Name: Understanding Redox Titrations
Uploaded: 2019-07-19
Duration: 11 min 41 s
Description: Redox titration is a great way to identify whether a reaction is completed or not, where we know that reaction is complete, and it helps us find the strength...

Question

A steel sample is to be analysed for

Cr

and

Mn

simultaneously. By suitable treatment the

Cr

is oxidized to

{Cr}_{2} O_{7}^{2 -}

and the

Mn

to

{MnO}_{4}^{-} . A 10.00 g

sample of steel is used to produce

250.0 mL

of a solution containing

{Cr}_{2} O_{7}^{2 -}

and

{MnO}_{4}^{-} . A 10.00 mL

portion of this solution is added to

{aBaCl}_{2}

solution and by proper adjustment of the acidity, the chromium is completely precipitated as

{BaCrO}_{4}; 0.0549 g

is obtained. A second

10.00 mL

portion of this solution requires exactly

15.95 mL

of

0.0750 M

standard

{Fe}^{2 +}

solution for its titration (in acid solution). Calculate the

%

of

Mn

and

%

of

Cr

in the steel sample.

Accepted Answer

Given: weight of the sample $= 10.00 gm$ , $[{Fe}^{2 +}] = 0.0750 M$ , the weight of ${BaCrO}_{4}$ $= 0.0549 gm$

$Moles = \frac{given wt .}{atomic wt .}$

Moles of ${BaCrO}_{4}$ $(n_{{BaCrO}_{4}}) = \frac{0.0549}{253}$ $= 0.0002169$

By Stoichiometry of compounds $n_{{BaCrO}_{4}} = n_{Cr}$

$15.95 mL$

Wt. of $Cr$ ${Fe}^{2 +}$ $n_{Cr} \times Atomic mass$

$\Rightarrow W_{Cr} = 0.0002196 \times 52 \times \frac{250}{10}$ (As an only $10 m l$ solution is used out of ${Cr}_{2} O_{7}^{2 -}$ )

$\Rightarrow W_{Cr} = 0.282 gm$

$%$ of $Cr$ $= \frac{0.282}{10} \times 100 = 2.82 %$

For ${Cr}_{2} O_{7}^{- 2}$ , by Stoichiometry of compounds

$n_{{Cr}_{2} {O_{7}}^{2 -}} = \frac{n_{Cr}}{2}$

$10.00 mL$ $($ Atomic wt. of $C r = 52)$

$gm Eq . = Normality \times volume$

$gm Eq . = mole \times n - factor$

As the chromium is completely precipitated as ${BaCrO}_{4}$ and solution is further titrated with ${Fe}^{2 +}$

So $gm Eq .$ of ${MnO}_{4}^{-}$ $= gm Eq .$ of ${Fe}^{2 +}$ $-$ $gm Eq .$ of ${Cr}_{2} O_{7}^{2 -}$

$(\overset{0}{Cr} \to_{}^{} {\overset{+ 6}{Cr}}_{2} {O_{7}}^{2 -}, n - factor = 6)$

$\begin{array}{l} = 0.075 \times 15.95 \times 10^{- 3} \end{array} \times \frac{250}{10} - 0.002711 \times 6$ $($ Only $10 m l$ solution is used $)$

$= 0.01364$

$Weight = \frac{gm Eq .}{n - factor} \times atomic weight$

$(\overset{+ 7}{Mn} {O_{4}}^{-} \to_{}^{} {Mn}^{2 +}, n - factor = 7 - 2 = 5)$

Wt. of $Mn$ $= \frac{0.01364}{5} \times 55 = 0.15 g$

$%$ of $Mn$ $= \frac{0.15}{10} \times 100 = 1.5 %$

Hence, $%$ of $Mn$ and $Cr$ are $1.5 %$ and $2.82 %$ respectively

Important Questions on Redox Reactions

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