1.

50 mLof a solution , containing 1 g each of Na_(2)CO_(3) ,NaHCO_(3) and NaOHwas titrated with N HCl . What will be the titre readings if only phenophthalein is used as indicator ?

Answer»

Solution :The titration reactions in this case are
`Na_(2)CO_(3) +HCl to NaHCO_(3) +NaCl`
and `NAOH +HCl to NaCl +H_(2)O `
Thus , we have
m.e of `Na_(2)CO_(3) + " m.e of NaOH = m.e of "v_(1) mL (say ) of N HCl `
`1/106 xx1000 +1/40 XX 1000 = 1 xx v_(1) :. v_(1) = 34.4 mL `


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