A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10-3 M CuNO3. Cu(I) forms complex ions with cyanide as follows: ![A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10<sup>-3</sup> M CuNO<sub>3</sub>. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN) <sub>3</sub><sup>2-</sup>] A) 3.33 M B) 1.67 M C) 2.0 × 10<sup>-3</sup> M D) 1.0 × 10<sup>-3</sup> M E) none of these](https://d2lvgg3v3hfg70.cloudfront.net/TB6422/11eaaf91_9e28_6ed1_892c_81ba02cfc46a_TB6422_00_TB6422_00_TB6422_00.jpg)
Calculate the following concentrations at equilibrium:
-[Cu(CN) 32-]
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