The amount of oxalic acid present in a solution can be determined by its titration with $KMnO_4$ solution in the presence of $H_2SO_4$. The titration gives an unsatisfactory result when carried out in the presence of $HCl$,because $HCl$:
- Agets oxidised by $KMnO_4$ to chlorine
- Bfurnishes $H^{+}$ ions in addition to those from oxalic acid
- Creduces permanganate to $Mn^{2+}$
- Doxidises oxalic acid to carbon dioxide and water
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Similar Questions
$10.0 \, mL$ of $Na_{2}CO_{3}$ solution is titrated against $0.2 \, M \, HCl$ solution. The following titre values were obtained in $5$ readings.
$4.8 \, mL, 4.9 \, mL, 5.0 \, mL, 5.0 \, mL$ and $5.0 \, mL$
Based on these readings,and convention of titrimetric estimation,the concentration of $Na_{2}CO_{3}$ solution is .... $mM$.
(Round off to the nearest integer)
$4.8 \, mL, 4.9 \, mL, 5.0 \, mL, 5.0 \, mL$ and $5.0 \, mL$
Based on these readings,and convention of titrimetric estimation,the concentration of $Na_{2}CO_{3}$ solution is .... $mM$.
(Round off to the nearest integer)
DifficultJEE MAIN 2021
View Solution$A$ solution containing $Na_2CO_3$ and $NaOH$ requires $300 \ mL$ of $0.1 \ N$ $HCl$ using phenolphthalein as an indicator. Methyl orange is then added to the above titrated solution when a further $25 \ mL$ of $0.2 \ N$ $HCl$ is required. The amount of $NaOH$ present in the solution is ................. $g$ $(NaOH = 40, Na_2CO_3 = 106)$.
Difficult
View SolutionConsider the titration of $NaOH$ solution versus $1.25\, M$ oxalic acid solution. At the end point,the following burette readings were obtained:
$(i)$ $4.5\, mL$ $\quad (ii)$ $4.5\, mL$ $\quad (iii)$ $4.4\, mL$
$(iv)$ $4.4\, mL$ $\quad (v)$ $4.4\, mL$
If the volume of oxalic acid taken was $10.0\, mL$,then the molarity of the $NaOH$ solution is .... $M$ (Rounded-off to the nearest integer).
$(i)$ $4.5\, mL$ $\quad (ii)$ $4.5\, mL$ $\quad (iii)$ $4.4\, mL$
$(iv)$ $4.4\, mL$ $\quad (v)$ $4.4\, mL$
If the volume of oxalic acid taken was $10.0\, mL$,then the molarity of the $NaOH$ solution is .... $M$ (Rounded-off to the nearest integer).
$20 \ mL$ of $HCl$ solution requires $19.85 \ mL$ of $0.01 \ M$ $NaOH$ solution for complete neutralization. The molarity of $HCl$ solution is
Easy
View SolutionIn the volumetric analysis of $HCl$,if we use phenolphthalein as an indicator,which base is unsuitable for titration?
Easy
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