Question

Question 7.67 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.

Class XI Equilibrium Page 230

Answer 1

(a) silver chromate
Ag2CrO4 <=> 2Ag+ + CrO42- ; Ksp = 1.1 × 10^-12

but solubility product (Ksp) = [2Ag+]² [CrO42-]
Let solubility of silver chromate is s
Ksp = [2s]² [s] = 4s³
s³ = Ksp/4 = 1.1 × 10^-12/4 = 2.75 × 10^-13
take log both sides,
3logs = log(2.75 × 10^-13) = -12.5607
s = 6.503 × 10^-5 M { using antilog concept }
so, [Ag+] = 2s = 1.3006 × 10^-5 M
[CrO4²-] = s = 6.503 × 10^-5 M


(b) BaCrO4 <=> Ba²+ + CrO4²- ; Ksp = 1.2 × 10^-10
Let solubility of BaCrO4 is s M

Ksp = [Ba2+][CrO4²-]
Ksp = s.s = s²
s = √Ksp = √{1.2 × 10^-10} = 1.1 × 10^-5 M
so, [Ba²+] = [CrO4²-] = 1.1 × 10^-5 M


(c) Fe(OH)3 <=> Fe³+ + 3OH- ;Ksp =10^-38
Let solubility of Fe(OH)3 is s M
Ksp = [Fe³+][3OH-]³
Ksp = s.27s³
27s⁴ = Ksp
s⁴ = Ksp/27 = 10^-38/27 = 0.037 × 10^-38
take log both sides,
4logs = log(0.037 × 10^-38) = -39.4318
logs = -9.8579 = 10(bar).1421
s = 1.387 × 10^-10 { by using antilog concept }
so, [Fe³+] = 1.387 × 10^-10 M
[OH-] = 3s = 3 × 1.387 × 10^-10 M
= 4.161 × 10^-10 M


(d) PbCl2 <=> Pb²+ + 2Cl- ;Ksp = 1.6 × 10^-5
Let solubility of PbCl2 is s M
Ksp = [Pb²+][2Cl-]²
Ksp = s.4s²
4s³ = Ksp
s³ = Ksp/4 = 1.6 × 10^-5/4 = 0.4 × 10^-5 M
3logs = log(4 × 10^-6) = -6 + 0.6021 = -5.3979
logs = -1.7993 = 2(bar).2007
take antilog both sides,
s = 1.585 × 10^-2 M
hence, [Pb²+] = 1.585 × 10^-2 M
[Cl-] = 2s = 2 × 1.585 × 10^-2 M = 3.17 × 10^-2 M

(e) Hg2I2 <=> Hg2²+ + 2I- ; Ksp = 4.5 × 10^-29
Let solubility of Hg2I2 is s M
Ksp = [Hg2²+][2I-]²
Ksp = s.4s² = 4s³
s³ = Ksp/4 = 4.5 × 10^-29/4 = 1.125 × 10^-29
3logs = log(1.125 × 10^-29) = -29+0.0512 =-28.9488
logs = -9.6496 = 10(bar).3504
s = 2.241 × 10^-10 M{by using antilog }
[Hg2²+] = s = 2.241 × 10^-10 M
[I-] = 2s = 2 × 2.241 × 10^-10 = 4.482 × 10^-10 M