Question

0.56g of lime (CaO) was treated with 100ml of 0.5M oxalic acid is give ppt of $Ca{C}_2{O}_4$. The filtrate required V ml of 0.2M NaOH for complete neutralization. The value of V is: (atomic Weight of Ca=40, O=16)

• A. 800ml
• B. 400ml
• C. 100ml
• D. none of these

Answer 1

Answer: (B)

400ml

$CaO+H_2{C}_2{O}_4\to Ca{C}_2{O}_4\downarrow H_{2}O$

the $NaOH$ would ne req to neutralize remaining oxalic acid

$\therefore$ initially

no. of moles of $CaO=\frac{0.56}{56}=0.01$mol

no. of moles of $H_2{C}_2{O}_4=0.5\times 0.1=0.05$ mol

$\therefore$ no. of moles of $H_2{C}_2{O}_4$ remaining after reaction $ =(0.05-0.01 )= 0.04$ mol

$H_2{C}_2{O}_4\left(s\right)+2NaOH\left(aq\right)=N{a}_2{C}_2{O}_4\left(aq\right)+2H_{2}O\left(l\right)$

$\therefore 0.04\times 2$mol $NaOH$ will be required.

$\therefore$ volume $=\frac{moles}{malarity}$

$=\frac{0.08}{0.2}$

$=0.4L=400$mL

Hence, the correct option is $B$