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Balancing Redox Reaction

10.0 mL of Na...

Question

$10.0 m L$ of $N a_{2} C O_{3}$ solution is titrated against $0.2 M H C l$ solution. The following litre values were obtained in 5 readings $4.8 m L, 4.9 m L, 5.0 m L, 5.0 m L and 5.0 m L$ . Based on these readings and the convention of titrimetric estimation the concentration of $N a_{2} C O_{3}$ solution is ______mM

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Solution

Answer: 50

From reading values we can conclude that $5 m L$ of $0.2 M H C l$ solution is required to neutralise $10 m L$ of the given $N a_{2} C O_{3}$ solution.
Thus,

$Equivalence of N a_{2} C O_{3} = Equivalence of H C l$

$M \times 10 \times n - f a c t o r (N a_{2} C O_{3}) = 0.2 \times 5 \times n - f a c t o r (H C l)$
$M \times 10 \times 2 = 0.2 \times 5 \times 1$
$M = 5 \times 10^{- 2} M = 50 \times 10^{- 3} M = 50 m M$

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