418.52 mL of x M acidified potassium permanganate solution can completely oxidize 523.15 × 10^–3 moles of Sn²⁺ ions to Sn⁴⁺ ions. What is the value of 10x?
$M n O_{4}^{-} + S n_{2 +} + H^{+} \to M n^{2 +} + S n^{4 +} + H_{2} O$

Question

418.52 mL of x M acidified potassium permanganate solution can completely oxidize 523.15 × 10^–3 moles of Sn²⁺ ions to Sn⁴⁺ ions. What is the value of 10x?

$M n O_{4}^{-} + S n_{2 +} + H^{+} \to M n^{2 +} + S n^{4 +} + H_{2} O$

TestHub · Accepted Answer

Solution:

The balanced redox reaction is $2 M n O_{4}^{-} + 5 S n^{2 +} + 16 H^{+} \to 2 M n^{2 +} + 5 S n^{4 +} + 8 H_{2} O$ .

Moles of $S n^{2 +}$ = $523.15 \times 1 0^{- 3}$ mol.

Moles of $M n O_{4}^{-}$ = $(523.15 \times 1 0^{- 3}) \times (2/5) = 209.26 \times 1 0^{- 3}$ mol.

Concentration $x = (209.26 \times 1 0^{- 3} m o l) / (0.41852 L) = 0.5$ M.

Therefore, $10 x = 10 \times 0.5 = 5$ .

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