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A solution contains `Na_(2)CO_(3)` and `NaHCO_(3). 10 mL` of the solution required `2.5 mL "of" 0.1M H_(2)SO_(4)` for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further `2.5 mL "of" 0.2M H_(2)SO_(4)`was required. The amount of `Na_(2)CO_(3)` and `NaHCO_(3)` in `1 "litre"` of the solution is:A. `5.3 g` and `4.2 g`B. `3.3 g` and `6.2 g`C. `4.2 g` and `5.3 g`D. `6.2 g` and `3.3 g` |
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Answer» Correct Answer - A For phenolphthalein: `(1)/(2)`Meq.of `Na_(2)CO_(3)= 2.5xx0.1xx2=0.5` For methyl orange: `(1)/(2)`Meq.of `Na_(2)CO_(3)+`Meq.of `NaHCO_(3)` `=2.5xx0.2xx2=1.0` `:.` Meq. of `NaHCO_(3)=0.5` and Meq. of `Na_(2)CO_(3)=1.0` `:. (w)/(84)xx10000=0.5` `(w)/(160//2)xx1000=1` `:. w=0.042 g "in" 10 mL` `=4.2 "in" 1 L` `:. w=0.053 g "in" 10 mL` `=5.3 "in" L` |
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