20 ml of the solution containg `Na_(2)CO_(3)` and `NaHCO_(3)` is titrated with `0.1MHCI` using Phenolphthalein indicator the end point was `10ml. 20ml` of the same solution is titrated with `0.1M HCI`, the end point was 25ml with Methylorange indicaor from the begining. `NaHCO_(3)+HCl rarr NaCl +H_(2)O +CO_(2)` `Na_(2)CO_(3)+2HCl rarr 2NaCl +CO_(2) +H_(2)O` What is amount of `NaHCO_(3)` present in 1 litre of solutionA. `2.1g`B. `1.05g`C. `8.1g`D. `0.844g`

20 ml of the solution containg `Na_(2)CO_(3)` and `NaHCO_(3)` is titra

1.	20 ml of the solution containg `Na_(2)CO_(3)` and `NaHCO_(3)` is titrated with `0.1MHCI` using Phenolphthalein indicator the end point was `10ml. 20ml` of the same solution is titrated with `0.1M HCI`, the end point was 25ml with Methylorange indicaor from the begining. `NaHCO_(3)+HCl rarr NaCl +H_(2)O +CO_(2)` `Na_(2)CO_(3)+2HCl rarr 2NaCl +CO_(2) +H_(2)O` What is amount of `NaHCO_(3)` present in 1 litre of solutionA. `2.1g`B. `1.05g`C. `8.1g`D. `0.844g`
Answer» Correct Answer - A Using methyl orange, `x +y = 25 xx 0.1 = 2.5, y = 2.5 - 2 = 0.5` Wt of `NaHCO_(3)` in 20ml `=(0.5)/(1000) xx 84 = 0.042g` wt of `NaHCO_(3)` in 1 lit `=(0.042)/(20) xx 1000 = 2.1g`

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