Copper(I) ions in aqueous solution react with NH3(aq) according to
Cu+(aq) + 2NH3(aq) -->Cu(NH3)2^+
Kf = 6.3x10^10
Calculate the solubility (in g·L–1) of CuBr(s) (Ksp = 6.3× 10–9) in 0.41 M NH3(aq).
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To calculate the solubility of CuBr(s) in 0.41 M NH3(aq), we need to determine the concentration of Cu+(aq) ions in the solution.
First, let's write the balanced chemical equation for the reaction between CuBr(s) and NH3(aq):
CuBr(s) + 2NH3(aq) → Cu(NH3)2^+ + Br^-
According to the equation, one mole of CuBr(s) reacts with two moles of NH3(aq) to form one mole of Cu(NH3)2^+.
Next, let's set up an ICE (Initial, Change, Equilibrium) table to calculate the concentration of Cu+(aq) ions:
CuBr(s) + 2NH3(aq) → Cu(NH3)2^+ + Br^-
Initial : 0 0.41 M 0 0
Change : -x -2x +x +2x
Equilibrium: x 0.41 M - 2x x 2x
Since the amount of CuBr(s) dissolved is equal to the concentration of Cu+(aq) ions, we can set up an expression for the solubility product (Ksp):
Ksp = [Cu+(aq)][Br^-]
6.3×10^(-9) = x * 2x^2
6.3×10^(-9) = 2x^3
Solving for x:
2x^3 = 6.3×10^(-9)
x^3 = (6.3×10^(-9)) / 2
x^3 = 3.15×10^(-9)
x ≈ 1.4×10^(-3)
The concentration of Cu+(aq) ions is approximately 1.4×10^(-3) M.
Finally, to calculate the solubility in g·L^(-1), we need to convert the concentration to mass. The molar mass of CuBr is 143.4 g·mol^(-1).
Solubility = (concentration) * (molar mass)
Solubility = 1.4×10^(-3) M * 143.4 g·mol^(-1)
Solubility ≈ 0.20 g·L^(-1)
Therefore, the solubility of CuBr(s) in 0.41 M NH3(aq) is approximately 0.20 g·L^(-1).
To calculate the solubility of CuBr(s) in 0.41 M NH3(aq), we need to determine the concentration of Cu+(aq) in the solution.
First, we can start by writing the equation for the dissolution of CuBr(s) in water:
CuBr(s) ⇌ Cu+(aq) + Br-(aq)
The solubility product constant, Ksp, for CuBr(s) is given as 6.3 × 10^–9, which represents the concentration of Cu+(aq) and Br-(aq) ions in equilibrium. From the balanced equation, we know that the concentration of Cu+(aq) is the same as the concentration of CuBr(s) that dissolves.
Let's assume that x g of CuBr(s) dissolves per liter. Therefore, the concentration of Cu+(aq) would also be x M.
Ksp = [Cu+(aq)] [Br-(aq)]
Substituting the known values:
6.3 × 10^–9 = (x)(x)
Simplifying the equation:
6.3 × 10^–9 = x^2
To solve for x, take the square root of both sides:
√(6.3 × 10^–9) = √(x^2)
x = 7.95 × 10^–5 M
Now that we have the concentration of Cu+(aq), we can use the given reaction to find the concentration of Cu(NH3)2^+, which is the same as the concentration of Cu+(aq) in this case:
Cu+(aq) + 2NH3(aq) ⇌ Cu(NH3)2^+
The formation constant, Kf, for this reaction is given as 6.3 × 10^10, which means:
Kf = [Cu(NH3)2^+]/([Cu+(aq)] [NH3]^2)
Substituting the known values:
6.3 × 10^10 = [Cu(NH3)2^+]/([7.95 × 10^–5] [0.41]^2)
Simplifying the equation:
6.3 × 10^10 = [Cu(NH3)2^+]/[2.729 × 10^–4]
Rearranging the equation:
[Cu(NH3)2^+] = (6.3 × 10^10)(2.729 × 10^–4)
[Cu(NH3)2^+] ≈ 1.717 g/L
Therefore, the solubility of CuBr(s) in 0.41 M NH3(aq) is approximately 1.717 g·L^–1.