1.

The Copper(I) ion forms a complex ion with CN− according to the following equation: Cu+(aq)+3CN−(aq)⇌[Cu(CN)3]2−(aq); K=1.0×1011 The concentration of CN− at equilibrium when 1.0 M NaCN (1 L) is added to a sufficient amount of CuBr(s) was found to be x×10−3 M. The value of x is Ksp=1×10−5

Answer»

The Copper(I) ion forms a complex ion with CN− according to the following equation:
Cu+(aq)+3CN−(aq)⇌[Cu(CN)3]2−(aq); K=1.0×1011
The concentration of CN− at equilibrium when 1.0 M NaCN (1 L) is added to a sufficient amount of CuBr(s) was found to be x×10−3 M. The value of x is

Ksp=1×10−5


Discussion

No Comment Found

Related InterviewSolutions