Arrange the following in increasing order of solubility in $H_{2} O$ .

$A g C l, A g B r, A g I, A g F$

A g I < A g B r < A g C l < A g F

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A g I < A g B r < A g F < A g C l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A g I < A C l < A g B r < A g F

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $A g I < A g B r < A g C l < A g F$
Solubility in water depends on the size. Smaller is the size of the anion, more it is solvated and more is the solubility.

AgI < AgBr < AgCl < AgF

Suggest Corrections

Similar questions

K_{s p} (A g C l) > K_{s p} (A g B r) > K_{s p} (A g I)

(K_{s p})

25^{\circ} C

K_{s p}

A g C l 1.8 \times 10^{- 10} A g I 1.1 \times 10^{- 16} P b C r O_{4} 1.8 \times 10^{- 14} A g B r 3.3 \times 10^{- 13}

[I^{-}] = [B r^{-}] = [C l^{-}] = 0.2 M

A g C l

A g B r

A g I

S a l t K_{s p} (m o l L^{- 1})^{2}

A g C l 1.8 \times 10^{- 10}

A g B r 5 \times 10^{- 13}

A g I 8.3 \times 10^{- 17}

Join BYJU'S Learning Program