1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 0.4 g mixture of NaOH and Na2CO3 and some inert impurities were first titrated with N10 HCl using phenolphthalein as an indicator. 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL of the same HCl was required for the next end point. The weight percentage of Na2CO3 in the mixture is ______. (Rounded off to the nearest integer)

Open in App
Solution

## 1st end point reaction, NaOH+HCl→NaCl+H2ONa2CO3+HCl→NaHCO3+NaCl ∴ Milliequivalents of HCl=Milliequivalents of NaOH+Milliequivalents of Na2CO3110×17.5×10−3=nNaOH×1+nNa2CO3×1nNaOH+nNa2CO3=1.75×10−3 ...(1) 2nd end point reaction, NaHCO3+HCl→NaCl+H2CO3 ∴ Milliequivalents of HCl=Milliequivalents of NaHCO3110×1.5×10−3=nNaHCO3×1nNaHCO3=0.15×10−3≡nNa2CO3nNa2CO3=0.15×10−3wNa2CO3=0.15×10−3×106=0.0159 Weight % of Na2CO3 in the mixture=0.01590.4×100≈4%

Suggest Corrections
11
Join BYJU'S Learning Program
Related Videos
Mole Concept
CHEMISTRY
Watch in App
Join BYJU'S Learning Program