Question

The following anions can be separated by precipitation as silver salts: Cl- , Br- , I- , CrO4 2-. If Ag is added to a solution containing the four anions, each at a concentration of 0.10 M, in what order would they precipitate?Compound:
a.AgCl ----ksp: 1.8x10^-10
b.Ag_2_CrO_4_-----ksp: 1.1 x 10 ^-12
c.AgBr----ksp: 5.4 x10 ^ -13
d.AgI-----ksp: 8.5 x 10 ^ -17

Answer 1

Answer:

AgI, AgBr, AgCl and Ag₂CrO₄

Explanation:

Ksp (product solubility constant) is defined as the equilibrium constant of the general reaction:

XₐYₙ(s) → aXⁿ⁺(aq) + nYᵃ⁻(aq)

Where X is cation and Y is anion.

Ksp = [aXⁿ⁺]ᵃ [nYᵃ⁻]ⁿ

The presence of XₐYₙ(s) produce ax moles of aXⁿ⁺ and nx moles of Yᵃ⁻. Where X is the solubility of the compound.

Replacing in Ksp:

Ksp = [ax]ᵃ [nx]ⁿ

Solving for x, Solubility (S) is defined as:

$S = \sqrt[n+a]{\frac{Ksp}{a^{a} n^n} }$

For AgCl, Ag₂CrO₄, AgBr and AgI solubilities are:

$S = \sqrt[2]{\frac{1.8x10^{-10}}{1} }$ = 1.34x10⁻⁵M

$S = \sqrt[3]{\frac{1.1x10^{-12}}{4} }$ = 6.50x10⁻⁵M

$S = \sqrt[2]{\frac{5.4x10^{-13}}{1} }$ = 7.35x10⁻⁷M

$S = \sqrt[2]{\frac{8.5^{-17}}{1} }$ = 9.22x10⁻⁹M

The lower solubility is the first compound in precipitate, thus, order of precipitation is:

AgI, AgBr, AgCl and Ag₂CrO₄