mL of 0.500 M holds 0.125 mol, not 125 mol.
• Forgetting $i$. NaCl gives 2 particles and $\text{CaCl}_2$ gives 3, so equal molalities do not give equal $\Delta T_f$.
• Ignoring the 2 : 1 ratio. One $\text{H}_2\text{SO}_4$ neutralizes two NaOH, so the acid needs half the moles of base.
• Splitting the solid. Only dissolved strong electrolytes are written as ions; a precipitate stays as one formula.

Solutions, Gases & Electrolysis

Part 2 of 3 — Gas Laws & Electrochemistry Stoichiometry

Ideal Gas Law

$PV = nRT$

$P$ = pressure (atm), $V$ = volume (L), $n$ = moles, $R = 0.0821$ L·atm/mol·K, $T$ = temperature (K)

Always convert temperature to Kelvin: $T(\text{K}) = T(°\text{C}) + 273$

Combined Gas Law (fixed $n$)

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Special Cases

Worked Example: Ideal Gas Law

Problem. 6.54 g of zinc ($M = 65.4$ g/mol) reacts with excess HCl, and the $\text{H}_2$ is collected dry at 27°C and 1.00 atm ($R = 0.0821$ L·atm/mol·K):

$\text{Zn} + 2\text{HCl} \to \text{ZnCl}_2 + \text{H}_2$

• (a) How many moles of Zn react?
• (b) How many moles of $\text{H}_2$ form?
• (c) What volume does that $\text{H}_2$ occupy?

(a) Moles of zinc. $n = 6.54/65.4 = 0.100$ mol

(b) Mole ratio. One $\text{H}_2$ forms per Zn, so the 0.100 mol from (a) gives 0.100 mol $\text{H}_2$.

(c) Volume. Convert to kelvin, $T = 27 + 273 = 300$ K, then with 0.100 mol from (b): $V = \frac{nRT}{P} = \frac{0.100 \times 0.0821 \times 300}{1.00} = \mathbf{2.46\text{ L}}$

Dalton's Law of Partial Pressures

$P_{\text{total}} = P_A + P_B + P_C + \cdots$

Each gas in a mixture exerts its partial pressure independently, and each contributes in proportion to its mole fraction: $P_A = \chi_A \times P_{\text{total}}$.

MCAT application — gas collected over water: the tube holds the gas plus water vapor, so $P_{\text{gas}} = P_{\text{atm}} - P_{\text{H}_2\text{O (vapor)}}$.

Faraday's Law: Electrochemical Stoichiometry

In electrolysis, electric current drives non-spontaneous redox reactions. The amount of product depends on the charge passed.

$\text{moles of electrons} = \frac{I \times t}{F}$

where $I$ = current (amperes), $t$ = time (seconds), $F = 96{,}485$ C/mol $\approx 96{,}500$ C/mol.

$\text{moles of product} = \frac{\text{moles of electrons}}{n}$

where $n$ = electrons required per atom/ion reduced or oxidized.

Worked Example — Electroplating Copper

Problem. A current of 2.00 A passes for 1930 s through $\text{CuSO}_4$ solution, plating copper: $\text{Cu}^{2+} + 2e^- \to \text{Cu}$ ($M_{\text{Cu}} = 63.5$ g/mol, $F \approx 96{,}500$ C/mol).

• (a) How many moles of electrons pass through the cell?
• (b) How many moles of Cu deposit?
• (c) What mass of Cu is that?

(a) Moles of electrons. $\text{mol } e^- = \frac{2.00 \times 1930}{96{,}500} = \frac{3860}{96{,}500} = 0.0400\text{ mol}$

(b) Moles of copper. Each $\text{Cu}^{2+}$ takes 2 electrons, so the 0.0400 mol from (a) gives: $\text{mol Cu} = \frac{0.0400}{2} = 0.0200\text{ mol}$

(c) Mass of copper. With 0.0200 mol from (b): $\text{mass Cu} = 0.0200 \times 63.5 = \mathbf{1.27\text{ g}}$

Gas Laws 🎯

Graham's Law of Effusion

The rate at which a gas escapes through a small orifice depends on its molar mass:

$\frac{\text{rate}_A}{\text{rate}_B} = \sqrt{\frac{M_B}{M_A}}$

Lighter gases effuse faster. MCAT application: Compare rates of diffusion/effusion for two gases at the same T and P.

Example: Which effuses faster — $\text{H}_2$ or $\text{O}_2$, and by what factor? $\text{rate}_{\text{H}_2}/\text{rate}_{\text{O}_2} = \sqrt{32/2} = \sqrt{16} = 4$, so $\text{H}_2$ effuses 4 times faster than $\text{O}_2$.

Real Gases vs. Ideal Gases

Ideal gas assumptions (particles have no volume, no intermolecular forces) deviate at:

• High pressure (volume of particles becomes significant)
• Low temperature (attractive forces between molecules become significant)

van der Waals equation: $\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$

• $a$ accounts for intermolecular attractions: they pull particles back from the walls, so the measured pressure is LOWER than the ideal-gas value
• $b$ accounts for molecular volume: the particles take up space, so the measured volume is LARGER than the ideal-gas value

On the MCAT, you don't calculate with van der Waals, but you must identify when gases deviate from ideal behavior and in which direction.

Faraday's Law & Graham's Law 🎯

Key Takeaways — Part 2

• Ideal gas law. $PV = nRT$ with $R = 0.0821$ L·atm/mol·K and $T$ in kelvin; for fixed $n$, $P_1V_1/T_1 = P_2V_2/T_2$.
• Dalton's law. Partial pressures add up to the total, and each gas contributes $P_i = \chi_i P_{\text{total}}$.
• Graham's law. $\text{rate}_A/\text{rate}_B = \sqrt{M_B/M_A}$: the lighter gas effuses faster.
• Faraday's law. mol $e^- = It/F$; divide by the electrons each ion gains or loses to get mol of product.

Common traps

• Celsius in gas laws. Gas-law temperatures are in kelvin: warming 27°C to 54°C takes 300 K to 327 K, not double.
• Wet gas pressure. For a gas collected over water, subtract the water's vapor pressure from the total first.
• Forgetting the square root. A 16-fold mass ratio gives only a 4-fold rate ratio, and the heavier gas is the slower one.
• Backwards real-gas deviations. Attractions make real $P$ lower than ideal, and particle volume makes real $V$ larger, not the reverse.

Solutions, Gases & Electrolysis

Part 3 of 3 — Mixed MCAT Review

This section mixes the lesson's topics the way an MCAT passage does. Each question may draw on solutions, colligative properties, gas laws, electrolysis, or several at once.

High-Yield Checklist Before You Practice

• ✅ Molarity $M = n/V(\text{L})$; dilution: $M_1V_1 = M_2V_2$
• ✅ Ideal gas law $PV = nRT$; $R = 0.0821$ L·atm/mol·K; $T$ in Kelvin
• ✅ Dalton's law $P_{\text{total}} = \sum P_i$
• ✅ Graham's law rate $\propto 1/\sqrt{M}$
• ✅ Colligative: $\Delta T_f = K_f m i$; $\Delta T_b = K_b m i$; $\Pi = iMRT$
• ✅ Faraday: mol $e^- = It/F$; $F \approx 96{,}500$ C/mol
• ✅ Net ionic equations: cancel spectator ions; know solubility rules

Mixed Review — MCAT-Style Questions 🎯

Solutions, Gases & Electrolysis — Complete Topic Summary

Part 1: Molarity, dilution ($M_1V_1 = M_2V_2$), solution stoichiometry, colligative properties, van 't Hoff factor, net ionic equations, solubility rules, $Q$ vs $K_{sp}$.

Part 2: Ideal gas law ($PV=nRT$), combined gas law, Dalton's partial pressures, Graham's law of effusion, real gas deviations, Faraday's law of electrolysis.

Part 3: Integrated MCAT practice combining all above topics.

Key Takeaways — Part 3

• Moles are the hub. $n = MV$ for solutions, $n = PV/RT$ for gases and mol $e^- = It/F$ for electrolysis all feed the same balanced-equation mole ratio.
• Count particles, not formulas. Colligative effects scale with $i \times m$ (or $i \times M$ for $\Pi$): $\text{CaCl}_2$ counts three times, glucose once.
• Kelvin and litres first. Every gas-law temperature is in kelvin and every $n = MV$ volume is in litres before any arithmetic.
• Precipitates need $Q > K_{sp}$. The solubility rules say what usually happens; when concentrations are given, the numbers decide.

Common traps

• °C in gas laws. 27°C is 300 K, not 27; using the Celsius value shrinks $nRT$ about elevenfold.
• Forgetting $i$. 1.0 m NaCl raises the boiling point by $2K_b$, not $K_b$; $\text{Al}_2(\text{SO}_4)_3$ counts five particles.
• Wet gas at full pressure. A gas collected over water is at $P_{\text{atm}} - P_{\text{H}_2\text{O}}$; subtract the vapor pressure before $PV = nRT$.
• Electrons are not atoms. Divide mol $e^-$ by the electrons each ion needs: 2 for $\text{Cu}^{2+}$, 3 for $\text{Al}^{3+}$, 1 for $\text{Ag}^+$.

Next up: the Thermodynamics lesson — enthalpy and Hess's law, entropy, Gibbs free energy, and phase changes.