Question

10 gm of impure sodium carbonate is dissolved in water and the solution is made up to 250 ml. To 50 ml of this made up solution, 50 ml of 0.1N, HCl is added and the mixture after shaking well required 10 ml of 0.16N, NaOH solution for complete titration. Calculate the % purity of the sample.

Answer 1

1.0g of (impure) $N{a}_{2}C{O}_3+H_{2}O$ →250mL

(50mLof $N{a}_{2}C{O}_3$ + 50mL of 0.1N NaOH) =10mLof0.16N NaOH

In this question, HCl is in excess.

⇒ Excess mEq of HCl=mEq of NaOH=0.16×10=1.6

mEq of HCl added to$N{a}_{2}C{O}_3$ =0.1×50=5

⇒ mEq of HCl used to neutralised $N{a}_{2}C{O}_3$ =5−1.6=3.4

So, mEq of $N{a}_{2}C{O}_3$ (pure) in 50mL=3.4 ≡ 17 (in 250 mL)

Weight/Ew×100=17

⇒Weight=17×(106/2)/1000=0.901g

($N{a}_{2}C{O}_3$ is diacidic base ⇒Ew=106/2)

So mass of pure $N{a}_{2}C{O}_3$ =0.9011×100=90.1%