For weak acids with λ∞Q−=λ∞Z−, the molar conductivity is λm=αλ∞.
From λmQ=301λmZ, we get αQ=30αZ.
Using Ka=α2C for weak acids: Ka(HQ)=αQ2×0.18=900αZ2×0.18 and Ka(HZ)=αZ2×0.02.
Therefore Ka(HQ)Ka(HZ)=0.180.02×900=100.
Thus pKa(HQ)−pKa(HZ)=log(100)=2.