The solubility of sparingly soluble salt A_3B_2 (Molar mass = ,'M… - Vidyadip

MCQ

The solubility of sparingly soluble salt $A_3B_2$ (Molar mass $=\,'M'\,\,g/mol$ ) in water is $x\,\,g/L.$ The ratio of solubility product of salt to the molar concentration of $B^{-3}$ ion is

A
$108\frac{{{x^5}}}{{{M^5}}}$
✓
$54\frac{{{x^4}}}{{{M^4}}}$
C
$108\frac{{{M^6}}}{{{x^6}}}$
D
$27 \times 16\,\frac{{{M^7}}}{{{x^7}}}$

✓

Answer

Correct option: B.

$54\frac{{{x^4}}}{{{M^4}}}$

b
$S=\frac{X}{M}$
$A_{3} B_{2} \Leftrightarrow \quad 3 A^{+2} \quad+2 B^{-3}$
$\left.B^{-3}\right]=2 S=2\left(\frac{X}{M}\right)$
$2 S$
$K_{s p}=\left[A^{+2}\right]^{3}\left[B^{-3}\right]^{2}$
$K_{s p}=108 S^{5}=108\left(\frac{X}{M}\right)^{5}$
$K_{s p}=54\left(\frac{X}{M}\right)^{4}\left(\frac{2 X}{M}\right)$
$K_{s p}=54\left(\frac{X}{M}\right)^{4}\left[B^{-3}\right]$

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