Answer:
the solubility at 25°c of cubr in pure water = 0.011 g/L
The solubility at 25°c of cubr in a 0.0030m cobr2 solution = 0.00014 g/L
Explanation:
Step 1: Data given
Ksp of CuBr = 6.27 × 10^-9
Molar mass CuBr = 143.45 g/mol
Step 2: Calculate the solubility at 25°c of cubr in pure water
Ksp = [Cu+][Br-]
The initial concentration
[Cu+] = 0M
[Br-] = 0M
The concentration at the equilibrium
[Cu+] = X M
[Br-] = X M
Ksp =6.27 * 10^-9 = X * X = X²
S = 7.9 *10^-5 mol /l
7.9 * 10^-5 mol/L * 143.45 g/mol = 0.011 g/L
Step 3: Calculate the solubility at 25°c of cubr in a 0.0030m cobr2 solution.
The balanced equation:
CoBr2(aq) → Co^2+(aq) + 2 Br⁻(aq)
For 1 mol CoBr2 we'll have 1 mol Co^2+ and 2 moles Br-
For 0.0030 M CoBr2 we'll have 0.0030 M Co^2+ and 0.0060 M Br-
The initial concentration
[Cu+] = 0M
[Br-] = 0.0060 M
At the equilibrium
[Cu+] = X
[Br-] = 0.0060 + X
Ksp = 6.27 * 10^-9 = X * (0.0060+X)
6.27 *10^-9 = 0.0060 X + X²
X = 1.0 * 10^-6 mol/L
1.0 * 10^-6 mol/L * 143.45 g/mol = 0.00014 g/L
