Question

A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains at equilibrium?

Which is the answer:

6.7 × 10^-28 M

3.8 × 10^-24 M

3.1 x 10^-24 M

2.9 × 10^-27 M

4.6 × 10^-25 M

Please show your work, thank you!

Answer 1

As LiCN is strong in nature, therefore, its dissociation is complete. The dissociation of LiCN is given as follows: LiCN 0.33M → Li* + CN 0.33M The formation of Cu (CN) is given as follows: O Cu²+ (aq) + 4CN (aq)=Cu(CN).- (aq) initial: 2.2x10-M 0.33M change: - -X equilibrium:2.2x10-M-x 0.33 M-x +x + + Cu(CN)2-1 K,"cu"][cn] But we only have 22.2x10-M in Cu?- so, it will react with 4CN to produce 2.2x 10-M s K, is 1.0 x1025 we can assume that the reaction goes for completion and almost all Cu?- concentration gets converted to Cu(CN). also 4x is negligible compared to 0.33 so, 0.33-4x is roughly 0.33) The dissociation of Cu( is given as follows: Cu(CN). (aq) = Cu2+ (aq) + initial: 2.2x10-'M 0 change: - +x equilibrium:2.2x10’M-x 4CN- (aq) 0.33M +4.x 0.33 M +4x

The K, is given as follows: K, - CuCNT ** [Cu(CN);] (1.0x102) Putting the values as follows: K. - [Cu][CN7 Cu (CN):) 1 _(x)(0.33+ 4x)* (1.0x1025) 2.2x10- - x x=1.9x10-26 The equilibrium concentration of Cu?+ is 1.9x10-261.