The solubility of $A_{2} B_{3}$ is $s$ $m o l L^{- 1}$ . Its solubility product is 108 $s^{n}$ . The value of $^{'} n^{'}$ will be:

Open in App

Solution

$A_{2} B_{3} ⇌ 2 A^{3 +} + 3 B^{2 -}$
$K s p = 108 S^{n}$
$108 S^{n} = {(2 S)}^{2} {(3 S)}^{3}$
$108 S^{n} = 4 S^{2} \times {27 S}^{3}$
$108 S^{n} = 108 S^{5}$
$∴ n = 5$

Suggest Corrections

Similar questions

{A s}_{2} S_{3}

S m o l

L^{- 1}

108 S^{n}

n

A_{2} B_{3}

1 \times 10^{- 4}

A X_{2}

3.2 \times 10^{- 11} .

s^{3}

(i n m o l L^{- 1})

C a (I O_{3})_{2}

6.5 \times 10^{- 6} .

(m o l L^{- 1})

(1.63)^{\frac{1}{3}} = 1.17

{L i}_{3} {N a}_{3} {(A l F_{6})}_{2}

S

m o l L^{- 1}

Join BYJU'S Learning Program