10.0 mL of 0.05 M KMnO$_4$ solution was consumed in a titration with 10.0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is ________ $\times 10^{-2}$ g/L. (Round off to the Nearest Integer).
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For redox titrations, using the equivalence formula $N_1V_1 = N_2V_2$ or $M_1V_1n_1 = M_2V_2n_2$ is often faster than using stoichiometry. Remember to correctly determine the n-factors (number of electrons transferred per molecule) for both the oxidizing and reducing agents.
Step 1: Write the balanced redox reaction.
In acidic medium, permanganate ion (MnO$_4^-$) is reduced and oxalate ion (C$_2$O$_4^{2-}$) is oxidized.
Reduction half-reaction: MnO$_4^-$ + 8H$^+$ + 5e$^-$ $\rightarrow$ Mn$^{2+}$ + 4H$_2$O (n-factor for KMnO$_4$ is 5)
Oxidation half-reaction: C$_2$O$_4^{2-}$ $\rightarrow$ 2CO$_2$ + 2e$^-$ (n-factor for oxalic acid is 2)
The balanced overall reaction is: 2MnO$_4^-$ + 5C$_2$O$_4^{2-}$ + 16H$^+$ $\rightarrow$ 2Mn$^{2+}$ + 10CO$_2$ + 8H$_2$O.
The stoichiometric ratio is 2 moles of KMnO$_4$ to 5 moles of oxalic acid.
Step 2: Use the titration equivalence formula.
At the equivalence point, M$_1$V$_1$n$_1$ (KMnO$_4$) = M$_2$V$_2$n$_2$ (Oxalic Acid).
Let M$_{ox}$ be the molarity of the oxalic acid solution.
(0.05 M)(10.0 mL)(5) = (M$_{ox}$)(10.0 mL)(2).
$0.05 \times 5 = M_{ox} \times 2$.
$0.25 = 2 \times M_{ox}$.
$M_{ox} = \frac{0.25}{2} = 0.125$ M.
Step 3: Calculate the strength in g/L.
Strength (g/L) = Molarity (mol/L) $\times$ Molar Mass (g/mol).
The solute is oxalic acid dihydrate, H$_2$C$_2$O$_4 \cdot$ 2H$_2$O.
Molar Mass = (2$\times$1.01) + (2$\times$12.01) + (4$\times$16.00) + 2$\times$(18.02) = 2.02 + 24.02 + 64.00 + 36.04 = 126.08 g/mol. We can approximate it as 126 g/mol.
Strength = 0.125 mol/L $\times$ 126 g/mol = 15.75 g/L.
Step 4: Express the answer in the required format.
The question asks for the strength in units of $\times 10^{-2}$ g/L.
Strength = 15.75 g/L = 1575 $\times 10^{-2}$ g/L.
The integer value is 1575.
