Answer
$V=300\ mL$
Work Step by Step
Initial percent ionization:
$Ka=[H_3O^+][A^-]/[HA]$
$[HA]=0.20\ M-[A^-]\approx0.20\ M$
$0.20Ka=[A^-]^2\rightarrow [A^-]=\sqrt{0.20Ka}$
$\%ion=[A^-]/[HA]=\sqrt{0.20Ka}/0.20$
Diluted percent ionization:
$[HA]=0.20\ M\times 100\ mL/(100 + V)\ mL-[A^-]$
$[HA]\approx20/(100+V)$
$Ka=[H_3O^+][A^-]/[HA]$
$Ka\times20/(100+V)=[A^-]^2\rightarrow [A^-]=\sqrt{20Ka/(100+V)}$
$\%ion=[A^-]/[HA]=\sqrt{20Ka/(100+V)}/(20/(100+V))$
Find volume for the percent ionization to double:
$\sqrt{20Ka/(100+V)}/(20/(100+V))=2\times\sqrt{0.20Ka}/0.20$
$1/\sqrt{(20/(100+V))}=2/\sqrt{0.20}$
$0.20=4\times20/(100+V)$
$100+V=400$
$V=300\ mL$
