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Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsDouble IndicatorHard2 minQB

ChemistryHardsingle choice

A volume of 25 ml of 0.1 M solution of sodium sesquicarbonate ( $N a_{2} C O_{3} \cdot NaHC O_{3} \cdot 2 H_{2} O$ ) is titrated against 0.1 M $HCl$ solution. $x$ ml of $HCl$ is used with phenolphthalein and $y$ ml with methyl orange as indicators in two separate titrations. Hence (y-x) is

Options:

Answer:

Solution:

Salt = 25 $x$ 0.1 = 2.5 mmol, i.e. 2.5 mmol $C O_{3}^{2 -}$ + 2.5 mmol $HC O_{3}^{-}$ . Phenolphthalein: $H^{+}$ = 2.5 mmol, $x$ = 25 ml. Methyl orange: $H^{+}$ = 2(2.5) + 2.5 = 7.5 mmol, $y$ = 75 ml. $y$ - $x$ = 50 ml, option (d).

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Double Indicator

⏱ 2mℹ️ Source: QB

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