(NH4)3PO4 get converted into NO3¯ and PH3 on reacting with KMnO4 in acidic medium. If 50 mL of 0.2 M (NH4)3PO4 solution reacts with 16 mL of KMnO4 solution, then normality of KMnO4 solution is

Engineering

Chemistry

Law of Chemical Equivalence

Question

(NH₄)₃PO₄ get converted into NO₃¯ and PH₃ on reacting with KMnO₄ in acidic medium. If 50 mL of 0.2 M (NH₄)₃PO₄ solution reacts with 16 mL of KMnO₄ solution, then normality of KMnO₄ solution is

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(NH₄)₃ PO₄ $\underset{16 ml}{\overset{{KMnO}_{4}}{\to}}$ NO₃¯ + PH₃
–3 +5 +5 –3
+24 – 8 = 16
50 × 0.2 M
n–factor of KMnO₄ in acidic medium = 5
n-factor of (NH₄)₃PO₄ = 16
Eq. of (NH₄)₃PO₄ = Eq. of KMnO₄
$\frac{0 .2 \times 50}{1000} \times 16 = \frac{N \times 16}{1000}$

N = 10