0.4g mixture of NaOH,Na2CO3 and some inert impurities was first titrated with ‌
N
10
HCl using phenolphthalein as an indicator, 17.5mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5mL of same HCl was required for the next end point. The weight percentage of Na2CO3 in the mixture is .............. (Rounded off to the nearest integer).
As given, NaOH and Na2CO3 is titrated with N∕10HCl. For NaOH, Equivalents of NaOH= Equivalents of HCl Equivalents of HCl= Normality × Volume (L) =0.1×‌
17.5
1000
Equivalent of HCl=1.75×10−3 Equivalents of NaOH=1.75×10−3 Weight of NaOH= Equivalent of NaOH× equivalent weight of =40×1.75×10−3=0.07g NaOH Now, weight % of NaOH =‌
0.07
0.4
×100=‌
70
4
=17.5% Similarly for Na2CO3, Equivalent of HCl= Equivalent of Na2CO3 Equivalent of Na2CO3=0.1×‌
1.5
1000
=0.15×10−3 Weight of Na2CO3= Equivalent of Na2CO3× equivalent weight of Na2CO3 =0.15×10−3×106=15.9×10−3g Weight \% of Na2CO3=‌