Question

26.
A solution of 0.075 M CoBr2 is saturated with H2S ( [H2S] = 0.10 M). What is the minimum pH at which CoS ( Ksp = 5.9 x 10 -21) will precipitate?
Ka1 =8.9 x 10 -8 & Ka2 = 1.2 x 10 -13 for H2S Ksp for CoS = 5.9 x 10 -21

Answer 1

If Ionic product of CoS > solubility product of CoS, CoS will start precipitation

K_SP = [Ca²⁺] [S^2-] =   5.9 x 10^-21

[Ca²⁺] of 0.075 M CoBr₂ = 7.5 x 10^-2 M

the minimum [S^2-] after which CoS will start precipitation is equal to K_SP/[Ca²⁺]

  [S^2-] = 5.9 x 10^-21/ 7.5 x 10^-2

= 7.86 x 10^-20 M

H₂S H₁⁺ + HS^-   K_a1 = 8.9 x 10^-8

^{Initial 0.1 M 0 0}

^{At equilibrium 0.1M - x x x}

K_a1 = [H₁⁺] [HS^-]/[H₂S]

8.9 x 10^-8 = x²/0.1-x

  x² + 8.9*10^-8x - 8.9*10^-9 = 0

After solving quadratic equation

x = 9.43*10^-5 M

[H₁⁺] = [HS^-] = 9.43*10^-5 M

  HS^- H₂⁺   + S^2-     K_a2 = 1.2 x 10^-13

intial   9.43*10^-5 M 0 0

Dissociation of HS^- results the equal concentration of [H₂⁺] and [S^2-]

the minimum [S^2-] after which CoS will start precipitation is 7.86 x 10^-20 M

But [H₂⁺] = [S^2-] = 7.86 x 10^-20 M at which CoS will start precipitation

[H⁺] = [H₁⁺] + [H₂⁺]

= 9.43*10^-5 + 7.86*10^-20

  [H₂⁺] is negligible compared with [H₁⁺]

[H⁺] = 9.43*10^-5

p^H = - log [H⁺]

= - log 9.43*10^-5

= 4.02

p^H = 4.02 at which CoS will start precipitate