Question

What is the value of Q when 100.0 mL of a 7.6 10-4 M Mg(NO3)2 is added to 100.0 mL of 7.1 10-4 M NaOH? (Ksp for Mg(OH)2 = 8.9 ✕ 10-12)

Answer 1

Q = 4.8x10⁻¹¹

Step-by-step explanation:

The solubility of Mg(OH)₂ is:

Mg(OH)₂(s) → Mg²⁺(aq) + 2OH⁻(aq)

And Ksp is:

Ksp = [Mg²⁺] [OH⁻]²

Where [] are concentrations in equilibrium

Thus, Q could be defined as:

Q = = [Mg²⁺] [OH⁻]²

Where [] are actual concentrations

[Mg(NO₃)₂] = [Mg²⁺]

And [NaOH] = [OH⁻]

Both solutions are diluted twice (Because 100mL of each solution are diluted to 200mL). The equilibrium concentrations are:

[Mg²⁺] = 7.6x10⁻⁴M / 2 = 3.8x10⁻⁴M

[OH⁻] = 7.1x10⁻⁴ / 2 = 3.55x10⁻⁴M

That means Q is:

Q = [3.8x10⁻⁴M] [3.55x10⁻⁴M]²

Q = 4.8x10⁻¹¹