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MMLU-Pro / 4188 / For the complex ion equilibrium for the dissociation of Cu(NH_3)^2_4^+ ,…

Problem

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question

For the complex ion equilibrium for the dissociation of Cu(NH_3)^2_4^+ , calculate (1) the ratio of Cu^2+ ion to Cu(NH_3)^2_4^+ in a solution 1 M in NH_3; (2) the equilibrium concentration of NH_3 needed to attain 50 % conversion of Cu^2+ to Cu(NH_3)^2_4^+. K_a for Cu(NH_3)^2_4^+ = 2 × 10^-13.

category

chemistry

options

  1. [Cu^2+] / [Cu(NH_3)^2_4^+] = 4 × 10^-13, [NH_3] = 7.0 × 10^-3 M
  2. [Cu^2+] / [Cu(NH_3)^2_4^+] = 3 × 10^-13, [NH_3] = 9.0 × 10^-3 M
  3. [Cu^2+] / [Cu(NH_3)^2_4^+] = 2 × 10^-12, [NH_3] = 6.0 × 10^-3 M
  4. [Cu^2+] / [Cu(NH_3)^2_4^+] = 1 × 10^-13, [NH_3] = 5.0 × 10^-3 M
  5. [Cu^2+] / [Cu(NH_3)^2_4^+] = 1 × 10^-14, [NH_3] = 4.5 × 10^-3 M
  6. [Cu^2+] / [Cu(NH_3)^2_4^+] = 2 × 10^-13, [NH_3] = 6.7 × 10^-3 M
  7. [Cu^2+] / [Cu(NH_3)^2_4^+] = 2 × 10^-14, [NH_3] = 5.7 × 10^-3 M
  8. [Cu^2+] / [Cu(NH_3)^2_4^+] = 2 × 10^-13, [NH_3] = 8.7 × 10^-3 M
  9. [Cu^2+] / [Cu(NH_3)^2_4^+] = 2 × 10^-12, [NH_3] = 7.7 × 10^-3 M
  10. [Cu^2+] / [Cu(NH_3)^2_4^+] = 5 × 10^-13, [NH_3] = 1.0 × 10^-2 M

src

stemez-Chemistry

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Source and history

upstream_id / Original problem ID
4188
task_id
43e48d16-bfb8-549d-8fd9-d02993f689a2
task_key
default--test--4188
task_revision_id
1
config / split
default / test

Official source

initial import