Question

Calculate the NH4+ concentration that is needed to prevent Mg(OH)2 from precipitating from a liter of solution which contains (3.00x10^-2) M ammonia and (3.3x10^-3) M of Mg2+ (aq). (Ka (NH4+) = 5.6 x 10–10, Ksp (Mg(OH)2) = 1.12 x 10–11)

Answer 1

Answer – Given, [NH₄⁺] = 0.03 M , [Mg²⁺] = 0.003 M,

Ka (NH₄⁺) = 5.6 x 10^–10, Ksp (Mg(OH)₂) = 1.12 x 10^–11

First we need to calculate the [OH^-] at equilibrium-

We know the Ksp expression for the

Mg(OH)₂ (s) <---> [Mg²⁺] [OH^-]

Ksp = [Mg²⁺][OH^-]²

1.12 x 10^–11 = 0.003 M *(2x)²

4x² = 1.12 x 10^–11 / 0.003 M

        = 3.73*10^-9

x = 3.06*10^-6 M

[OH^-] = 2x = 2*3.06*10^-6 M

                   = 6.11*10^-5 M

Now we need to calculate the [NH₄⁺]

   NH₃ +H₂O <------> NH₄⁺ + OH^-

I   0.03                 0             0

C   -x                   6.11*10^-5 +x

E 0.03-x             6.11*10^-5 +x   6.11*10^-5

We know,

At equilibrium, [NH₃] = 0.03-x

                               = 0.03 -6.11*10^-5

                              = 0.0299 M

Kb = 1.0*10^-14 / 5.6 x 10^–10

      = 1.8*10^-5

Kb = [NH₄⁺][OH^-] / [NH₃]

1.8*10^-5= (x+6.11*10^-5) (6.11*10^-5) / (0.0299)

1.8*10^-5*0.0299 = (x+6.11*10^-5) (6.11*10^-5)

5.4*10^-7 = 6.11*10^-5 x + 3.7*10^-9

5.4*10^-7 -3.7*10^-9 = 6.11*10^-5 x

5.35*10^-7 = 6.11*10^-5 x

So, x = 0.00876 M

So, the NH4+ concentration that is needed to prevent Mg(OH)2 from precipitating = 6.11*10^-5 +x

                     = 0.00876+ 6.11*10^-5

                     = 0.00882 M