Answer:
The order of solubility is AgBr < Ā AgāCOā < AgCl
Explanation:
The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:
Ksp = (A) (B) where A and B are the molar solubilities = sĀ² Ā (for compounds with 1:1 Ā ratio).
It follows then Ā that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:
Comparing AgBr: Ksp = 5.4 x 10ā»Ā¹Ā³ with AgCl: Ksp = 1.8 x 10ā»Ā¹ā°, AgCl will be more soluble.
Comparing Ag2CO3: Ksp = 8.0 x 10ā»Ā¹Ā² Ā with AgCl Ksp = AgCl: Ksp = 1.8 x 10ā»Ā¹ā° we have the complication of Ā the ratio of ions 2:1 in Ag2CO3, Ā so the answer is not obvious. But since we know that
Ag2CO3 ā 2 Agāŗ + COāĀ²ā
Ksp Ag2CO3 Ā = 2s x s = 2 sĀ² = Ā 8.0 x 10-12
s = 4 x 10ā»12 ā“ s= 2 x 10ā»ā¶
And for AgCl
AgCl Ā ā Agāŗ + Clā»
Ksp = sĀ² = 1.8 x 10ā»Ā¹ā° Ā ā“ s = ā 1.8 x 10ā»Ā¹ā° Ā = 1.3 x 10ā»āµ
Therefore, AgCl is more soluble than AgāCOā
The order of solubility is AgBr < Ā AgāCOā < AgCl
Answer:
5.4 x 10^-13
Explanation:
Edmentum says it is correct
