Question

How to calculate solubility product of Li3K3(AlF6) ?

Answer 1

dissociation of $Li_3K_3(AlF_6)_2$ is .....

$Li_3K_3(AlF_6)_2\Leftrightarrow 3Li^++3K^++2[AlF_6]^{3-}$

so, $[Li^+]$ = 3s

$[K^+]$ = 3s

and $[[AlF_6]^{3-}]$ = 2s

solubility product = $[Li^+]^3[K^+]^3[[AlF_6]^{3-}]^2$

= (3s)³ (3s)³ (2s)²

= 27s³ × 27s³ × 4s²

= 2916 $s^8$

hence, solubility product of $Li_3K_3(AlF_6)_2$ is 2916 $s^8$