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Mole Concept

500 ml of 0.1...

Question

500 ml of 0.150 M $A g N O_{3}$ solution is mixed with 500 ml of 1.09 M $F e^{2 +}$ solution and the reaction is allowed to reach equilibrium at 300K $F e^{2 +} + A g^{+} ⇌ F e^{3 +} + A g$ For 25 ml of solution , 30 ml of 0.833 M $K M n O_{4}$ were required for oxidation.
Calculate the equilibrium constant for reaction at 300 K

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Solution

Initial no of mmoles $A g^{+}$ and $F e^{2 +}$ are 75 and 545 respectively
Now let x mmoles of the reactant consumed at equilibrium then 545-x mmoles of $F e^{2 +}$ are left and only it will be oxidized
for 25 ml of such solution $\frac{545 - x}{1000} \times 25 = 30 \times 0.0833 \times 5$
hence x = 45 mmol
$K_{c} = \frac{45 \times 1000}{30 \times 500} = 3$

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