Question

Calculate the equilbrium concentration of Zn² (aq) in a solution that is initially 0.150 M Zn(NO₃)₂ and 0.800 M NaCN. The formation constant for [Zn(CN)₄]^2- (aq) is K_f = 2.1 x 10¹⁹.

Please show all of your work and include any formulas you use! Thank you!

Answer 1

For the reaction,

Zn ²⁺ (aq) + 4 CN^- (aq) <-------------> [Zn(CN)₄]^2- (aq)

Initial concentration of Zn²⁺ = 0.150 M

Initial concentration of CN^- = 0.800 M

Equilibrium concentration of Zn²⁺ = ( 0.150 - x ) M

Equilibrium concentration of CN^- = ( 0.150 - 4x ) M

K_f = 2.1 * 10¹⁹

K_f = [ [Zn(CN)₄]^2-_(aq) ] / ( [Zn²⁺_(aq)] [CN^-_(aq)]⁴ )

2.1 * 10¹⁹ = (x) / [ (0.150-x) (0.800-4x)⁴ ]

x = (2.1 * 10¹⁹ ) (0.150-x) (0.800-4x)⁴

Equilibrium concentration of Zn²⁺  = 0.150 - [ (2.1 * 10¹⁹ ) (0.150-x) (0.800-4x)⁴ ] M

Solve for x and you get the final answer....