Calculate the solubility at 25°C of CuBr in pure water and in 0.30M NaCN . You'll probably find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits.

Question

Calculate the solubility at 25°C

 of CuBr

 in pure water and in 0.30M

 NaCN

. You'll probably find some useful data in the ALEKS Data resource.

Round your answer to 

2

 significant digits.

Answer 1

Not sure what ALEKS DATA resource is, but the only useful data required that isn't provided would be the Ksp for CuBr, which is 6.27×10⁻⁹ according to Wikipedia, and the Kf for Cu(CN)₃^2- which is 1x10¹¹

CuBr(s) <==> Cu⁺(aq) + Br^-(aq) ... Ksp = 6.27x10^-9

Ksp = [Cu⁺][Br^-]

6.27×10⁻⁹ = (x)(x) = x²

x = 7.9x10^-5 M = solubility of CuBr in pure water

The solubility in 0.30 M NaCN will be greater than that in pure water because of the formation of the soluble complex ion Cu(CN)3^2-, as follows:

Cu⁺(aq) + 3CN^-(aq) ==> Cu(CN)₃^2-(aq) ... Kf = 1x10¹¹

Combining the reactions we have...

CuBr(s) <==> Cu⁺(aq) + Br^-(aq)

Cu⁺(aq) + 3CN^-(aq) ==> Cu(CN)₃^2-(aq)

______________________________________

CuBr(s) + 3CN^-(aq) ==> Cu(CN)₃^2-(aq) + Br^-(aq) ... K = (6.27x10^-9)(1x10¹¹) = 627

K = 627 = [Cu(CN)₃^2-][Br^-] / [CN^-]³

627 = (x)(x) / (0.3 + x)³ and assuming x is << 0.3 we simplify to...

627 = x² / (0.3)³

x² = 16.9

x = 4.1 M = solubility CuBr in 0.30 M NaCN

(be sure to check all of the calculations)