Q. Ionic strength of $0.2 M N a_{3} P O_{4}$ is

87 156 Electrochemistry Report Error

A

$2.8$

B

$1.2$

C

$1.0$

D

$1.8$

Solution:

$[N a^{\oplus}] = 0.6 M [P O_{4}^{3 -}] = 0.2$
$μ = \frac{1}{2} [0.6 (1)^{2} + 0.2 (3)^{2}] = 1.2$

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