Question

Cd2+(aq) + 4CH3NH2(aq)≤≥ [Cd(CH3NH2)4]
2+(aq) Kstab = 3.6 × 106 equilibrium I

The concentration of Cd2+(aq) is 1.00×10^-4 moldm-3
Calculate the concentration of CH3NH2(aq) needed to reduce the concentration of Cd2+(aq) in this dilute solution by a factor of one thousand.

Answer 1

x mol/dm³ of CH3NH2(aq) was added

1*10^-7 mol/dm³ Cd²⁺(aq) left

9.99*10^-5 mol/dm³ of Cd²⁺(aq) and 3.996*10^-4 mol/dm³ of CH₃NH₂(aq) reacted

9.99*10^-5 mol/dm³ of [Cd(CH₃NH₂)₄]²⁺(aq) formed

(x-3.996*10^-4) mol/dm³ of CH₃NH₂(aq) left

3*10⁶ = (9.99*10^-5)/(1*10^-7*(x-3.996*10^-4)⁴)

x-3.996*10^-4 = 0.1351

x=0.1355 mol/dm³ - concentration of CH₃NH₂(aq) needed