Chemistry - Dissolving Precipitates

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SUMMARY

The discussion focuses on the dissolution of CuCl in the presence of NaCN, forming the complex ion [Cu(CN)2]-. The net ionic equation presented is CuCl + 2NaCN → [Cu(CN)2]- + 2NaCl. The equilibrium constants involved are Ksp for CuCl, which is 1.9 x 10^-7, and Kf for the formation of [Cu(CN)2]-, which is 1.0 x 10^16. The participant seeks clarification on how to calculate the overall equilibrium constant (K) for the reaction and the role of Ksp in this context.

PREREQUISITES
  • Understanding of net ionic equations
  • Familiarity with solubility product constant (Ksp)
  • Knowledge of formation constants (Kf)
  • Basic principles of chemical equilibria
NEXT STEPS
  • Study the calculation of equilibrium constants from Ksp and Kf values
  • Learn about complex ion formation and its significance in solubility
  • Explore the concept of net ionic equations in detail
  • Investigate the role of stoichiometry in chemical reactions involving precipitates
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Chemistry students, educators, and professionals involved in inorganic chemistry, particularly those studying solubility and complex ion formation.

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Homework Statement



Write an overall net ionic equation and calculate K for the reaction where CuCl (Ksp = 1.9 x 10^-7) is dissolved by NaCN to from [Cu(CN)2]- Kf = 1.0 x 10^16).

Homework Equations





The Attempt at a Solution



The net ionic equation I've come up with is:

CuCl + Na(CN)2- --> Cu(CN)2- + NaCl

The equilibrium equations to figure K that I came with are:

CuCl --> Cu+ + Cl-
2CN- + Cu+ --> [Cu(CN)2]-

Am I on the right track here, or am I completely off? How would I then determine K from this? How do I know when it's 1/Ksp or just Ksp?
 
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Jeann25 said:
The net ionic equation I've come up with is:

CuCl + Na(CN)2- --> Cu(CN)2- + NaCl

NaCl in net ionic?



 

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