Answer
$63.7\ mL$
Work Step by Step
By the Henderson-Hasselbalch equation:
$pH=pKa+\log([HC_2O_4^-]/[H_2C_2O_4])$
$pKa=4.19$
$4.70=4.19+\log([HC_2O_4^-]/[H_2C_2O_4])$
$\log([HC_2O_4^-]/[H_2C_2O_4])=0.51$
$[HC_2O_4^-]/[H_2C_2O_4]=3.236$
$[H_2C_2O_4]=(0.100\ M\times 0.100\ L - 0.120\ M\times V)\div(0.100\ L+V)$
$[HC_2O_4^-]=0.120\ M\times V\div(0.100\ L+V)$
$[HC_2O_4^-]/[H_2C_2O_4]=0.120\ M\times V\div(0.100\ M\times 0.100\ L - 0.120\ M\times V)$
$3.236(0.01-0.12V)=0.12V$
$0.032=0.508V$
$V=0.064\ L = 63.7\ mL$
