Question

$x$ mole of $BaCO_{3}$ are soluble in $1 \,L$ then $x$ is related to its solubility product by the expression

• A. $x=K_{s p}$
• B. $x^{2}=\sqrt{K_{s p}}$
• C. $x=\sqrt{K_{s p}}$
• D. $x=\frac{K_{s p}}{2}$

Answer 1

Answer: (C)

$BaCO _{3} \rightarrow Ba ^{2+}+ CO _{3}^{2-}$
If solubility $= x$, then
$K_{s p}=x^{2}$
$x=\sqrt{K_{s p}}$