Question

Solubility product of $ BaCl_{2}$ is $ 4\times 10^{-9}$ . Its solubility would be:

• A. $ 1\times 10^{-27}$
• B. $ 1\times 10^{-3}$
• C. $ 1\times 10^{-7}$
• D. $ 1\times 10^{-2}$

Answer 1

Answer: (B)

$BaCl _{2}$ is a ternary electrolyte.
$BaCl _{2}= \underset{s}{Ba^{2+}}+\underset{2s}{2 Cl ^{-}}$
$K_{s p} =\left[ Ba ^{2+}\right]\left[ Cl ^{-}\right]^{2}$
$=[s][2 s ]^{2}$
$K_{s p} =4 s^{3}$
$K_{s p} =4 s^{3}$ where s=molar solubility
$4 s^{3} =4 \times 10^{-9}$(given)
or $s=10^{-3} M$