Solubility product of BaCl2 is 4× 10-9 . Its solubility would be:

Question

Solubility product of BaCl2 is 4× 10-9 . Its solubility would be: (A)  1× 10-27 (B)  1× 10-3 (C)  1× 10-7 (D)  1× 10-2

Tardigrade Admin · Accepted Answer

BaCl 2 is a ternary electrolyte. BaCl 2= undersetsBa2++ underset2s2 Cl - Ks p =[ Ba 2+][ Cl -]2 =[s][2 s ]2 Ks p =4 s3 Ks p =4 s3 where s=molar solubility 4 s3 =4 × 10-9(given) or s=10-3 M

Q. Solubility product of $B a C l_{2}$ is $4 \times 1 0^{- 9}$ . Its solubility would be:

$1 \times 1 0^{- 27}$

$1 \times 1 0^{- 3}$

$1 \times 1 0^{- 7}$

$1 \times 1 0^{- 2}$

$B a C l_{2}$ is a ternary electrolyte.
$B a C l_{2} = s B a^{2 +} + 2 s 2 C l^{-}$
$K_{s p} = [B a^{2 +}] [C l^{-}]^{2}$
$= [s] [2 s]^{2}$
$K_{s p} = 4 s^{3}$
$K_{s p} = 4 s^{3}$ where s=molar solubility
$4 s^{3} = 4 \times 1 0^{- 9}$ (given)
or $s = 1 0^{- 3} M$

Q. Solubility product of BaCl2​ is 4×10−9 . Its solubility would be:

1×10−27

1×10−3

1×10−7

1×10−2