
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
In this case, direct interaction between the determinable substance $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and the titrant $\mathrm{KMnO}_{4}$ is impossible because they are both strong oxidizers and their oxidation-reduction potential values are close (1.36 and 1.51 V respectively). Therefore, the method of back titration (titration by residue), consisting of two stages, is used. In the first stage, two oxidation-reduction reactions occur:

1) Interaction of a portion of Mohr's salt taken in excess with potassium dichromate in the analyte solution:

$$
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} \leftrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+7 \mathrm{H}_{2} \mathrm{O}
$$

2) Interaction of the unreacted Mohr's salt (residue) with the titrant:

$$
\mathrm{MnO}_{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+}+5 \mathrm{e} \leftrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_{2} \mathrm{O}
$$

The volume used for titration is $\mathrm{V}_{\text{residual}}(\mathrm{KMnO}_{4})=20.00$ ml.

In the second stage, the entire amount of Mohr's salt taken is titrated without the addition of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$, so only the second reaction occurs. The volume used for titration is $\mathrm{V}_{1}(\mathrm{KMnO}_{4})=25.50$ ml.

Hence, the number of equivalents of potassium dichromate in the solution is determined by the difference in potassium permanganate equivalents that reacted in the first and second titration stages. Thus, the mass of chromium in the analyte solution can be calculated by the formula:

$$
\begin{gathered}
\mathrm{m}(\mathrm{Cr})=\left[\mathrm{C}\left(1 / \mathrm{z}_{1} \mathrm{KMnO}_{4}\right) \cdot \mathrm{V}_{1}\left(\mathrm{KMnO}_{4}\right) \cdot 10^{-3}-\mathrm{C}\left(1 / \mathrm{z}_{1} \mathrm{KMnO}_{4}\right) \cdot \mathrm{V}_{\text{ocr.}}\left(\mathrm{KMnO}_{4}\right) \cdot 10^{-3}\right] \cdot \\
\mathrm{M}\left(1 / \mathrm{z}_{2} \mathrm{Cr}\right) \\
\text{or} \\
\mathrm{m}(\mathrm{Cr})=\mathrm{C}\left(1 / \mathrm{z}_{1} \mathrm{KMnO}_{4}\right) \cdot\left[\mathrm{V}_{1}\left(\mathrm{KMnO}_{4}\right) \cdot 10^{-3}-\mathrm{V}_{\text{ocr.}}\left(\mathrm{KMnO}_{4}\right)\right] \cdot 1 \mathrm{M}\left(1 / \mathrm{z}_{2} \mathrm{Cr}\right) \cdot 10^{-3}, \text{g}
\end{gathered}
$$

The molar mass of chromium equivalent is $1 / 3$ of the atomic mass of chromium, as 6 electrons take part in the half-reaction of the reduction of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$:

$$
\begin{gathered}
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O} \\
1 / \mathrm{z}_{2}=2 / 6=1 / 3 \\
\mathrm{M}\left(1 / \mathrm{z}_{2} \mathrm{Cr}\right)=1 / \mathrm{z}_{2} \cdot \mathrm{M}(\mathrm{Cr})=1 / 3 \cdot 52.00 \text{g/mol}=17.33 \text{g/mol.}
\end{gathered}
$$

Substitute the known values and calculate the mass of chromium:

$$
\mathrm{m}(\mathrm{Cr})=0.1247 \text{mol/l} \cdot(25.50-20.00) \text{ml} \cdot 17.33 \text{g/mol} \cdot 10^{-3} \text{l/ml}=0.01189 \text{g}
$$

The mass fraction of chromium in steel:

$$
\mathrm{w}(\mathrm{Cr})=[\mathrm{m}(\mathrm{Cr}) / \mathrm{m}(\mathrm{steel})] \cdot 100 \%=[0.0119 \text{g} / 2.597 \text{g}] \cdot 100 \%=0.4578 \%
$$

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