Lessons

25.0 mL of 0.100 M CaCl₂(aq) is mixed with 20.0 mL of 0.150 M Na₂CO₃(aq). A white precipitate forms. What mass of calcium carbonate forms, which reactant limits the reaction, and which ions remain in solution?

30.0 mL of 0.100 M CaCl₂(aq) is mixed with 25.0 mL of 0.100 M Na₃PO₄(aq). A white precipitate of calcium phosphate forms. What mass of Ca₃(PO₄)₂ forms, which reactant limits the reaction, and which ions are left in solution? Watch the coefficients: the balanced reaction consumes calcium and phosphate in a 3-to-2 ratio, so more moles does not mean more room to react.

In a gravimetric analysis, 25.0 mL of 0.100 M ZnCl₂(aq) is mixed with 20.0 mL of 0.150 M Na₂CO₃(aq); the zinc carbonate precipitate is filtered, dried, and weighed. 0.276 g of ZnCO₃ is recovered. What mass should form in theory, which reactant limits it, and what percent yield does the recovered solid represent?

20.0 mL of 0.100 M HCl(aq) is mixed with 30.0 mL of 0.0500 M NaOH(aq). The acid's H+ and the base's OH- combine into water, leaving dissolved sodium chloride. How much water forms, which reactant runs out first, and what stays in solution?

A 3.269 g strip of zinc metal is dropped into 120.0 mL of 1.00 M hydrochloric acid. The zinc dissolves, bubbling hydrogen gas, which is collected at 1.00 atm and 25.00 °C. Which reactant runs out first, and what volume of hydrogen forms?

A burner burns 0.80215 g of methane (CH₄) in 3.83976 g of oxygen. The reaction goes to completion, and the heat released is measured. Which reactant runs out first, and how much heat does this burn actually release?

Water is the most familiar molecule there is, and almost everyone first pictures it as a straight line: H–O–H. It isn't. Water is bent — and that bend is the reason it dissolves salt, climbs up plant stems, and boils at 100 °C instead of somewhere below −80 °C like a nonpolar molecule its size. We'll build H₂O one accounting step at a time — count its valence electrons, place them into bonds and lone pairs, read the shape off the electron domains, and let the shape decide the polarity — and watch where the 'straight line' picture breaks.

Carbon dioxide has two carbon–oxygen bonds, and oxygen pulls electrons hard, so each bond is polar. It seems obvious the molecule should be polar too. It isn't — CO₂ is perfectly nonpolar, and it comes down to shape. Water, built from the same 'one central atom, two outer atoms' recipe, is bent and polar; CO₂ is straight and nonpolar. We'll build CO₂ one accounting step at a time and find exactly where the two stories split — and why polar bonds do not add up to a polar molecule.

Methane (CH₄), ammonia (NH₃), and water (H₂O) are almost exactly the same size — 16, 17, and 18 grams per mole, three atoms of the second row wrapped around hydrogens. Yet methane boils at −161 °C, ammonia at −33 °C, and water not until +100 °C. That is a spread of more than 250 degrees between molecules of the same mass. If size isn't doing it, what is? The answer is the force acting between the molecules — and it lines up exactly with the boiling point.

Dissolve enough acetic acid — the acid in vinegar — to make the solution $0.100\,\text{M}$. Acetic acid is a weak acid: put it in water and only a small fraction of its molecules hand a proton to a water molecule, so the ionization $\mathrm{HC_2H_3O_2} \rightleftharpoons \mathrm{H^+} + \mathrm{C_2H_3O_2^-}$ settles at an equilibrium short of completion. What is the pH? The move is the ICE table — the very same species ledger as any reaction, initial → change → equilibrium — but instead of running to a limiting reagent, the extent stops where the forward and reverse rates balance: where the reaction quotient equals $K_a$.

Dissolve enough ammonia to make the solution $0.100\,\text{M}$. Ammonia is a weak base: it has no $\mathrm{OH^-}$ of its own — instead it pulls a proton off a water molecule, $\mathrm{NH_3} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_4^+} + \mathrm{OH^-}$, and only a small fraction reacts, so the equilibrium sits far to the left. What is the pH? It is the ICE table once more — the same species ledger, initial → change → equilibrium — but with two new twists. First, water is the pure solvent: its activity is 1, so it never enters the equilibrium expression $K_b = [\mathrm{NH_4^+}][\mathrm{OH^-}]/[\mathrm{NH_3}]$ — exactly as a pure solid drops out of a $K_{sp}$. Second, the extent gives $[\mathrm{OH^-}]$, not $[\mathrm{H^+}]$; the bridge to pH is the ion-product of water, $[\mathrm{H^+}][\mathrm{OH^-}] = K_w = 1.0\times10^{-14}$, which is why $\mathrm{pH} + \mathrm{pOH} = 14.00$.

Dissolve $0.100\,\text{M}$ acetic acid and $0.100\,\text{M}$ sodium acetate in the same beaker. The acetate ion $\mathrm{C_2H_3O_2^-}$ from the salt is the acid's own conjugate base — so the solution now holds both halves of the equilibrium $\mathrm{HC_2H_3O_2} \rightleftharpoons \mathrm{H^+} + \mathrm{C_2H_3O_2^-}$ from the start. This is a buffer. It is still the ICE table — the same species ledger, the same solver — but the change column starts from $[\mathrm{C_2H_3O_2^-}]_0 = 0.100\,\text{M}$, not zero. That pre-loaded product is a common ion: by Le Chatelier it pushes the ionization left, so hardly any acid ionizes and the pH lands near $\mathrm{p}K_a$. What is it?

Phosphoric acid, $\mathrm{H_3PO_4}$, has three acidic protons — it is triprotic. But it does not give them up all at once: it ionizes in stages, each with its own equilibrium constant, and each $K_a$ is about $10^5$ smaller than the last ($K_{a1} = 7.5\times10^{-3}$, $K_{a2} = 6.2\times10^{-8}$, $K_{a3} = 4.2\times10^{-13}$). So the first ionization is the only one that releases an appreciable amount of $\mathrm{H^+}$ — and it is just the weak-acid ICE table again. Each later stage is the same solve, run on the concentrations the stage before it left behind.

Calcium fluoride — the mineral fluorite — is called “insoluble,” but no salt is perfectly insoluble: a tiny amount dissolves until the solution is saturated, $\mathrm{CaF_2(s)} \rightleftharpoons \mathrm{Ca^{2+}(aq)} + 2\,\mathrm{F^-(aq)}$. How much? It is the ICE table again — but the dissolving species is a pure solid, whose activity is 1, so it never appears in the equilibrium expression: $K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^-}]^2$. The extent of this reaction is the molar solubility $s$ — solve $K_{sp}$ for it.

In the last lesson, $\mathrm{CaF_2}$ dissolved in pure water. Now dissolve it in water that is already **0.10 M in $\mathrm{F^-}$** — a solution of sodium fluoride, $\mathrm{NaF}$, which dissociates completely. The $\mathrm{Na^+}$ is a spectator, but the $\mathrm{F^-}$ is shared with the salt: it is a common ion. The very same equilibrium, $\mathrm{CaF_2(s)} \rightleftharpoons \mathrm{Ca^{2+}(aq)} + 2\,\mathrm{F^-(aq)}$, now starts with $\mathrm{F^-}$ already on the product side. By Le Chatelier that extra $\mathrm{F^-}$ pushes the dissolution backward — so how much less dissolves? It is the ICE table again, with one column no longer starting at zero.

Add sodium hydroxide, a strong base, drop by drop to 25.0 mL of 0.100 M acetic acid and watch the pH. Each drop neutralizes a little acid, $\mathrm{HC_2H_3O_2 + OH^- \rightarrow C_2H_3O_2^- + H_2O}$ — but the pH at every moment is still set by the acetic-acid equilibrium $\mathrm{HC_2H_3O_2 \rightleftharpoons H^+ + C_2H_3O_2^-}$. So the whole curve is just the ICE table marched: a weak acid at the start, a buffer through the middle (flattest right where half the acid is neutralized — there $\mathrm{pH} = \mathrm{p}K_a$), a jump through the equivalence point (where all the acid has become acetate, a weak base — so the equivalence pH is above 7), and excess strong base after.

Two clear solutions sit on the bench — calcium nitrate, Ca(NO₃)₂, and sodium fluoride, NaF. Neither holds a speck of solid; both are perfectly transparent. Pour 40.0 mL of 0.010 M Ca(NO₃)₂ into 60.0 mL of 0.010 M NaF and does anything happen? Calcium fluoride is the sparingly soluble pair here — $\mathrm{CaF_2(s)} \rightleftharpoons \mathrm{Ca^{2+}(aq)} + 2\,\mathrm{F^-(aq)}$ — but nothing says a solid must appear. The test is the reaction quotient: mix, dilute each ion into the combined 100.0 mL, evaluate $Q = [\mathrm{Ca^{2+}}][\mathrm{F^-}]^2$ at that instant, and compare it to $K_{sp}$. If $Q > K_{sp}$ the solution is supersaturated and $\mathrm{CaF_2}$ crashes out; if $Q < K_{sp}$ it stays clear. No ICE table to solve — one comparison decides it.

Milk of magnesia is a suspension of solid magnesium hydroxide, Mg(OH)₂ — the textbook “insoluble” base. So surely mixing a magnesium salt with a hydroxide makes the solid appear? Stir 50.0 mL of 0.00010 M Mg(NO₃)₂ (magnesium nitrate) into 50.0 mL of 0.00010 M NaOH — both perfectly clear. The dissolution equilibrium is $\mathrm{Mg(OH)_2(s)} \rightleftharpoons \mathrm{Mg^{2+}(aq)} + 2\,\mathrm{OH^-(aq)}$, and whether a solid forms is decided by the reaction quotient: dilute each ion into the combined 100.0 mL, evaluate $Q = [\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2$, and compare it to $K_{sp}$. If $Q < K_{sp}$ the solution is unsaturated and stays clear — however “insoluble” the compound is called.

A bottle of hydrogen peroxide slowly falls apart on the shelf: 2 H₂O₂ → 2 H2O + O2. The reaction is first order in H₂O₂ — its rate is proportional to how much is left, $\text{rate} = k[\mathrm{H_2O_2}]$, with $k = 3.21\times10^{-5}\ \mathrm{s^{-1}}$. That one fact fixes the whole future: the amount left follows the integrated rate law $[\mathrm{H_2O_2}] = [\mathrm{H_2O_2}]_0\,e^{-kt}$, a smooth exponential decay from 1.000 M. Its half-life — the time to lose half — is $t_{1/2} = \ln 2 / k$, and for a first-order reaction it does not depend on how much you start with. This is the species ledger again, but with the extent marching in time.

Two butadiene molecules snap together into a ring: 2 C₄H₆ → C8H12. The reaction is second order in butadiene — its rate depends on two C₄H₆ molecules colliding, $\text{rate} = k[\mathrm{C_4H_6}]^2$, with $k = 5.76\times10^{-2}\ \mathrm{M^{-1}\,min^{-1}}$. The amount left follows the second-order integrated law $\frac{1}{[\mathrm{C_4H_6}]} = \frac{1}{[\mathrm{C_4H_6}]_0} + kt$ (it is $1/[\mathrm{C_4H_6}]$, not $[\mathrm{C_4H_6}]$, that climbs in a straight line). Its half-life is $t_{1/2} = 1/(k[\mathrm{C_4H_6}]_0)$ — and because $[\mathrm{C_4H_6}]_0$ sits in the denominator, the half-life grows as the reactant thins out: the second half takes longer than the first. Same species ledger, extent marching in time — but a very different clock from first order.

On a hot tungsten wire, ammonia breaks down: 2 NH₃ → N2 + 3 H2. The surface is saturated with NH₃, so speeding it up by adding more does nothing — the reaction is zero order: $\text{rate} = k$, a constant, with $k = 1.3\times10^{-6}\ \mathrm{M\,s^{-1}}$. The amount left falls in a straight line, $[\mathrm{NH_3}] = [\mathrm{NH_3}]_0 - kt$, so unlike an exponential decay it reaches exactly zero at a finite time $t = [\mathrm{NH_3}]_0/k$. Its half-life is $t_{1/2} = [\mathrm{NH_3}]_0/(2k)$ — proportional to how much is present, so as the ammonia runs low the half-life shrinks: the last half disappears faster than the first. The species ledger again, extent marching in time — at a steady pace that never slackens.

Dip a strip of zinc into a zinc-sulfate solution and a strip of copper into a copper-sulfate solution, then connect the two metals with a wire and bridge the beakers with a salt bridge. A current flows — the cell does electrical work — and the reason is redox: one metal gives up electrons and the other takes them. This is the species ledger again, but now the quantity that moves is electrons. Which way do they go? Track the oxidation numbers: zinc goes from 0 in the metal to +2 in solution (it is oxidized — it loses electrons, at the anode), while copper(II) goes from +2 to 0 (it is reduced — it gains electrons, at the cathode). The two standard reduction potentials decide the direction and the voltage, and the free energy follows from $\Delta G^\circ = -nFE^\circ$.