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30 " mL of " a solution of mixture of `Na_2CO_3` and `NaHCO_3` required 12 " mL of " 0.05 M `H_2SO_4` using phenolphthalein as indicator. With methyl orange 30 " mL of " the same solution required 40 " mL of " same `H_2SO_4`. Calculate the amount of `Na_2CO_3` and `NaHCO_3` per litre in the mnixture. |
Answer» With phenolphthalein indicator `m" Eq of "H_(2)SO_(4) used =12xx0.05xx2` (n-factor) `=1.2` Let a and b m" Eq of "`NaHCO_(3)` and `Na_(2)CO_(3)`, respectively. `therefore(b)/(2)=1.2` With methyl orange indicator `m" Eq of "H_(2)SO_(4)` used from the begining `=4.0xx0.05xx2` `=4.0` `a+b=4.0` By equations (i) and (ii) we get `a(m" Eq of "NaHCO_(3))=1.6` Weight of `NaHCO_(3)=1.6xx10^(-3)xx84` `=(0.1344g)/(30mL)` `=(0.1344xx1000)/(30)gL^(-1)` `=4.48 g L^(-1)` `b(m" Eq of "Na_(2)CO_(3))=2.4` Weight of `Na_(2)CO_(3)=2.4xx10^(-3)xx(106)/(2)` `=(0.1272g)/(30mL)` `=(0.1272xx1000)/(30)gL^(-1)` `=4.24 g L^(-1)` |
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