For a weak acid HA that dissociates partially, HA(aq)⇌H+(aq)+A−(aq), the acid-dissociation constant is Ka=[HA][H+][A−]; for a weak base BOH, BOH(aq)⇌B+(aq)+OH−(aq), the base-dissociation constant is Kb=[BOH][B+][OH−]. A larger Ka or Kb means a stronger (more dissociated) weak acid or base. Ostwald's dilution law connects this constant to the degree of dissociation α (the fraction of the electrolyte that has split into ions) and the initial concentration c: starting 1 mol of acid in V dm3 of solution and substituting the equilibrium amounts into the Ka expression gives the exact relation Ka=1−αα2c, which for a genuinely weak acid (small α, so 1−α≈1) simplifies to the very useful approximate form Ka≈α2c, i.e. α≈Ka/c. The same pair of relations, Kb=1−αα2c≈α2c, holds for a weak base with Kb. This is the single relation that lets a chemist move freely between three quantities -- the dissociation constant, the degree/percent of dissociation, and the concentration -- given any two of them, and it is the basis for essentially every acid/base numerical problem in this chapter (percent dissociation, [H3O+] in a weak acid, pH from percent dissociation, and so on). The law also predicts, and confirms, that dilution INCREASES the degree of dissociation (alpha rises as c falls) even though Ka itself, being a true equilibrium constant, stays fixed at a given temperature.
For a weak monobasic acid the small-α form of Ostwald's dilution law, Ka=α2c, applies directly once the percent dissociation is converted to a fraction.
✓Final answer
Ka=α2c=(5×10−4)2×2×10−2=5×10−9 -- identical to the textbook's printed final.
With α=0.05/100=5×10−4 and c=0.02 M, Ka=α2c=5×10−9.
Step 1. For a weak monobasic acid HA, Ostwald's dilution law in its small-α form (Eq. 3.7) gives Ka=α2c, where α is the degree of dissociation (a fraction) and c the molar concentration.
Step 2. Convert the data: α=0.05%=1000.05=5×10−4; c=0.02M=2×10−2 M.
Step 4. So Ka=5×10−9 (dimensionless as the book reports it -- no unit printed).
✓Final answer
Ka=50×10−10=5×10−9 -- digit-for-digit the textbook's printed final.
Convert the percent dissociation to the fraction alpha, then apply the small-alpha Ostwald relation Ka = alpha^2 c (the (1-alpha) correction is negligible at alpha = 5 x 10^-4).
Using 0.05 directly as alpha instead of dividing by 100 (0.05% means alpha = 5 x 10^-4).
Squaring slip: (5 x 10^-4)^2 = 25 x 10^-8, not 25 x 10^-16 or 5 x 10^-8.
Reaching for the full alpha^2 c/(1-alpha) form unnecessarily -- at alpha this small, 1-alpha = 0.9995 changes nothing at this precision.