Question

The conductivity of saturated solution of Ag3po4 is 9 ^ 10^-6 s/m and equivalent conductivity is1.5^10^-4 Sm^2/equivalent the Ksp is

Answer 1

The conductivity of saturated solution of Ag3po4 is 9 × 10^-6 s/m and equivalent conductivity is1.5 ×10^-4 Sm^2/equivalent the Ksp is....

using formula,

$\bf{solubility}=\frac{\kappa\times1000}{\Lambda}$

where $\kappa$ is conductivity of saturated solution , $\Lambda$ is molar conductivitance.

given, $\kappa$ = 9 × 10^-6 s/m

= 9 × 10^-6 s/(100cm)

= 9 × 10^-8 s/cm

$\Lambda$ = 1.5 × 10^-4 Sm²/eq

relation between equivalent conductance and molar conductance is $\Lambda=\left(\frac{1}{n^+\times n^-}\right)\Lambda_m$

where n+ is number of cations and n- is number of anions and $\Lambda_m$ is molar conductivity.

here, n+ = 3 and n- = 1

so, molar conductivity, $\Lambda_m$ = (3 × 1) × 1.5 × 10^-4 Sm²/eq

= 4.5 × 10^-4 S(100cm)²/mol

= 4.5 Scm²/mol

so, solubility = 9 × 10^-8 × 1000/4.5 = 2 × 10^-5

now dissociation of Ag3(PO4) is ...

$Ag_3PO_4\rightarrow 3Ag^++PO_4^{3-}$

so, Ksp = [3Ag^+]³[PO4³-]

= (3 × 2 × 10^-5)² × (2 × 10^-5)

= 72 × 10^-15

= 7.2 × 10^-14

Answer 2

Answer:

Explanation:

Hope it is useful.