Redox and Equivalent ConceptHard
Question
What volume of 0.40 M-Na2S2O3 would be required to react with the I2 liberated by adding excess of KI to 50 ml of 0.20 M CuSO4?
Options
A.12.5 ml
B.25 ml
C.50 ml
D.2.5 ml
Solution
$Cu^{2 +} + I^{-} \rightarrow Cu^{+} + I_{2}$
$n_{eq}Na_{2}S_{2}O_{3} = n_{eq}I_{2} = n_{eq}Cu^{2 +}$
Or $\frac{V \times 0.4}{1000} \times 1 = \frac{50 \times 0.2}{1000} \times 1 \Rightarrow V_{Na_{2}S_{2}O_{3}} = 25\text{ ml}$
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