Unit 4: Chemical Reactions

20 terms. Say what each one means, then open the card to check.

• Balanced equation

  A chemical equation with the same number of each kind of atom, and the same total charge, on both sides. Only coefficients change when you balance; subscripts never do.

  Example. In 2Na(s) + Cl₂(g) → 2NaCl(s), there are 2 sodium and 2 chlorine atoms on each side.

• Coefficient

  The number in front of a formula in an equation. It counts particles or moles of that substance, and the ratio between coefficients is the mole ratio.

  Example. In 2H₂O₂ → 2H₂O + O₂, the coefficients 2 : 2 : 1 mean 2 mol of peroxide give 1 mol of oxygen gas.

• Net ionic equation

  An equation that shows only the ions and compounds that actually change in a reaction in water. Ions that stay dissolved and unchanged are left out.

  Example. Mixing aqueous Pb(NO₃)₂ and KI has the net ionic equation Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).

• Spectator ion

  An ion that is present in the solution before and after the reaction, unchanged, so it cancels out of the net ionic equation.

  Example. When aqueous KOH neutralizes HNO₃, the K⁺ and NO₃⁻ ions are spectators; the net change is H⁺ + OH⁻ → H₂O.

• Precipitation reaction

  A reaction in which two dissolved ionic compounds swap partners and one product is an insoluble solid, the precipitate.

  Example. Aqueous AgNO₃ and KBr form a pale solid, AgBr, while K⁺ and NO₃⁻ stay dissolved.

• Always-soluble ions

  Ionic compounds that contain Na⁺, K⁺, NH₄⁺ or NO₃⁻ dissolve in water, so these ions are never part of a precipitate.

  Example. Of the two possible products PbSO₄ and NaNO₃, only PbSO₄ can be the precipitate, because NaNO₃ contains always-soluble ions.

• Acid-base (proton transfer) reaction

  A reaction in which an acid gives a hydrogen ion (proton) to a base. The acid is the proton donor and the base is the proton acceptor.

  Example. In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, water donates a proton, so it is the acid, and ammonia accepts it, so it is the base.

• Oxidation number

  A bookkeeping charge for an atom: free elements are 0, a monatomic ion equals its charge, oxygen is usually −2 and hydrogen +1, and the numbers in a species add to its charge.

  Example. In SO₄²⁻, oxygen is −2 four times (−8), so sulfur is +6 to give the 2− charge.

• Oxidation-reduction (redox) reaction

  A reaction in which electrons move from one species to another. The species that loses electrons is oxidized (its oxidation number rises); the one that gains them is reduced.

  Example. In Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s), iron goes from 0 to +2 (oxidized) and each silver ion from +1 to 0 (reduced).

• Chemical change

  A change that makes new substances by breaking and forming chemical bonds. A physical change, such as melting, changes the form of a substance but not its identity.

  Example. Burning candle wax makes CO₂ and water (chemical); a puddle of melted candle wax hardening is a physical change.

• Mole ratio

  The ratio of coefficients between two substances in a balanced equation. It converts moles of one substance into moles of another.

  Example. In 2SO₂ + O₂ → 2SO₃, the ratio of O₂ to SO₃ is 1 : 2, so 0.90 mol O₂ makes 1.80 mol SO₃.

• Gravimetric analysis

  Finding how much of a substance is in a sample by turning it into a solid of known formula, then filtering, drying and weighing that solid.

  Example. A water sample treated with excess AgNO₃ gives dry AgCl; the AgCl mass gives the moles of chloride in the water.

• Solution stoichiometry

  Using molarity times volume in liters to get moles of a dissolved reactant, then using the mole ratio as in any other stoichiometry problem.

  Example. A 0.0300 L sample of 0.400 M KOH holds 0.0120 mol of OH⁻.

• Limiting reactant

  The reactant that runs out first. It sets the greatest amount of product that can form, and the other reactants are left over.

  Example. Each sandwich takes 2 slices of bread and 1 slice of cheese; with 10 slices of bread and 8 of cheese, the bread runs out after 5 sandwiches.

• Excess reactant

  A reactant that is not used up when the reaction stops. The amount left is what you started with minus what the limiting reactant used.

  Example. If 2.0 mol of O₂ meet 1.0 mol of CH₄ in CH₄ + 2O₂ → CO₂ + 2H₂O, neither is left; with 2.5 mol of O₂, 0.5 mol O₂ is left over.

• Theoretical yield

  The amount of product the limiting reactant could make if it reacted completely, found from the balanced equation.

  Example. Burning 0.40 mol of Mg in excess O₂ by 2Mg + O₂ → 2MgO has a theoretical yield of 0.40 mol of MgO, about 16 g.

• Percent yield

  The actual amount of product collected divided by the theoretical yield, times 100. Losses keep it below 100 percent; a wet or impure product can push it above.

  Example. A theoretical yield of 4.00 g and an actual yield of 3.40 g give a percent yield of 85.0 percent.

• Titration

  Adding a solution of known concentration from a buret to a measured sample until the reaction is just complete, then using the volume used to find the amount in the sample.

  Example. A student adds 0.150 M HCl from a buret to 10.00 mL of an ammonia solution until the indicator changes color.

• Equivalence point

  The point in a titration where the moles of titrant added are exactly the amount the balanced equation needs to react with the sample.

  Example. For HNO₃ titrated with KOH (1 : 1), the equivalence point comes when the moles of KOH added equal the moles of HNO₃ in the flask.

• End point and indicator

  The end point is where an indicator changes color during a titration. A good indicator is chosen so the end point lands very close to the equivalence point.

  Example. Phenolphthalein turns pink just past the point where a strong acid is fully neutralized by a strong base.