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Grade 12Physical Chemistry

arun sir i need your help..please help me..
please please answer this question please please sir
how many ml of 0.150 M Na2CrO4 will be required to oxidize 40 ml of 0.5 M Na2S2O3
Cr2O42- + S2O32- ------> Cr(OH)4- + So42-
a. 225 ml
b. 335 ml
c. 455 ml
d.555 ml
please sir …..request
and thanks for help

Profile image of Mrunal Sonawane
7 Years agoGrade 12
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1 Answer

Profile image of Arun
6 Years ago
Here,
CrO₄²⁻ ⇔ Cr(OH)₄⁻
Change in oxidation number of Cr = oxidation number of Cr in CrO₄²⁻ - oxidation number of Cr in Cr(OH)₄⁻
= +6 - (+3) = +3
∴ n - factor = 3
 
Similarly, S₂O₃²⁻ ⇔ 2SO₄²⁻
change in oxidation number of S = oxidation number of S in S₂O₃²⁻ - oxidation number of S in SO₄²⁻
= +4 - (+12) = -8
∴ n- factor = 8
 
Now, equivalent of CrO₄²⁻ = equivalent of S₂O₃²⁻
N₁V₁ = N₂V₂
∵ normality = n-factor × molarity
so, n₁M₁V₁ = n₂M₂V₂
⇒3 × 1.54 × V₁ = 8 × 0.246 × 40
⇒V₁ = 320 × 0.246/(3 × 1.54) = 17.038 ml
 
Hence, volume of CrO₄²⁻ = 17.0.38 mL
 
Hope you understand