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

Hello There!

In order to calculate the solubility of CuBr aka Copper I Bromide, in pure water and in 0.30 M NaCN aka Sodium Cyanide, you're going to need to find the solubility constant (Ksp) first.

Step 1:

Write your solubility equation

CuBr (s) ↔ Cu⁺ (aq) + Br^- (aq)

From that you get your Ksp expression

Ksp = [Cu⁺][Br^-]

Now, from the literature we know that Ksp = 1.0 x 10^-12

In pure water, the copper and bromide ions are equal, meaning the solubility (S) is the concentration of each ion at equilibrium.

Therefore, Ksp = S²

Solving for S → S = √Ksp = √(1.0 x 10^-12) = 1.0 x 10^-6 M

However, when NaCN is added to the solution, we get cyanide ions, which can form bonds with Copper, increasing the solubility of CuBr by shifting the equilibrium to the right.

Cu⁺ + CN^- ↔ Cu(CN)₂^-

Here we can assume this reaction is highly favored causing most of the copper to become bonded to the cyanide.

Thus, in the presence of the 0.30 M NaCN, the solubility of CuBr can be determined by the amount of bromide ions in solution (due to the copper being accounted for as it bond with cyanide).

This means the concentration of copper ions will be equal to the concentration of bromide ions at equilibrium.

If you want to determine exactly what this number is you'll need the formation constant (Kf). for Cu(CN)₂

Hope this helps!