Respuesta :
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
Solubility in 0.0120 M CoBr₂ (S')
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
The solubility of CuBr in pure water is 0.2583 g/L and the solubility in CoBr2 is 3.46*10^-5 g/L
Data;
- CuBr2
- CoBr2
- Ksp = 6.27*10^-9
Solubility of CuBr
The equation of reaction is given as
[tex]CuBr_2 \to Cu^2^+ + 2Br^-[/tex]
In pure water, the Ksp of the solution is 6.27 * 10^-9
[tex]Ksp = [Cu^2^+][Br^-]^2\\Ksp = x^3[/tex]
Let's substitute and solve
[tex]x^3 = 6.27*10^-^9\\x = \sqrt[3]{6.27*10^-^9} \\x = 0.0018[/tex]
The molecular weight of CuBr2 is 143.5 g/mol
The solubility of CuBr2 is
[tex]0.0018 * 143.5 = 0.2583 g/L[/tex]
The solubility in 0.012M of CoBr2
The equation of dissociation is given as
[tex]CoBr_2 \to Co^2^+ + 2Br^-\\[/tex]
0.012 0 0
0 0.120 2 * 0.0120 = 0.024M
The dissociation of copper bromide is
[tex]CuBr_2 \to Cu^2^+ + Br^-\\[/tex]
c 0 0.024
c - s s s + 0.024
The Ksp of this reaction is
[tex][Ksp] = [Co^2^+][Br^-][/tex]
The Ksp of CuBr2 = 6.27 * 10^-9
[tex]6.27*10^-^9 = s(s + 0.024)\\s + 0.024 = 0.024\\0.024 > > > s\\6.27*10^-^9 = s * 0.024\\s = 2.41*10^-^7 mol/l[/tex]
Let's convert this into g/L
[tex]s = 2.411 * 10^-^7 * 143.5 = 3.46*10^-6 g/L[/tex]
The solubility of CuBr in pure water is 0.2583 g/L and the solubility in CoBr2 is 3.46*10^-5 g/L
Learn more on solubility here;
https://brainly.com/question/16903071
