Question

estion 1 For the aqueous (Hg(NH3)4]2+ complex has a Ky - 1,8 x 1019 at 25°C. Suppose equal volumes of 0.0082 M Hg(NO3)2 solution and 0.72 M NH3 solution were mixed. Calculate the equilibrium concentration (M) of aqueous (Please enter your answer is scientific notation. For example if you answer was 3.15 x 10. please enter 3.1585) Answer should be in 2 significant figures.

Calculate the molar concentration of Hg+2 ion.

Answer 1

Initial concentration of Hg(NO₃)₂ = 0.0082M

Initial concentration of NH₃ = 0.72M

Equilibrium constant, K = 1.8*10¹⁹

^{Hg(NO3) + 4NH3 [Hg(NH3)4]2+}

Concentration	Hg(NO3)2	NH3	[Hg(NH3)4]2+
Initial	0.0082M	0.72M	0
Change	-x	-x	+x
Equilibrium	0.0082-x	0.72-x	x

K = [Hg(NH₃)4]²⁺/[Hg(NO₃)₂][NH₃]⁴

  1.8*10¹⁹   = x / (0.0082) (0.72)⁴ =  x / (0.0082) (0.5184)

x = (1.8*10¹⁹ ) (0.0082) (0.5184)

Equilibrium concentration of Hg²⁺ ion = 7.651E16