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

The first part of this question is pretty straightforward, using the solubility product since CuBr is not very soluble.

cuBr (s) <===> Cu⁺ + Br^-

K_sp = [Cu⁺] * [Br^-] / [CuBr] = [Cu⁺] * [Br^-] , (when the solid salt is in excess).

We get the solubility product from a handbook table (or the all knowing Google) as 6.27 E-09 at room temp.

Let [Cu⁺] = x.

Since the [Cu⁺] and [Br^-] in solution are the same, [Br^-] = x. Therefore,

6.27 E-09 = x * x = x²

Then, x = sqrt(6.27 E-09) = sqrt(62.7 E-10) = 7.92 E-05

[Cu⁺] = 7.9 E-05 M

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The second part of the question is a bit more complicated.

If cuperous bromide,CuBr, powder is added to a NaCN solution, a reaction takes place. As trace amounts of CuBr dissolve, the CuCN formed, which is also only slightly soluble in water, rapidly reacts further.

Again from a Wiki reference for the solubility of CuCN,

"Copper cyanide is insoluble in water but rapidly dissolves in solutions containing CN⁻ to form [Cu(CN)₃]²⁻ and [Cu(CN)₄]³⁻ ..."

The balanced equations for the system in the tricyano case are

CuBr (s) + CN^- (sol) ==> CuCN + Br- (1)

CuCN + 2 CN^- <==> Cu(CN)₃^-2 (sol) (2)

Then, assuming an excess of CuBr, the equilibrium constant for equation (1) is

K_sp(CuBr) = [CuCN] * [Br^-] = [CuCN] * [Br^-] (3)

[CuBr]_s * [CN^-] [CN^-]

The equilibrium constant for equation (2) is

K_sp(CuCN) = [Cu(CN)₃^-2] (4)

[CuCN] * [CN^-]²

Since the CuCN is an intermediate species, even though insoluble, we can rearrange equ (4) and substitute into equ (3).

[CuCN] = [Cu(CN)₃^-2]

K_sp(CuCN) * [CN^-]²

K_sp(CuBr) = [Cu(CN)₃^-2] * [Br^-] = [Cu(CN)₃^-2] * [Br^-] (5)

K_sp(CuCN) * [CN^-]² [CN^-] K_sp(CuCN) * [CN^-]³

Rearranging(5), we have

[Cu(CN)₃^-2] * [Br^-] = K_sp(CuBr) * K_sp(CuCN) * [CN^-]³

Note that the copper complex concentration and the bromide concentration must be the same.

Letting x = [Cu(CN)₃^-2] gives

x*x = x² = K_sp(CuBr) * K_sp(CuCN) * [CN^-]³ (6)

From reference tables,

K_sp(CuBr) = 6.27 E-09

K_sp(CuCN) = 3.47 E-20 , and

[CN^-]³ = 0.030 M , from the given concentration of the NaCN solution.

Thus, x2 = (6.27 E-09) * (3.47 E-20) * (0.30)³ = 5.874 E-30

x = sqrt(5.874 E-30) = 2.42 E-15

[Cu(CN)₃^-2] = 2.4 E-15 M

We can do a similar calculation for the tetracyano complex, [Cu(CN)₄^-3]. The result is a tetracyano copper concentration of [Cu(CN)₄^-3]. = 1.3 E-15 M.

Since no information is given which complex is favored, or in what ratio, we conclude that the copper complex concentration is in the range (1.3 - 2.4) E-15 M.

This is much less than the solubility in pure water, and is similar to the common ion method for suppression of solubility.