During the estimation of oxalic acid $V/S$ $KMnO_4$,the self-indicator is:

  • A
    $KMnO_4$
  • B
    Oxalic acid
  • C
    $K_2SO_4$
  • D
    $MnSO_4$

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Similar Questions

Given below are two statements :
Statement $I$ : Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution.
Statement $II$ : In this titration,phenolphthalein can be used as an indicator.
In the light of the above statements,choose the most appropriate answer from the options given below:

Given below are two statements:
Statement $I$: In the oxalic acid vs $KMnO_4$ (in the presence of dil. $H_2SO_4$) titration,the solution needs to be heated initially to $60^{\circ}C$,but no heating is required in Ferrous Ammonium Sulphate $(\text{FAS})$ vs $KMnO_4$ titration (in the presence of dil. $H_2SO_4$).
Statement $II$: In oxalic acid vs $KMnO_4$ titration,the initial formation of $MnSO_4$ takes place at high temperature,which then acts as a catalyst for the further reaction. In the case of $\text{FAS}$ vs $KMnO_4$,heating oxidizes $Fe^{2+}$ into $Fe^{3+}$ by oxygen of air and error may be introduced in the experiment.
In the light of the above statements,choose the correct answer from the options given below:

In the titration of potassium permanganate $(KMnO_4)$ against ferrous ammonium sulphate $(FAS)$ solution,dilute sulphuric acid $(H_2SO_4)$ but not nitric acid $(HNO_3)$ is used to maintain an acidic medium,because:

In a titration experiment,$10 \, mL$ of an $FeCl_{2}$ solution consumed $25 \, mL$ of a standard $K_{2}Cr_{2}O_{7}$ solution to reach the equivalence point. The standard $K_{2}Cr_{2}O_{7}$ solution is prepared by dissolving $1.225 \, g$ of $K_{2}Cr_{2}O_{7}$ in $250 \, mL$ water. The concentration of the $FeCl_{2}$ solution is closest to $..... \, N$
[Given : molecular weight of $K_{2}Cr_{2}O_{7} = 294 \, g \, mol^{-1}$]

An aqueous solution of oxalic acid dihydrate contains $6.3 \ g$ in $250 \ mL$. The volume of $0.1 \ N \ NaOH$ required to completely neutralize $10 \ mL$ of this solution is $.............. \ mL$.

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