Question

The $K _{ sp }$ for bismuth sulphide $\left( Bi _{2} S _{3}\right)$ is $1.08 \times 10^{-73}$. The solubility of $Bi _{2} S _{3}$ in $mol\, L ^{-1}$ at $298\, K$ is

• A. $1.0 \times 10^{-15}$
• B. $2.7 \times 10^{-12}$
• C. $3.2 \times 10^{-10}$
• D. $4.2 \times 10^{-8}$

Answer 1

Answer: (A)

$ k _{ sp } =(2 s )^{2}(3 s )^{3} $
$=4 s ^{2} \times 27( s )^{3} $
$=108( s )^{5} $
$( s )^{5} =\frac{1.08 \times 10^{-73}}{108} $
$ \Rightarrow s =10^{-15}$