1.3Radicals and Rational Exponents

Graham's law is a ratio of square roots (rate A/rate B = √M_B/√M_A); simplifying √32/√2 to √16=4 or squaring (1.66)² to isolate an unknown molar mass requires fluency with radical and rational-exponent manipulation, not just formula substitution.

u_rms=√(3RT/M) is a square root of a quotient, and deriving Graham's law from it means squaring both sides to isolate M (M=3RT/u²_rms) — the same radical-manipulation skill §8.5 needs, now embedded inside a formula with three other variables.

Finding an atomic or ionic radius from an edge length runs the Pythagorean relation a² + a² = (4r)² backward into a radical, r = √[(a²+a²)/16], and finding density from an edge length runs V = l³ forward while end-of-chapter problems run it backward as a cube root — both worked examples stall without radical and rational-exponent manipulation.

For a 1:1 salt like CuBr, Ksp = x² is solved by x = √Ksp — a direct radical evaluation that is the entire content of the calculation once the ICE table is set up.

For an AB2 salt like Ca(OH)2, Ksp = (x)(2x)² = 4x³ collapses to a cube, so solving for molar solubility means x = ∛(Ksp/4) — a rational-exponent solve the student cannot skip or approximate away.

Complex-ion dissociation reduces to x³ = 0.10/(4×Kf), solved as x = ∛(...) — the same cube-root solve as an AB2 solubility problem, now applied to a formation-constant expression instead of Ksp.

The Al(OH)3-in-water molar solubility is computed as [Al³⁺] = (2×10⁻³²/27)^(1/4), a fourth-root (rational-exponent) solve arising from the 1:3 dissolution stoichiometry — one exponent higher than the cube-root case in 15.2, so the same radical-exponent skill has to generalize.

The photographic-fixer example rearranges the combined-equilibrium K expression and solves for [S2O3²⁻] with a square root, [S2O3²⁻] = √([Ag(S2O3)2³⁻][Br⁻]/K) — another radical-equation solve, this time on a two-step coupled-equilibrium expression rather than a simple Ksp.

The Check Your Learning answer [F] = (k₁[F₂]/k₋₁)^(1/2) comes from taking a square root of both sides of k₁[F₂] = k₋₁[F]² — a rational-exponent move that surfaces whenever a pre-equilibrium step is bimolecular in the intermediate rather than the reactant, and a student who forgets the square root leaves [F] squared instead of linear.