Acid-base (neutralization) and redox reactions

There are three big families of aqueous reactions: precipitation (week 3), acid-base and redox. This page covers the last two. On the acid-base side the calculation question is almost always a titration: measuring the concentration of one solution with another solution of known concentration. On the redox side two skills are required: finding the oxidation number and balancing the equation with the half-reaction method.

Golden rules:

1. At the equivalence point, mol $\mathrm{H^+}$ = mol $\mathrm{OH^-}$. Pay attention to how many $\mathrm{H^+}$ the acid gives.
2. Weak acids and bases are written as molecules in the net ionic equation.
3. Oxidation = loss of electrons, reduction = gain of electrons. If one exists, the other exists too.
4. A redox equation must be balanced in both atoms and charge.

Acids, bases and neutralization

By the Arrhenius definition, an acid gives $\mathrm{H^+}$ (really $\mathrm{H_3O^+}$) in water, and a base gives $\mathrm{OH^-}$ in water. The Brønsted-Lowry definition is broader: an acid is a proton donor, a base is a proton acceptor. That is why $\mathrm{NH_3}$ is a base: $\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}$.

Treat acids not on the list (HF, $\mathrm{CH_3COOH}$, $\mathrm{H_3PO_4}$, $\mathrm{H_2CO_3}$) as weak.

Neutralization: acid + base → salt + water.

$\mathrm{HNO_3(aq) + KOH(aq) \rightarrow KNO_3(aq) + H_2O(l)}$

For a strong acid with a strong base the net ionic equation is always the same: $\mathrm{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}$. With a weak acid, the acid is written as a molecule.

Titration

In a titration, a solution of known concentration (the titrant) is added drop by drop from a buret. The moment of exact neutralization is called the equivalence point; the point where an indicator (like phenolphthalein) changes color (the end point) is very close to it.

The calculation recipe:

1. Write the balanced equation (H2SO4 + 2 NaOH !)
2. The known solution: mol = M × V(L)
3. Moles of the other one from the coefficient ratio
4. The wanted: M = mol / V   or   g = mol × M

Example 1 (standardization). To find the concentration of a NaOH solution, 0.4085 g of KHP ($\mathrm{KHC_8H_4O_4}$, 204.22 g/mol, monoprotic) is titrated. 20.00 mL of NaOH is used.

$\frac{0.4085\ \text{g}}{204.22\ \text{g/mol}} = 2.000\times10^{-3}\ \text{mol KHP} = \text{mol NaOH} \qquad M = \frac{2.000\times10^{-3}}{0.02000\ \text{L}} = 0.1000\ \text{M}$

Example 2 (diprotic acid). $\mathrm{H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O}$. One mole of sulfuric acid consumes two moles of NaOH. Someone who memorizes the formula as “mol acid = mol base” gets burned in this question.

Example 3 (base with two hydroxides). 15.00 mL of a $\mathrm{Ba(OH)_2}$ solution is neutralized by 22.40 mL of 0.1000 M HCl. $\mathrm{Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O}$.

mol HCl     = 0.02240 L × 0.1000 M = 2.240×10^-3 mol
mol Ba(OH)2 = 2.240×10^-3 / 2      = 1.120×10^-3 mol
M Ba(OH)2   = 1.120×10^-3 / 0.01500 L = 0.07467 M

PITFALL · $M_1V_1 = M_2V_2$ in titration. This formula is for dilution. In titration it works only for 1:1 reactions. If the ratio is 1:2, you need to include the coefficient. The safe way is always to go through moles.

Oxidation numbers

The oxidation number is the charge an atom would carry if all bonds were ionic. It is a bookkeeping tool: it lets you track where the electrons go. The rules (the earlier ones take priority):

Example 4. Mn in $\mathrm{KMnO_4}$: $+1 + x + 4(-2) = 0 \Rightarrow x = +7$. C in $\mathrm{C_2O_4^{2-}}$: $2x + 4(-2) = -2 \Rightarrow x = +3$.

Redox: who oxidizes whom?

• Oxidation: loss of electrons, the oxidation number increases.
• Reduction: gain of electrons, the oxidation number decreases.
• Oxidizing agent: the species that is itself reduced and oxidizes the other.
• Reducing agent: the species that is itself oxidized.

Mnemonic: “OIL RIG” (Oxidation Is Loss, Reduction Is Gain).

$\mathrm{\overset{0}{Mg}(s) + 2\overset{+1}{H}Cl(aq) \rightarrow \overset{+2}{Mg}Cl_2(aq) + \overset{0}{H_2}(g)}$

Mg goes 0 → +2: it was oxidized, it is the reducing agent. H goes +1 → 0: it was reduced; HCl is the oxidizing agent. Cl stayed at −1 and did not take part in the redox.

Is a reaction redox?

Write the oxidation number of every element on both sides. If at least one changes, it is redox.

Every reaction that contains a free element (like $\mathrm{O_2}$, Zn, $\mathrm{Cl_2}$ on its own) is redox, because the free element has number 0 and it changes when it enters a compound.

The activity series of metals

A metal reduces the ion of a metal below it in the series from solution. An abbreviated series (from active to inactive): Li > K > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Au.

• Zn reduces $\mathrm{Cu^{2+}}$, which is below it: $\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$ takes place.
• Cu cannot reduce $\mathrm{Zn^{2+}}$: no reaction.
• Metals above H (Mg, Zn, Fe) release $\mathrm{H_2}$ from acid; Cu, Ag, Au do not.

The half-reaction method

In redox equations, simple trial and error often gets stuck, because the charge must also balance. In acidic solution the steps are:

1. Split the reaction into oxidation and reduction half-reactions.
2. In each half, balance the atoms other than O and H.
3. Balance O by adding $\mathrm{H_2O}$.
4. Balance H by adding $\mathrm{H^+}$.
5. Balance the charge by adding electrons ($e^-$).
6. Multiply the halves by suitable numbers to equalize the number of electrons.
7. Add them, cancel the species on both sides, check the atoms and the charge.

Example 5. In acid: $\mathrm{MnO_4^- + C_2O_4^{2-} \rightarrow Mn^{2+} + CO_2}$

Reduction:    MnO4−                       → Mn2+
              MnO4−                       → Mn2+ + 4 H2O        (O)
              MnO4− + 8 H+                → Mn2+ + 4 H2O        (H)
              MnO4− + 8 H+ + 5 e−         → Mn2+ + 4 H2O        (charge: +7 → +2)
Oxidation:    C2O4 2−                     → 2 CO2
              C2O4 2−                     → 2 CO2 + 2 e−        (charge: −2 → 0)
equalize e−:  reduction ×2 (10 e−), oxidation ×5 (10 e−)
Total:        2 MnO4− + 5 C2O4 2− + 16 H+ → 2 Mn2+ + 10 CO2 + 8 H2O
Charge check: −2 − 10 + 16 = +4      right: 2(+2) = +4

In basic solution

First balance as if in acid. Then for every $\mathrm{H^+}$ add one $\mathrm{OH^-}$ to both sides; the $\mathrm{H^+ + OH^-}$ pairs become water, and the extra waters are canceled.

Example 6. In base: $\mathrm{MnO_4^- + I^- \rightarrow MnO_2 + I_2}$

In acid:  2 MnO4− + 6 I− + 8 H+ → 2 MnO2 + 3 I2 + 4 H2O
+8 OH−:   2 MnO4− + 6 I− + 8 H2O → 2 MnO2 + 3 I2 + 4 H2O + 8 OH−
Cancel:   2 MnO4− + 6 I− + 4 H2O → 2 MnO2 + 3 I2 + 8 OH−

Redox titration

The calculation is the same as an acid-base titration; you take the ratio from the balanced redox equation. For example, for 0.02000 M $\mathrm{KMnO_4}$ titrated against 0.2500 g of $\mathrm{Na_2C_2O_4}$ (134.00 g/mol): $0.2500/134.00 = 1.866\times10^{-3}$ mol oxalate, $\times \tfrac{2}{5} = 7.463\times10^{-4}$ mol $\mathrm{MnO_4^-}$, volume $= 37.31$ mL. Permanganate is its own indicator: when the purple color becomes permanent, equivalence has been reached.

Common mistakes