Question

What is the volume (in $mL$ ) of $20vol{H}_2{O}_2$ required to completely react with $500mL$ of $0.02M$ acidified $KMn{O}_4$ solution?

• A. 14.0
• B. 7.0
• C. 28.0
• D. 42.0

Answer 1

Answer: (A)

Given,

Volume of acidified $KMn{O}_4$ solution $=500mL$

Molarity of acidified $KMn{O}_4$ solution $=0.02M$

Volume strength of $H_2{O}_2=20vol$

$\because$ Normality $=\frac{\text{ Volume strength }}{\text{ Equivalent weight }}$

$\therefore$ Normality for $H_2{O}_2=\frac{20}{5.6}=3.57N$

$\because$ For $KMn{O}_4$, reaction is in acidic medium, thus valence factor is $5$.

$M{n}^{7+}+5e^{-}⟶M{n}^{2+}$

Thus, normality for $KMn{O}_4=$ Molarity $\times 5$

Now, applying normality equation,

$N_1{V}_1\left(H_2{O}_2\right)=N_2{V}_2\left(KMn{O}_4\right)$

$3.57\times V_1=0.02\times 5\times 500$

$V_1=\frac{50}{3.57}=14.0mL$